EjsTeóricos.lyx

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eqs-within-sections
figs-within-sections
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\begin_body

\begin_layout Title
Ejercicios pedidos por el tp
\end_layout

\begin_layout Section*
Ejercicios de la Guía 2
\end_layout

\begin_layout Subsection*
Ejercicio 7
\end_layout

\begin_layout Subsubsection*
a
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\hat{\overline{y}}^{*}=\begin{cases}
\overline{y}_{\pi}-C & y_{1}\notin s;y_{N}\in s\\
\overline{y}_{\pi}+C & y_{1}\in s;y_{N}\notin s\\
\overline{y}_{\pi} & y_{1}\in s;y_{N}\in s
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Proof
\begin_inset Formula $\hat{\overline{y}}^{*}$
\end_inset


\begin_inset Foot
status collapsed

\begin_layout Plain Layout
Favor de leer 
\begin_inset Quotes fld
\end_inset

Y media sombrero estrella
\begin_inset Quotes frd
\end_inset

.
\end_layout

\end_inset

 es insesgado.
\end_layout

\begin_layout Proof
\begin_inset Formula $S_{A}=\left\{ s\in\Omega/y_{1}\notin s;y_{N}\in s\right\} $
\end_inset


\end_layout

\begin_layout Proof
\begin_inset Formula $S_{B}=\left\{ s\in\Omega/y_{1}\in s;y_{N}\notin s\right\} $
\end_inset


\end_layout

\begin_layout Proof
\begin_inset Formula $S_{C}=\overline{S_{A}\cup S_{B}}=\left\{ s\in\Omega/\left(y_{1}\in s\wedge y_{N}\in s\right)\vee\left(y_{1}\notin s\wedge y_{N}\notin s\right)\right\} $
\end_inset


\end_layout

\begin_layout Proof
\begin_inset Formula $\bigcup_{\left\{ A;B;C\right\} }S_{i}=\Omega$
\end_inset


\end_layout

\begin_layout Proof
\begin_inset Formula 
\[
\mathbb{E}\left(\hat{\overline{y}}^{*}\left(S\right)\right)=\sum_{s\in\Omega}p\left(s\right)\cdot\hat{\overline{y}}^{*}\left(s\right)
\]

\end_inset


\end_layout

\begin_layout Proof
\begin_inset Formula $\Omega=S_{A}+S_{B}+S_{C}$
\end_inset


\end_layout

\begin_layout Proof
\begin_inset Formula 
\[
\mathbb{E}\left(\hat{\overline{y}}^{*}\left(S\right)\right)=\sum_{s\in S_{A}}p\left(s\right)\cdot\hat{\overline{y}}^{*}\left(s\right)+\sum_{s\in S_{B}}p\left(s\right)\cdot\hat{\overline{y}}^{*}\left(s\right)+\sum_{s\in S_{C}}p\left(s\right)\cdot\hat{\overline{y}}^{*}\left(s\right)
\]

\end_inset


\end_layout

\begin_layout Proof
\begin_inset Formula 
\[
\mathbb{E}\left(\hat{\overline{y}}^{*}\left(S\right)\right)=\sum_{s\in S_{A}}p\left(s\right)\cdot\left(\overline{y}_{\pi}\left(s\right)-C\right)+\sum_{s\in S_{B}}p\left(s\right)\cdot\left(\overline{y}_{\pi}\left(s\right)+C\right)+\sum_{s\in S_{C}}p\left(s\right)\cdot\overline{y}_{\pi}\left(s\right)
\]

\end_inset


\end_layout

\begin_layout Proof
\begin_inset Formula 
\[
\mathbb{E}\left(\hat{\overline{y}}^{*}\left(S\right)\right)=\sum_{s\in S_{A}}p\left(s\right)\cdot\overline{y}_{\pi}\left(s\right)-\sum_{s\in S_{A}}p\left(s\right)\cdot C+\sum_{s\in S_{B}}p\left(s\right)\cdot\overline{y}_{\pi}\left(s\right)+\sum_{s\in S_{B}}p\left(s\right)\cdot C+\sum_{s\in S_{C}}p\left(s\right)\cdot\overline{y}_{\pi}\left(s\right)
\]

\end_inset


\end_layout

\begin_layout Proof
\begin_inset Formula 
\[
\mathbb{E}\left(\hat{\overline{y}}^{*}\left(S\right)\right)=\left[\sum_{s\in S_{A}}p\left(s\right)\cdot\overline{y}_{\pi}\left(s\right)+\sum_{s\in S_{B}}p\left(s\right)\cdot\overline{y}_{\pi}\left(s\right)+\sum_{s\in S_{C}}p\left(s\right)\cdot\overline{y}_{\pi}\left(s\right)\right]-\sum_{s\in S_{A}}p\left(s\right)\cdot C+\sum_{s\in S_{B}}p\left(s\right)\cdot C
\]

\end_inset


\end_layout

\begin_layout Proof
\begin_inset Formula 
\[
\mathbb{E}\left(\hat{\overline{y}}^{*}\left(S\right)\right)=\mathbb{E}\left(\overline{y}_{\pi}\left(s\right)\right)-C\cdot\sum_{s\in S_{A}}p\left(s\right)+C\cdot\sum_{s\in S_{B}}p\left(s\right)
\]

\end_inset


\end_layout

\begin_layout Proof
\begin_inset Formula $\sum_{s\in S_{A}}p\left(s\right)=p\left(S_{A}\right)=p\left(\underset{1}{\mathbb{I}}=1;\underset{N}{\mathbb{I}}=0\right)=\pi_{1}-\pi_{1N}$
\end_inset


\end_layout

\begin_layout Proof
Como estamos en MSA, todos los 
\begin_inset Formula $\pi_{i}$
\end_inset

 y 
\begin_inset Formula $\pi_{ij}$
\end_inset

 son del mismo valor, con 
\begin_inset Formula $\forall i:\pi_{i}=\frac{n}{N}=f$
\end_inset

 y 
\begin_inset Formula $\pi_{ij}=\frac{n\left(n-1\right)}{N\left(N-1\right)}$
\end_inset

.
\end_layout

\begin_layout Proof
\begin_inset Formula $\sum_{s\in S_{B}}p\left(s\right)=p\left(S_{B}\right)=p\left(\underset{1}{\mathbb{I}}=0;\underset{N}{\mathbb{I}}=1\right)=\pi_{N}-\pi_{1N}=\pi_{1}-\pi_{1N}$
\end_inset


\end_layout

\begin_layout Proof
\begin_inset Formula 
\[
\mathbb{E}\left(\hat{\overline{y}}^{*}\left(S\right)\right)=\mathbb{E}\left(\overline{y}_{\pi}\left(s\right)\right)-C\cdot\left(\pi_{1}-\pi_{1N}\right)+C\cdot\left(\pi_{1}-\pi_{1N}\right)
\]

\end_inset


\end_layout

\begin_layout Proof
\begin_inset Formula 
\[
\mathbb{E}\left(\hat{\overline{y}}^{*}\left(S\right)\right)=\mathbb{E}\left(\overline{y}_{\pi}\left(s\right)\right)
\]

\end_inset


\end_layout

\begin_layout Proof
Ya probamos que el 
\begin_inset Formula $\pi$
\end_inset

-estimador de la media es insesgado:
\end_layout

\begin_layout Proof
\begin_inset Formula 
\[
\mathbb{E}\left(\hat{\overline{y}}^{*}\left(S\right)\right)=\overline{y}_{U}
\]

\end_inset


\end_layout

\begin_layout Subsubsection*
b Verificar la fórmula de la varianza.
\end_layout

\begin_layout Standard
Se hace evidente que vamos a necesitar desarrollar de alguna forma la ecuación
 e intentar aproximarla
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V\left(\hat{\overline{y}}^{*}\left(S\right)\right)=\mathbb{E}\left[\left(\hat{\overline{y}}^{*}\right)^{2}\left(S\right)\right]-\mathbb{E}^{2}\left[\hat{\overline{y}}^{*}\left(S\right)\right]
\]

\end_inset


\end_layout

\begin_layout Subsection*
Buscamos 
\begin_inset Formula $\mathbb{E}\left[\left(\hat{\overline{y}}^{*}\right)^{2}\left(S\right)\right]$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\mathbb{E}\left[\left(\hat{\overline{y}}^{*}\right)^{2}\left(S\right)\right]=\sum_{s\in\Omega}p\left(s\right)\cdot\left(\hat{\overline{y}}^{*}\right)^{2}\left(s\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\mathbb{E}\left(\hat{\overline{y}}^{*}\left(S\right)\right)=\sum_{s\in S_{A}}p\left(s\right)\cdot\left(\overline{y}_{\pi}\left(s\right)-C\right)^{2}+\sum_{s\in S_{B}}p\left(s\right)\cdot\left(\overline{y}_{\pi}\left(s\right)+C\right)^{2}+\sum_{s\in S_{C}}p\left(s\right)\cdot\overline{y}_{\pi}^{2}\left(s\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\mathbb{E}\left(\hat{\overline{y}}^{*}\left(S\right)\right)=\sum_{s\in S_{A}}p\left(s\right)\cdot\left(\overline{y}_{\pi}^{2}\left(s\right)+2C\overline{y}_{\pi}\left(s\right)+C^{2}\right)+\sum_{s\in S_{B}}p\left(s\right)\cdot\left(\overline{y}_{\pi}^{2}\left(s\right)-2C\overline{y}_{\pi}\left(s\right)+C^{2}\right)+\sum_{s\in S_{C}}p\left(s\right)\cdot\overline{y}_{\pi}^{2}\left(s\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align*}
\mathbb{E}\left(\hat{\overline{y}}^{*}\left(S\right)\right) & =\sum_{s\in S_{A}}p\left(s\right)\cdot\overline{y}_{\pi}^{2}\left(s\right)-2C\cdot\sum_{s\in S_{A}}p\left(s\right)\overline{y}_{\pi}\left(s\right)+\sum_{s\in S_{A}}p\left(s\right)\cdot C^{2}\\
 & +\sum_{s\in S_{B}}p\left(s\right)\cdot\overline{y}_{\pi}^{2}\left(s\right)+2C\sum_{s\in S_{B}}p\left(s\right)\cdot\overline{y}_{\pi}\left(s\right)+C^{2}\sum_{s\in S_{B}}p\left(s\right)+\sum_{s\in S_{C}}p\left(s\right)\cdot\overline{y}_{\pi}^{2}\left(s\right)
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align*}
\mathbb{E}\left(\hat{\overline{y}}^{*}\left(S\right)\right) & =\left[\sum_{s\in S_{A}}p\left(s\right)\cdot\overline{y}_{\pi}^{2}\left(s\right)+\sum_{s\in S_{B}}p\left(s\right)\cdot\overline{y}_{\pi}^{2}\left(s\right)+\sum_{s\in S_{C}}p\left(s\right)\cdot\overline{y}_{\pi}^{2}\left(s\right)\right]\\
 & +2C\cdot\sum_{s\in S_{A}}p\left(s\right)\overline{y}_{\pi}\left(s\right)-2C\sum_{s\in S_{B}}p\left(s\right)\cdot\overline{y}_{\pi}\left(s\right)+C^{2}\cdot\sum_{s\in S_{A}}p\left(s\right)+C^{2}\cdot\sum_{s\in S_{B}}p\left(s\right)
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align*}
\mathbb{E}\left(\hat{\overline{y}}^{*}\left(S\right)\right) & =\mathbb{E}\left(\overline{y}_{\pi}^{2}\right)\\
 & -2C\cdot\sum_{s\in S_{A}}p\left(s\right)\overline{y}_{\pi}\left(s\right)+2C\sum_{s\in S_{B}}p\left(s\right)\cdot\overline{y}_{\pi}\left(s\right)+C^{2}\cdot\left(\pi_{1}-\pi_{N1}\right)+C^{2}\cdot\left(\pi_{1}-\pi_{N1}\right)
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align*}
\mathbb{E}\left(\hat{\overline{y}}^{*}\left(S\right)\right) & =\mathbb{E}\left(\overline{y}_{\pi}^{2}\right)-2C\cdot\left(\sum_{s\in S_{A}}p\left(s\right)\overline{y}_{\pi}\left(s\right)-\sum_{s\in S_{B}}p\left(s\right)\cdot\overline{y}_{\pi}\left(s\right)\right)+2C^{2}\cdot\left(\pi_{1}-\pi_{N1}\right)
\end{align*}

\end_inset


\end_layout

\begin_layout Paragraph*
Volvemos a la fórmula 
\end_layout

\begin_layout Standard
Considerando que conocemos 
\begin_inset Formula $\mathbb{E}\left(\overline{y}_{\pi}\right)=\overline{y}_{U}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V\left(\hat{\overline{y}}^{*}\left(S\right)\right)=\mathbb{E}\left[\left(\hat{\overline{y}}^{*}\right)^{2}\left(S\right)\right]-\mathbb{E}^{2}\left[\hat{\overline{y}}^{*}\left(S\right)\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V\left(\hat{\overline{y}}^{*}\left(S\right)\right)=\left[\mathbb{E}\left(\overline{y}_{\pi}^{2}\right)+2C\cdot\left(\sum_{s\in S_{A}}p\left(s\right)\overline{y}_{\pi}\left(s\right)-\sum_{s\in S_{B}}p\left(s\right)\cdot\overline{y}_{\pi}\left(s\right)\right)+2C^{2}\cdot\left(\pi_{1}-\pi_{N1}\right)\right]-\mathbb{E}^{2}\left[\hat{\overline{y}}^{*}\left(S\right)\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V\left(\hat{\overline{y}}^{*}\left(S\right)\right)=\mathbb{E}\left(\overline{y}_{\pi}^{2}\right)-\mathbb{E}^{2}\left[\hat{\overline{y}}^{*}\left(S\right)\right]-2C\cdot\left[\left(\sum_{s\in S_{A}}p\left(s\right)\overline{y}_{\pi}\left(s\right)-\sum_{s\in S_{B}}p\left(s\right)\cdot\overline{y}_{\pi}\left(s\right)\right)-C\cdot\left(\pi_{1}-\pi_{N1}\right)\right]
\]

\end_inset


\end_layout

\begin_layout Standard
Bajo MSA, la probabilidad de cada muestra es la misma: 
\begin_inset Formula $\forall s\in\Omega:p\left(s\right)=\nicefrac{1}{\binom{N}{n}}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V\left(\hat{\overline{y}}^{*}\left(S\right)\right)=V\left(\overline{y}_{\pi}^{2}\right)-2C\left[p\left(s\right)\cdot\left(\sum_{s\in S_{A}}\overline{y}_{\pi}\left(s\right)-\sum_{s\in S_{B}}\overline{y}_{\pi}\left(s\right)\right)-C\cdot\left(\pi_{1}-\pi_{N1}\right)\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\overline{y}_{\pi}\left(s\right)=\sum_{i=1}^{N}e_{i}\cdot s$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\forall s\in S_{A}:\exists j/e_{j}\cdot s=N$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\forall s\in S_{B}:\exists j/e_{j}\cdot s=1$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $N_{A}=N_{B}=\binom{N-2}{n-1}$
\end_inset

, ya que son las colas que se arman cuando se fija uno de los elementos
 atípicos y se excluye el otro.
\end_layout

\begin_layout Standard
En un punto aparte vamos a desarrollar la idea de que 
\begin_inset Formula $\sum_{s\in S_{A}}\overline{y}_{\pi}\left(s\right)-\sum_{s\in S_{B}}\overline{y}_{\pi}\left(s\right)=N_{A}\cdot\left(\frac{y_{N}}{n}-\frac{y_{1}}{n}\right)$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V\left(\hat{\overline{y}}^{*}\left(S\right)\right)=V\left(\overline{y}_{\pi}^{2}\right)-2C\left[\frac{p\left(s\right)}{n}\cdot\left(\sum_{s\in S_{A}}\sum_{i=1}^{N_{A}}e_{i}\cdot s-\sum_{s\in S_{B}}\sum_{i=1}^{N_{B}}e_{i}\cdot s\right)-C\cdot\left(\pi_{1}-\pi_{N1}\right)\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V\left(\hat{\overline{y}}^{*}\left(S\right)\right)=V\left(\overline{y}_{\pi}^{2}\right)-2C\cdot\left[\frac{p\left(s\right)}{n}\cdot\left(N_{A}\cdot y_{N}-N_{A}\cdot y_{1}\right)-C\cdot\left(\pi_{1}-\pi_{N1}\right)\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V\left(\hat{\overline{y}}^{*}\left(S\right)\right)=V\left(\overline{y}_{\pi}^{2}\right)-2C\left[\frac{1}{\binom{N}{n}}\cdot\binom{N-2}{n-1}\left(y_{N}-y_{1}\right)-C\cdot\left(\pi_{1}-\pi_{N1}\right)\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\frac{1}{\binom{N}{n}}\cdot\binom{N-2}{n-1}=\frac{\left(N-2\right)!}{\left(n-1\right)!\left(\left(N-2\right)-\left(n-1\right)\right)!}\cdot\frac{n!\left(N-n\right)!}{N!}=\frac{\left(N-n\right)!}{\left(N-n-1\right)!}\cdot\frac{n!}{\left(n-1\right)!}\cdot\frac{\left(N-2\right)!}{N!}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\frac{1}{\binom{N}{n}}\cdot\binom{N-2}{n-1}=\frac{\left(N-n\right)\cdot n}{N\left(N-1\right)}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V\left(\hat{\overline{y}}^{*}\left(S\right)\right)=V\left(\overline{y}_{\pi}^{2}\right)-2C\cdot\left[\frac{1}{n}\cdot\frac{\left(N-n\right)\cdot n}{N\left(N-1\right)}\cdot\left(y_{N}-y_{1}\right)-C\cdot\left(\pi_{1}-\pi_{N1}\right)\right]
\]

\end_inset


\end_layout

\begin_layout Standard
Además, 
\begin_inset Formula $\pi_{1}-\pi_{N1}=\frac{n}{N}-\frac{n\left(n-1\right)}{N\left(N-1\right)}=\frac{n\left(N-1\right)-n\left(n-1\right)}{N\left(N-1\right)}=\frac{nN-n^{2}}{N\left(N-1\right)}=\frac{n\left(N-n\right)}{N\left(N-1\right)}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V\left(\hat{\overline{y}}^{*}\left(S\right)\right)=V\left(\overline{y}_{\pi}^{2}\right)-2C\cdot\left[\frac{\left(N-n\right)}{N\left(N-1\right)}\cdot\left(y_{N}-y_{1}\right)-C\cdot\frac{n\left(N-n\right)}{N\left(N-1\right)}\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V\left(\hat{\overline{y}}^{*}\left(S\right)\right)=V\left(\overline{y}_{\pi}^{2}\right)-2C\frac{\left(N-n\right)}{N\left(N-1\right)}\cdot\left[\left(y_{N}-y_{1}\right)-nC\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V\left(\hat{\overline{y}}^{*}\left(S\right)\right)=V\left(\overline{y}_{\pi}^{2}\right)-2C\cdot\frac{N-n}{N}\cdot\frac{1}{N-1}\cdot\left[\left(y_{N}-y_{1}\right)-nC\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V\left(\hat{\overline{y}}^{*}\left(S\right)\right)=V\left(\overline{y}_{\pi}^{2}\right)-\left(1-f\right)\cdot\frac{2C}{N-1}\cdot\left[\left(y_{N}-y_{1}\right)-nC\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $V\left(\overline{y}_{\pi}^{2}\right)$
\end_inset

 es conocida.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V\left(\hat{\overline{y}}^{*}\left(S\right)\right)=\frac{1-f}{n}\cdot S_{yU}^{2}-\left(1-f\right)\cdot\frac{2C}{N-1}\cdot\left[\left(y_{N}-y_{1}\right)-nC\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V\left(\hat{\overline{y}}^{*}\left(S\right)\right)=\left(1-f\right)\cdot\left[\frac{S_{yU}^{2}}{n}-\frac{2C}{N-1}\cdot\left(y_{N}-y_{1}-nC\right)\right]
\]

\end_inset


\end_layout

\begin_layout Paragraph*
Identidad usada:
\end_layout

\begin_layout Standard
\begin_inset Formula $\sum_{s\in S_{A}}\overline{y}_{\pi}\left(s\right)-\sum_{s\in S_{B}}\overline{y}_{\pi}\left(s\right)=N_{A}\cdot\left(\frac{y_{N}}{n}-\frac{y_{1}}{n}\right)$
\end_inset


\end_layout

\begin_layout Standard
La idea coloquial a la que acudimos acá es que cada conjunto tiene todas
 las muestras donde está su elemento característico pero no está el otro
 y por lo demás sus 
\begin_inset Quotes fld
\end_inset

colas
\begin_inset Quotes frd
\end_inset

 son iguales.
\end_layout

\begin_layout Standard
La media está fuertemente emparentada con la suma de los totales.
 Es claro que cada conjunto de 
\begin_inset Formula $S_{A}$
\end_inset

 tiene un análogo en 
\begin_inset Formula $S_{B}$
\end_inset

 que es igual en todos sus elementos excepto en 
\begin_inset Formula $1/N$
\end_inset

.
 Entonces, es fácil ver que la diferencia entre estos dos es 
\begin_inset Formula $y_{N}-y_{1}$
\end_inset

.
 Podemos agrupar así los 
\begin_inset Formula $N_{A}=\binom{N-2}{n-1}$
\end_inset

 conjuntos obteniendo siempre la misma diferencia.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\sum_{s\in S_{A}}\overline{y}_{\pi}\left(s\right)-\sum_{s\in S_{B}}\overline{y}_{\pi}\left(s\right)=\sum_{a_{i}\in S_{A}}\sum_{j=1}^{n}\frac{a_{ij}}{n}-\sum_{b_{i}\in S_{B}}\sum_{j=1}^{n}\frac{b_{ij}}{n}
\]

\end_inset


\end_layout

\begin_layout Standard
Reordenamos 
\begin_inset Formula $S_{B}$
\end_inset

 para agrupar los conjuntos similares:
\end_layout

\begin_layout Standard
\begin_inset Formula $S_{B}'=\left\{ b'_{ij}/b'_{ij}\in S_{B}\wedge b'_{ij}\setminus a_{ij}=y_{1}\wedge a_{ij}\setminus b'_{ij}=y_{N}\right\} $
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $S_{B}'=S_{B}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\sum_{s\in S_{A}}\overline{y}_{\pi}\left(s\right)-\sum_{s\in S_{B}}\overline{y}_{\pi}\left(s\right)=\frac{1}{n}\left[\sum_{s_{i}\in S_{A}}\sum_{j=1}^{n}a_{ij}-\sum_{s_{i}\in S_{B}'}\sum_{j=1}^{n}b'_{ij}\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\#S_{B}=\#S_{A}=\binom{N-2}{n-1}=N_{A}=N_{B}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\sum_{s\in S_{A}}\overline{y}_{\pi}\left(s\right)-\sum_{s\in S_{B}}\overline{y}_{\pi}\left(s\right)=\frac{1}{n}\left[\sum_{i=1}^{N_{A}}\sum_{j=1}^{n}a_{ij}-\sum_{i=1}^{N_{A}}\sum_{j=1}^{n}b'_{ij}\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\sum_{s\in S_{A}}\overline{y}_{\pi}\left(s\right)-\sum_{s\in S_{B}}\overline{y}_{\pi}\left(s\right)=\frac{1}{n}\left[\sum_{i=1}^{N_{A}}\sum_{j=1}^{n}\left(a_{ij}-b'_{ij}\right)\right]
\]

\end_inset


\end_layout

\begin_layout Standard
Estamos comparando siempre los conjuntos que sólo se diferencian en los
 elementos 
\begin_inset Formula $y_{N}$
\end_inset

 e 
\begin_inset Formula $y_{1}$
\end_inset

, ya que 
\begin_inset Formula $b'_{ij}\setminus a_{ij}=y_{1}\wedge a_{ij}\setminus b'_{ij}=y_{N}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\sum_{s\in S_{A}}\overline{y}_{\pi}\left(s\right)-\sum_{s\in S_{B}}\overline{y}_{\pi}\left(s\right)=\frac{1}{n}\sum_{i=1}^{N_{A}}\left(y_{N}-y_{1}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\sum_{s\in S_{A}}\overline{y}_{\pi}\left(s\right)-\sum_{s\in S_{B}}\overline{y}_{\pi}\left(s\right)=\frac{N_{A}}{n}\left(y_{N}-y_{1}\right)
\]

\end_inset


\end_layout

\begin_layout Subsubsection*
c ¿Para qué valores de 
\begin_inset Formula $C$
\end_inset

 el nuevo estimador es más preciso que el 
\begin_inset Formula $\pi$
\end_inset

-estimador?
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{V\left(\hat{\overline{y}}^{*}\left(S\right)\right)}{V\left(\overline{y}_{\pi}\left(S\right)\right)}=\frac{\left(1-f\right)\cdot\left[\frac{S_{yU}^{2}}{n}-\frac{2C}{N-1}\cdot\left(y_{N}-y_{1}-nC\right)\right]}{\left(1-f\right)\cdot\frac{S_{yU}^{2}}{n}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{V\left(\hat{\overline{y}}^{*}\left(S\right)\right)}{V\left(\overline{y}_{\pi}\left(S\right)\right)}=1-\frac{2C\cdot\left(\nicefrac{y_{N}}{n}-\nicefrac{y_{1}}{n}-C\right)}{\left(N-1\right)S_{yU}^{2}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\frac{V\left(\hat{\overline{y}}^{*}\left(S\right)\right)}{V\left(\overline{y}_{\pi}\left(S\right)\right)}<1\Rightarrow\frac{2C\cdot\left(\nicefrac{y_{N}}{n}-\nicefrac{y_{1}}{n}+C\right)}{\left(N-1\right)S_{yU}^{2}}>0$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{2C\cdot\left(\nicefrac{y_{N}}{n}-\nicefrac{y_{1}}{n}-C\right)}{\left(N-1\right)S_{yU}^{2}}>0
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\forall N;\Omega:\left(N-1\right)S_{yU}^{2}>0$
\end_inset


\end_layout

\begin_layout Standard
Busquemos el caso límite:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
2C\cdot\left(\nicefrac{y_{N}}{n}-\nicefrac{y_{1}}{n}-C\right)>0
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
2C\cdot\left(\nicefrac{y_{N}}{n}-\nicefrac{y_{1}}{n}-C\right)>0
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $C>0$
\end_inset


\end_layout

\begin_layout Standard
Si 
\begin_inset Formula $2C=0$
\end_inset

, estamos ante el estimador 
\begin_inset Formula $\overline{y}_{\pi}$
\end_inset

.
 No analizamos este caso.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{y_{N}}{n}-\nicefrac{y_{1}}{n}-C>0
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{y_{N}}{n}-\nicefrac{y_{1}}{n}>C
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
y_{N}-y_{1}>nC
\]

\end_inset


\end_layout

\begin_layout Itemize
\begin_inset Formula 
\[
C<\frac{y_{N}-y_{1}}{n}
\]

\end_inset


\end_layout

\begin_layout Standard
Esto condice con lo que sería intuitivamente esperable: El indicador disminuye
 la varianza respecto al tradicional siempre y cuando 
\begin_inset Formula $C$
\end_inset

 esté 
\begin_inset Quotes fld
\end_inset

sustrayendo
\begin_inset Quotes frd
\end_inset

 la diferencia introducida por los atípicos 
\begin_inset Formula $y_{1}$
\end_inset

 e 
\begin_inset Formula $y_{N}$
\end_inset

, es decir si es menor que el aporte que hace esta diferencia.
 
\end_layout

\begin_layout Subsection*
13 Suponer que 
\begin_inset Formula $\forall k\in U:y_{k}=k$
\end_inset

.
 Demostrar que 
\begin_inset Formula $V_{SIST}\leq V_{MSA}$
\end_inset

 del estimador de la media para una muestra de tamaño 
\begin_inset Formula $n$
\end_inset

 del total 
\begin_inset Formula $N$
\end_inset

 de elementos de 
\begin_inset Formula $U$
\end_inset

 con 
\begin_inset Formula $a=\frac{N}{n}$
\end_inset

 entero.
\end_layout

\begin_layout Standard
Es inmediato que hay una alta heterogeneidad entre muestras.
\end_layout

\begin_layout Standard
Probar lo pedido equivale a mostrar que el efecto de diseño tiene un valor
 mayor a 1.
 Este indicador tiene una forma conocida que se desarrolló detaladamente
 en Särndall.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\text{efdis}\left(MS;\hat{t}_{\pi}\right)=\frac{V_{SIST}}{V_{MSA}}=1+\frac{n-1}{1-f}\cdot\delta
\]

\end_inset


\end_layout

\begin_layout Standard
Para comprobar lo pedido hay que observar si 
\begin_inset Formula $\delta<0$
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\delta=1-\frac{N-1}{N-a}\cdot\frac{SCD}{SCT}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
1-\frac{N-1}{N-a}\cdot\frac{SCD}{SCT}<0
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
-\frac{N-1}{N-a}\cdot\frac{SCD}{SCT}<-1
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{N-1}{N-a}\cdot\frac{SCD}{SCT}>1
\]

\end_inset


\end_layout

\begin_layout Subsubsection*
\begin_inset Formula $SCD=\sum_{r=1}^{a}\sum_{s_{r}}\left(y_{k}-\overline{y}_{sr}\right)^{2}$
\end_inset


\end_layout

\begin_layout Standard
Se observa en este caso que las muestras 
\begin_inset Formula $s_{r}$
\end_inset

 son de la forma 
\begin_inset Formula $\left\{ s_{ir}\right\} _{i=0}^{n}/s_{ir}=ia+r$
\end_inset

 con 
\begin_inset Formula $r$
\end_inset

 una constante fija para cada muestra, 
\begin_inset Formula $1<r<a$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula $\overline{y}_{sr}=\frac{1}{n}\cdot\sum_{i=0}^{n-1}\left(i\cdot a+r\right)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\overline{y}_{sr}=\frac{n}{n}r+\frac{a}{n}\cdot\sum_{i=1}^{n-1}i$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\overline{y}_{sr}=r+\frac{a}{\cancel{n}}\cdot\frac{\cancel{n}\left(n-1\right)}{2}i$
\end_inset


\end_layout

\begin_layout Standard
La suma de esta serie finita es conocida.
\end_layout

\begin_layout Standard
\begin_inset Formula $\overline{y}_{sr}=r+a\frac{n-1}{2}$
\end_inset


\end_layout

\begin_layout Itemize
\begin_inset Formula 
\[
\overline{y}_{sr}=r+a\cdot\frac{n-1}{2}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
SCD=\sum_{r=1}^{a}\sum_{s_{r}}\left(y_{k}-\overline{y}_{sr}\right)^{2}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
SCD=\sum_{r=1}^{a}\sum_{i=0}^{n-1}\left(\left(i\cdot a+r\right)-r-a\cdot\frac{n-1}{2}\right)^{2}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
SCD=\frac{a^{2}}{4}\sum_{r=1}^{a}\sum_{i=0}^{n-1}\left(2i-\left(n-1\right)\right)^{2}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
SCD=\frac{a^{2}}{4}\sum_{r=1}^{a}\sum_{i=0}^{n-1}\left[4i^{2}-4i\left(n-1\right)+\left(n-1\right)^{2}\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\sum_{k=1}^{N}k=\frac{n\cdot\left(n+1\right)}{2}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\sum_{k=1}^{N}k^{2}=\frac{n\cdot\left(n+1\right)\left(2n+1\right)}{6}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
SCD=\frac{a^{2}}{4}\sum_{r=1}^{a}\left[4\frac{\left(n-1\right)\cdot n\left(2n-1\right)}{6}-4\frac{n\cdot\left(n-1\right)}{2}\left(n-1\right)+n\left(n-1\right)^{2}\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
SCD=\frac{a^{2}}{12}n\cdot\left(n-1\right)\sum_{r=1}^{a}\left[2\left(2n-1\right)-6\left(n-1\right)+3\left(n-1\right)\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
SCD=\frac{a^{2}}{12}n\cdot\left(n-1\right)\sum_{r=1}^{a}\left(4n-2-6n+6+3n-3\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
SCD=\frac{a^{2}}{12}n\cdot\left(n-1\right)\sum_{r=1}^{a}\left(n+1\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
SCD=\frac{a^{2}}{12}n\cdot\left(n-1\right)a\left(n+1\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $N=an$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
SCD=\frac{1}{12}N\cdot\left(N-a\right)\left(N+a\right)
\]

\end_inset


\end_layout

\begin_layout Subsubsection*
\begin_inset Formula $SCT=\sum_{U}\left(y_{k}-\overline{y}_{U}\right)^{2}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\overline{y}_{U}=\frac{1}{N}\sum_{i=1}^{N}i$
\end_inset


\end_layout

\begin_layout Standard
Esta es la misma serie finita que nos cruzamos en 
\begin_inset Formula $\overline{y}_{sr}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula $\overline{y}_{U}=\frac{1}{N}\frac{N\left(N+1\right)}{2}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\overline{y}_{U}=\frac{N+1}{2}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\sum_{U}\left(y_{k}-\overline{y}_{U}\right)^{2}=\sum_{k=1}^{N}\left(k-\overline{y}_{U}\right)^{2}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\sum_{i=1}^{n}\left(X_{i}-\overline{X}\right)^{2}=\sum_{i=1}^{n}X_{i}^{2}-n\overline{X}^{2}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\sum_{U}\left(y_{k}-\overline{y}_{U}\right)^{2}=\left(\sum_{k=1}^{N}k^{2}\right)-N\overline{y}_{U}^{2}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\sum_{k=1}^{N}k^{2}=\frac{n\cdot\left(n+1\right)\left(2n+1\right)}{6}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\sum_{U}\left(y_{k}-\overline{y}_{U}\right)^{2}=\frac{N\cdot\left(N+1\right)\left(2N+1\right)}{6}-N\overline{y}_{U}^{2}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\overline{y}_{U}=\frac{N+1}{2}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\sum_{U}\left(y_{k}-\overline{y}_{U}\right)^{2}=\frac{N\cdot\left(N+1\right)\left(2N+1\right)}{6}-N\left(\frac{N+1}{2}\right)^{2}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\sum_{U}\left(y_{k}-\overline{y}_{U}\right)^{2}=\left(N+1\right)\left[\frac{N\cdot\left(2N+1\right)}{6}-N\frac{N+1}{4}\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\sum_{U}\left(y_{k}-\overline{y}_{U}\right)^{2}=\frac{1}{12}N\cdot\left(N+1\right)\left(N-1\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\sum_{U}\left(y_{k}-\overline{y}_{U}\right)^{2}=\frac{1}{12}N\cdot\left(N+1\right)\left(N-1\right)
\]

\end_inset


\end_layout

\begin_layout Standard
Podemos armar el cociente
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{SCD}{SCT}\cdot\frac{N-1}{N-a}=\frac{\frac{1}{12}N\cdot\left(N-a\right)\left(N+a\right)}{\frac{1}{12}N\cdot\left(N+1\right)\left(N-1\right)}\cdot\frac{N-1}{N-a}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{SCD}{SCT}\cdot\frac{N-1}{N-a}=\frac{N+a}{N+1}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $a>1$
\end_inset


\end_layout

\begin_layout Standard
Ya tenemos lo que necesitamos.
 Esta condición es razonable ya que si 
\begin_inset Formula $a=1$
\end_inset

 estamos en realidad ante un MSA.
\end_layout

\begin_layout Itemize
\begin_inset Formula 
\[
\delta\leq1
\]

\end_inset


\end_layout

\begin_layout Standard
Empleando las relaciones antes citadas, es evidente que 
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{V_{MS}}{V_{MSA}}\leq1
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V_{MS}\leq V_{MSA}
\]

\end_inset


\end_layout

\begin_layout Section*
Ejercicios de la Guía 3
\end_layout

\begin_layout Subsection*
3 Considerando el ejercicio 13 de la práctica 2, para la población de 
\begin_inset Formula $N$
\end_inset

 unidades dividida en 
\begin_inset Formula $H=n$
\end_inset

 estratos de cada uno de los cuales se selecciona un elemento, probar que
 
\begin_inset Formula $V_{EST}\leq V_{SIS}$
\end_inset


\end_layout

\begin_layout Quote
13-2: 
\begin_inset Quotes fld
\end_inset

Suponer que 
\begin_inset Formula $\forall k\in U:y_{k}=k$
\end_inset

.
 Demostrar que 
\begin_inset Formula $V_{SIST}\leq V_{MSA}$
\end_inset

 del estimador de la media para una muestra de tamaño 
\begin_inset Formula $n$
\end_inset

 del total 
\begin_inset Formula $N$
\end_inset

 de elementos de 
\begin_inset Formula $U$
\end_inset

 con 
\begin_inset Formula $a=\frac{N}{n}$
\end_inset

 entero.
\begin_inset Quotes frd
\end_inset


\end_layout

\begin_layout Standard
En este caso, no desarrollamos una fórmula especial para el efecto diseño,
 así que conviene comparar directamente las dos expresiones de varianza.
 
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V_{SIS}\left(\hat{t}_{\pi}\right)=a\sum_{r=1}^{a}\left(t_{s_{r}}-\overline{t}\right)^{2}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
V_{EST}\left(\hat{t}_{\pi}\right)=\sum_{h=0}^{H-1}N_{h}^{2}\frac{1-f_{h}}{n_{h}}S_{yU_{h}}^{2}
\]

\end_inset


\end_layout

\begin_layout Itemize

\lang spanish
Se aclaró que los 
\begin_inset Formula $N_{h}$
\end_inset

 son todos iguales.
 Además, se toma un elemento por estrato.
\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula $H=n=\frac{N}{a}$
\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula $\forall1<h<H:\left(N_{h};n_{h}\right)=\left(a;1\right)$
\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
V_{EST}\left(\hat{t}_{\pi}\right)=\sum_{h=0}^{H-1}N_{h}^{2}\frac{1-\frac{1}{N_{h}}}{1}S_{yU_{h}}^{2}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
V_{EST}\left(\hat{t}_{\pi}\right)=\sum_{h=0}^{H-1}N_{h}\frac{N_{h}-1}{1}S_{yU_{h}}^{2}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
V_{EST}\left(\hat{t}_{\pi}\right)=\sum_{h=0}^{n-1}a\left(a-1\right)S_{yU_{h}}^{2}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
V_{EST}\left(\hat{t}_{\pi}\right)=a\left(a-1\right)\sum_{h=0}^{n-1}S_{yU_{h}}^{2}
\]

\end_inset


\end_layout

\begin_layout Subsubsection*

\lang spanish
Buscamos 
\begin_inset Formula $S_{yU_{h}}^{2}$
\end_inset

 y 
\begin_inset Formula $t_{s_{r}}$
\end_inset

 y 
\begin_inset Formula $\overline{t}$
\end_inset

.
\end_layout

\begin_layout Standard

\lang spanish
Cada uno de los 
\begin_inset Formula $n$
\end_inset

 estratos tiene 
\begin_inset Formula $a$
\end_inset

 elementos.
 Por la expresión elegida, 
\begin_inset Formula $y_{hi}=ah+i$
\end_inset

, los 
\begin_inset Formula $n$
\end_inset

 estratos están numerados con etiquetas 
\begin_inset Formula $\left\{ h\right\} _{0}^{n-1}$
\end_inset

.
\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
S_{yU_{h}}^{2}=\frac{\sum_{i=1}^{a}\left(y_{i+h}-\overline{y}_{h}\right)^{2}}{a-1}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula $\overline{y}_{hEST}=\frac{\sum_{i=1}^{a}ah+i}{a}$
\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula $\overline{y}_{hEST}=ah+\frac{\frac{a\left(a+1\right)}{2}}{a}$
\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula $\overline{y}_{hEST}=ah+\frac{a+1}{2}$
\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
S_{yU_{h}}^{2}=\frac{\sum_{i=1}^{a}\left[ah+i-\left(ah+\frac{a+1}{2}\right)\right]^{2}}{a-1}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
S_{yU_{h}}^{2}=\frac{\sum_{i=1}^{a}\left(i-\frac{a+1}{2}\right)^{2}}{a-1}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
S_{yU_{h}}^{2}=\frac{1}{4}\frac{\sum_{i=1}^{a}\left[2i-\left(a+1\right)\right]^{2}}{a-1}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
S_{yU_{h}}^{2}=\frac{1}{4}\frac{\sum_{i=1}^{a}\left[4i^{2}-4i\left(a+1\right)+\left(a+1\right)^{2}\right]}{a-1}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\sum_{k=1}^{N}k=\frac{n\cdot\left(n+1\right)}{2}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\sum_{k=1}^{N}k^{2}=\frac{n\cdot\left(n+1\right)\left(2n+1\right)}{6}$
\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
S_{yU_{h}}^{2}=\frac{1}{4}\frac{4\frac{a\cdot\left(a+1\right)\left(2a+1\right)}{6}-4\frac{a\cdot\left(a+1\right)}{2}\left(a+1\right)+a\left(a+1\right)^{2}}{a-1}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
S_{yU_{h}}^{2}=\frac{a\left(a+1\right)}{12}\frac{2\cdot\left(2a+1\right)-6\cdot\left(a+1\right)+3\left(a+1\right)}{a-1}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
S_{yU_{h}}^{2}=\frac{a\left(a+1\right)}{12}\frac{4a+2-6a-6+3a+3}{a-1}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
S_{yU_{h}}^{2}=\frac{a\left(a+1\right)}{12}\frac{a-1}{a-1}
\]

\end_inset


\end_layout

\begin_layout Itemize

\lang spanish
\begin_inset Formula 
\[
S_{yU_{h}}^{2}=\frac{a\left(a+1\right)}{12}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\overline{y}_{sr}=r+a\frac{n-1}{2}$
\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
t_{s_{r}}=n\overline{y}_{sr}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
t_{s_{r}}=nr+a\frac{\left(n-1\right)n}{2}
\]

\end_inset


\end_layout

\begin_layout Itemize

\lang spanish
\begin_inset Formula 
\[
t_{s_{r}}=n\left(r+\frac{N-a}{2}\right)
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
\overline{t}=\frac{t}{a}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula $t=\sum_{i=1}^{N}i=\frac{N\left(N+1\right)}{2}$
\end_inset


\end_layout

\begin_layout Itemize

\lang spanish
\begin_inset Formula 
\[
\overline{t}=\frac{N\left(N+1\right)}{2a}
\]

\end_inset


\end_layout

\begin_layout Subsubsection*

\lang spanish
Determinamos ambas varianzas con los datos hallados
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V_{SIS}\left(\hat{t}_{\pi}\right)=a\sum_{r=1}^{a}\left(t_{s_{r}}-\overline{t}\right)^{2}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V_{SIS}\left(\hat{t}_{\pi}\right)=a\sum_{r=1}^{a}\left(n\left(r+\frac{N-a}{2}\right)-\frac{N\left(N+1\right)}{2a}\right)^{2}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $na=N$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V_{SIS}\left(\hat{t}_{\pi}\right)=an^{2}\sum_{r=1}^{a}\left(r+\frac{N-a}{2}-\frac{N+1}{2}\right)^{2}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V_{SIS}\left(\hat{t}_{\pi}\right)=an^{2}\sum_{r=1}^{a}\left(r-\frac{a+1}{2}\right)^{2}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V_{SIS}\left(\hat{t}_{\pi}\right)=n^{2}\frac{a}{4}\sum_{r=1}^{a}\left[2r-\left(a+1\right)\right]^{2}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V_{SIS}\left(\hat{t}_{\pi}\right)=n^{2}\frac{a}{4}\sum_{r=1}^{a}\left[4r^{2}-4r\left(a+1\right)+\left(a+1\right)^{2}\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\sum_{k=1}^{N}k=\frac{n\cdot\left(n+1\right)}{2}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\sum_{k=1}^{N}k^{2}=\frac{n\cdot\left(n+1\right)\left(2n+1\right)}{6}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V_{SIS}\left(\hat{t}_{\pi}\right)=n^{2}\frac{a}{4}\left[4\frac{a\cdot\left(a+1\right)\left(2a+1\right)}{6}-4\frac{a\cdot\left(a+1\right)}{2}\left(a+1\right)+a\left(a+1\right)^{2}\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V_{SIS}\left(\hat{t}_{\pi}\right)=n^{2}a^{2}\frac{a+1}{4}\left[\frac{2}{3}\left(2a+1\right)-2\cdot\left(a+1\right)+\left(a+1\right)\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V_{SIS}\left(\hat{t}_{\pi}\right)=n^{2}a^{2}\frac{a+1}{12}\left[4a+2-6\cdot\left(a+1\right)+3\left(a+1\right)\right]
\]

\end_inset


\end_layout

\begin_layout Itemize
\begin_inset Formula 
\[
V_{SIS}\left(\hat{t}_{\pi}\right)=n^{2}a^{2}\frac{a+1}{12}\left(a-1\right)
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
V_{EST}\left(\hat{t}_{\pi}\right)=a\left(a-1\right)\sum_{h=0}^{n-1}S_{yU_{h}}^{2}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
V_{EST}\left(\hat{t}_{\pi}\right)=a\left(a-1\right)\sum_{h=0}^{n-1}\frac{a\left(a+1\right)}{12}
\]

\end_inset


\end_layout

\begin_layout Itemize

\lang spanish
\begin_inset Formula 
\[
V_{EST}\left(\hat{t}_{\pi}\right)=a\left(a-1\right)n\frac{a\left(a+1\right)}{12}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
\frac{V_{EST}\left(\hat{t}_{\pi}\right)}{V_{SIS}\left(\hat{t}_{\pi}\right)}=\frac{a\left(a-1\right)n\frac{a\left(a+1\right)}{12}}{n^{2}a^{2}\frac{a+1}{12}\left(a-1\right)}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
\frac{V_{EST}\left(\hat{t}_{\pi}\right)}{V_{SIS}\left(\hat{t}_{\pi}\right)}=\frac{ana}{n^{2}a^{2}}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
\frac{V_{EST}\left(\hat{t}_{\pi}\right)}{V_{SIS}\left(\hat{t}_{\pi}\right)}=\frac{1}{n}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
V_{EST}\left(\hat{t}_{\pi}\right)\cdot n=V_{SIS}\left(\hat{t}_{\pi}\right)
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula $n\geq1$
\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
V_{EST}\left(\hat{t}_{\pi}\right)<V_{SIS}\left(\hat{t}_{\pi}\right)
\]

\end_inset


\end_layout

\begin_layout Subsection*

\lang spanish
Ejercicio 4
\end_layout

\begin_layout Subsubsection*
a Sea una población dividida en dos estratos.
 Se define 
\begin_inset Formula $\phi$
\end_inset

 a la razón entre 
\begin_inset Formula $\nicefrac{n_{1}}{n_{2}}$
\end_inset

 y 
\begin_inset Formula $\nicefrac{n_{1Ney}}{n_{2Ney}}$
\end_inset

 que surjen de la adjudicación de Neyman.
 El objetivo es estimar el total usando un diseño MESA.
 Probar que la varianza obtenida con la adjudicación de Neyman es menor.
 
\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
V_{EST}\left(\hat{t}_{\pi}\right)=N_{1}^{2}\frac{1-f_{1}}{n_{1}}S_{yU_{1}}^{2}+N_{2}^{2}\frac{1-f_{2}}{n_{2}}S_{yU_{2}}^{2}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula $N_{h}^{2}\frac{1-f_{h}}{n_{h}}S_{yU_{h}}^{2}=N_{h}^{2}\left(\frac{1}{n_{h}}-\frac{1}{N_{h}}\right)S_{yU_{h}}^{2}$
\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula $N_{h}\gg0\Rightarrow\frac{1}{N_{h}}\approx0$
\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula $N_{h}^{2}\frac{1-f_{h}}{n_{h}}S_{yU_{h}}^{2}=\frac{N_{h}^{2}}{n_{h}}S_{yU_{h}}^{2}$
\end_inset


\end_layout

\begin_layout Itemize

\lang spanish
\begin_inset Formula 
\[
V_{EST}\left(\hat{t}_{\pi}\right)\approx\frac{N_{1}^{2}}{n_{1}}S_{yU_{1}}^{2}+\frac{N_{2}^{2}}{n_{2}}S_{yU_{2}}^{2}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
V_{EST,Ney}\left(\hat{t}_{\pi}\right)=\frac{N_{1}^{2}}{n_{1Ney}}S_{yU_{1}}^{2}+\frac{N_{2}^{2}}{n_{2Ney}}S_{yU_{2}}^{2}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
V_{EST,Ney}\left(\hat{t}_{\pi}\right)=\frac{N_{1}^{2}}{n\frac{N_{1}S_{yU_{1}}}{N_{1}S_{yU_{1}}+N_{2}S_{yU_{2}}}}S_{yU_{1}}^{2}+\frac{N_{2}^{2}}{n\frac{N_{2}S_{yU_{2}}}{N_{1}S_{yU_{1}}+N_{2}S_{yU_{2}}}}S_{yU_{2}}^{2}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
V_{EST,Ney}\left(\hat{t}_{\pi}\right)=\frac{1}{n}\left(N_{1}S_{yU_{1}}+N_{2}S_{yU_{2}}\right)\frac{1}{n}\left(N_{1}S_{yU_{1}}+N_{2}S_{yU_{2}}\right)
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
V_{EST,Ney}\left(\hat{t}_{\pi}\right)=\frac{1}{n^{2}}\left(N_{1}S_{yU_{1}}+N_{2}S_{yU_{2}}\right)^{2}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
Planteamos la diferencia:
\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
V_{EST}\left(\hat{t}_{\pi}\right)-V_{EST,Ney}\left(\hat{t}_{\pi}\right)=\frac{N_{1}^{2}}{n_{1}}S_{yU_{1}}^{2}+\frac{N_{2}^{2}}{n_{2}}S_{yU_{2}}^{2}-\frac{1}{n^{2}}\left(N_{1}S_{yU_{1}}+N_{2}S_{yU_{2}}\right)^{2}
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\[
V_{EST}\left(\hat{t}_{\pi}\right)-V_{EST,Ney}\left(\hat{t}_{\pi}\right)=\frac{N_{1}^{2}}{n_{1}}S_{yU_{1}}^{2}+\frac{N_{2}^{2}}{n_{2}}S_{yU_{2}}^{2}-\frac{1}{n^{2}}\left(N_{1}^{2}S_{yU_{1}}^{2}+N_{2}^{2}S_{yU_{2}}^{2}+2N_{1}S_{yU_{1}}N_{2}S_{yU_{2}}\right)
\]

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
Adecuamos el primer término para poder usar un factor común.
\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula $\frac{N_{1}^{2}}{n_{1}}S_{yU_{1}}^{2}+\frac{N_{2}^{2}}{n_{2}}S_{yU_{2}}^{2}=\frac{1}{n}\left(\frac{N_{1}^{2}}{\frac{n_{1}}{n_{1}+n_{2}}}S_{yU_{1}}^{2}+\frac{N_{2}^{2}}{\frac{n_{2}}{n_{1}+n_{2}}}S_{yU_{2}}^{2}\right)$
\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula $\frac{N_{1}^{2}}{n_{1}}S_{yU_{1}}^{2}+\frac{N_{2}^{2}}{n_{2}}S_{yU_{2}}^{2}=\frac{1}{n}\left[N_{1}^{2}\left(n_{2}+1\right)S_{yU_{1}}^{2}+N_{2}^{2}\left(n_{1}+1\right)S_{yU_{2}}^{2}\right]$
\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula $\frac{N_{1}^{2}}{n_{1}}S_{yU_{1}}^{2}+\frac{N_{2}^{2}}{n_{2}}S_{yU_{2}}^{2}=\frac{1}{n^{2}}\left[N_{1}^{2}n\left(n_{2}+1\right)S_{yU_{1}}^{2}+N_{2}^{2}n\left(n_{1}+1\right)S_{yU_{2}}^{2}\right]$
\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\begin{align*}
V_{EST}\left(\hat{t}_{\pi}\right)-V_{EST,Ney}\left(\hat{t}_{\pi}\right) & =\frac{1}{n^{2}}\left[N_{1}^{2}n\left(n_{2}\cancel{+1}\right)S_{yU_{1}}^{2}+N_{2}^{2}n\left(n_{1}\cancel{+1}\right)S_{yU_{2}}^{2}\right]\\
 & -\frac{1}{n^{2}}\left(\cancel{N_{1}^{2}S_{yU_{1}}^{2}}\cancel{+N_{2}^{2}S_{yU_{2}}^{2}}+2N_{1}S_{yU_{1}}N_{2}S_{yU_{2}}\right)
\end{align*}

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\begin{align*}
V_{EST}\left(\hat{t}_{\pi}\right)-V_{EST,Ney}\left(\hat{t}_{\pi}\right) & =\frac{1}{n^{2}}\left[N_{1}^{2}nn_{2}S_{yU_{1}}^{2}+N_{2}^{2}nn_{1}S_{yU_{2}}^{2}\right]\\
 & -\frac{1}{n^{2}}\left(2N_{1}S_{yU_{1}}N_{2}S_{yU_{2}}\right)
\end{align*}

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\begin{align*}
V_{EST}\left(\hat{t}_{\pi}\right)-V_{EST,Ney}\left(\hat{t}_{\pi}\right) & =\frac{1}{n^{2}}\left[N_{1}^{2}nn_{2}S_{yU_{1}}^{2}+N_{2}^{2}nn_{1}S_{yU_{2}}^{2}-2N_{1}S_{yU_{1}}N_{2}S_{yU_{2}}\right]
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
Dada la existencia de estratos, es razonable asumir
\end_layout

\begin_layout Standard
\begin_inset Formula $n\geq2$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $n_{h}>1$
\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula $nn_{h}>1$
\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\begin{align*}
V_{EST}\left(\hat{t}_{\pi}\right)-V_{EST,Ney}\left(\hat{t}_{\pi}\right) & >\frac{1}{n^{2}}\left[N_{1}^{2}S_{yU_{1}}^{2}+N_{2}^{2}S_{yU_{2}}^{2}-2N_{1}S_{yU_{1}}N_{2}S_{yU_{2}}\right]
\end{align*}

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\begin{align*}
V_{EST}\left(\hat{t}_{\pi}\right)-V_{EST,Ney}\left(\hat{t}_{\pi}\right) & >\frac{1}{n^{2}}\left(N_{1}^{2}S_{yU_{1}}^{2}-N_{2}^{2}S_{yU_{2}}^{2}\right)^{2}
\end{align*}

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\begin{align*}
V_{EST}\left(\hat{t}_{\pi}\right)-V_{EST,Ney}\left(\hat{t}_{\pi}\right) & >0
\end{align*}

\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula 
\begin{align*}
V_{EST}\left(\hat{t}_{\pi}\right) & >V_{EST,Ney}\left(\hat{t}_{\pi}\right)
\end{align*}

\end_inset


\end_layout

\begin_layout Subsubsection*
b Probar que 
\begin_inset Formula $\frac{V_{Ney}}{V_{EST}}\geq\frac{4\phi}{\left(1+\phi\right)^{2}}$
\end_inset


\begin_inset Foot
status open

\begin_layout Plain Layout

\series bold
Ayuda:
\series default
 
\begin_inset Formula $\frac{x+a}{x+b}\geq\frac{a}{b}$
\end_inset

 si 
\begin_inset Formula $x\geq0$
\end_inset

 y 
\begin_inset Formula $a\geq b$
\end_inset

.
\end_layout

\end_inset

.
\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula $V_{EST}\left(\hat{t}_{\pi}\right)\approx\frac{N_{1}^{2}}{n_{1}}S_{yU_{1}}^{2}+\frac{N_{2}^{2}}{n_{2}}S_{yU_{2}}^{2}$
\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula $V_{EST,Ney}\left(\hat{t}_{\pi}\right)=\frac{N_{1}^{2}}{n_{1Ney}}S_{yU_{1}}^{2}+\frac{N_{2}^{2}}{n_{2Ney}}S_{yU_{2}}^{2}$
\end_inset


\end_layout

\begin_layout Subsubsection*

\lang spanish
Bucamos alguna relación entre los 
\begin_inset Formula $n_{hNey}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
n_{hNey}=n\frac{N_{h}S_{yU_{h}}}{\sum_{h=1}^{H}N_{h}S_{yU_{h}}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
n_{hNey}=n\frac{N_{h}S_{yU_{h}}}{N_{1}S_{yU_{1}}+N_{2}S_{yU_{2}}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{n_{1Ney}}{n_{2Ney}}=\frac{N_{1}S_{yU_{1}}}{N_{2}S_{yU_{2}}}
\]

\end_inset


\end_layout

\begin_layout Subsubsection*

\lang spanish
Buscamos Expresiones para Ambas Varianzas
\end_layout

\begin_layout Standard

\lang spanish
Recordando la hipótesis de que los 
\begin_inset Formula $N_{h}$
\end_inset

 son 
\begin_inset Quotes cld
\end_inset

muy grandes
\begin_inset Quotes crd
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V_{MESA}\left(\hat{t}_{\pi}\right)=\sum N_{h}^{2}\frac{\left(1-\cancel{f}\right)}{n_{h}}S_{h}^{2}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V_{MESA}\left(\hat{t}_{\pi}\right)=\frac{N_{1}^{2}S_{1}^{2}}{n_{1}}+\frac{N_{2}^{2}S_{2}^{2}}{n_{2}}
\]

\end_inset


\end_layout

\begin_layout Standard
Aplicamos la asignación neyman para encontrar la otra expresión:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V_{MESA_{Ney}}\left(\hat{t}_{\pi}\right)=\frac{N_{1}^{2}S_{1}^{2}}{n\left(\frac{N_{1}S_{1}}{N_{1}S_{1}+N_{2}S_{2}}\right)}+\frac{N_{2}^{2}S_{2}^{2}}{n\left(\frac{N_{2}S_{2}}{N_{1}S_{1}+N_{2}S_{2}}\right)}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V_{MESA_{Ney}}\left(\hat{t}_{\pi}\right)=\left(N_{1}S_{1}+N_{2}S_{2}\right)\left(\frac{N_{1}S_{1}}{n}+\frac{N_{2}S_{2}}{n}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V_{MESA_{Ney}}\left(\hat{t}_{\pi}\right)=\frac{\left(N_{1}S_{1}+N_{2}S_{2}\right)^{2}}{n}
\]

\end_inset


\end_layout

\begin_layout Subsubsection*

\lang spanish
Buscamos Algunas identidades para 
\begin_inset Formula $\phi$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\phi=\frac{\nicefrac{n_{1}}{n_{2}}}{\nicefrac{n_{1Ney}}{n_{2Ney}}}$
\end_inset


\end_layout

\begin_layout Itemize
\begin_inset Formula $\phi=\frac{n_{1}n_{2Ney}}{n_{2}n_{1Ney}}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\nicefrac{n_{2Ney}}{n_{1Ney}}=\frac{n\left(\frac{N_{2}S_{2}}{N_{1}S_{1}+N_{2}S_{2}}\right)}{n\left(\frac{N_{1}S_{1}}{N_{1}S_{1}+N_{2}S_{2}}\right)}=\frac{N_{2}S_{yU_{2}}}{N_{1}S_{yU_{1}}}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\phi=\frac{n_{1}}{n_{2}}\frac{N_{2}S_{yU_{2}}}{N_{1}S_{yU_{1}}}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\phi\frac{n_{2}}{n_{1}}=\frac{N_{2}S_{yU_{2}}}{N_{1}S_{yU_{1}}}$
\end_inset


\end_layout

\begin_layout Subsubsection*

\lang spanish
Trabajamos con la Razón pedida entre las varianzas.
 
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{V_{MESA_{Ney}}\left(\hat{t}_{\pi}\right)}{V_{EST}\left(\hat{t}_{\pi}\right)}=\frac{\frac{\left(N_{1}S_{1}+N_{2}S_{2}\right)^{2}}{n}}{\frac{N_{1}^{2}S_{1}^{2}}{n_{1}}+\frac{N_{2}^{2}S_{2}^{2}}{n_{2}}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{V_{MESA_{Ney}}\left(\hat{t}_{\pi}\right)}{V_{EST}\left(\hat{t}_{\pi}\right)}=\frac{1}{n}\frac{N_{1}^{2}S_{1}^{2}\left(1+\frac{N_{2}S_{2}}{N_{1}S_{1}}\right)^{2}}{N_{1}^{2}S_{1}^{2}\left[\frac{1}{n_{1}}+\frac{1}{n_{2}}\left(\frac{N_{2}S_{2}}{N_{1}S_{1}}\right)^{2}\right]}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{V_{MESA_{Ney}}\left(\hat{t}_{\pi}\right)}{V_{EST}\left(\hat{t}_{\pi}\right)}=\frac{1}{n}\frac{\left(1+\frac{N_{2}S_{2}}{N_{1}S_{1}}\right)^{2}}{\frac{1}{n_{1}}+\frac{1}{n_{2}}\left(\frac{N_{2}S_{2}}{N_{1}S_{1}}\right)^{2}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{V_{MESA_{Ney}}\left(\hat{t}_{\pi}\right)}{V_{EST}\left(\hat{t}_{\pi}\right)}=\frac{\left(1+\frac{N_{2}S_{2}}{N_{1}S_{1}}\right)^{2}}{\left(n_{1}+n_{2}\right)\left[\frac{1}{n_{1}}+\frac{1}{n_{2}}\left(\frac{N_{2}S_{2}}{N_{1}S_{1}}\right)^{2}\right]}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\phi\frac{n_{2}}{n_{1}}=\frac{N_{2}S_{yU_{2}}}{N_{1}S_{yU_{1}}}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{V_{MESA_{Ney}}\left(\hat{t}_{\pi}\right)}{V_{EST}\left(\hat{t}_{\pi}\right)}=\frac{1}{n}\frac{\left(1+\phi\frac{n_{2}}{n_{1}}\right)^{2}}{\left[\frac{1}{n_{1}}+\frac{1}{n_{2}}\left(\phi\frac{n_{2}}{n_{1}}\right)^{2}\right]}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{V_{MESA_{Ney}}\left(\hat{t}_{\pi}\right)}{V_{EST}\left(\hat{t}_{\pi}\right)}=\frac{1}{n}\frac{\left(\frac{n_{1}+\phi n_{2}}{n_{1}}\right)^{2}}{\frac{n_{2}+n_{1}\left(\phi\frac{n_{2}}{n_{1}}\right)^{2}}{n_{1}n_{2}}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{V_{MESA_{Ney}}\left(\hat{t}_{\pi}\right)}{V_{EST}\left(\hat{t}_{\pi}\right)}=\frac{1}{n}\cdot\frac{n_{1}n_{2}}{n_{1}^{2}}\cdot\frac{\left(n_{1}+\phi n_{2}\right)^{2}}{n_{2}+n_{1}\left(\phi\frac{n_{2}}{n_{1}}\right)^{2}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{V_{MESA_{Ney}}\left(\hat{t}_{\pi}\right)}{V_{EST}\left(\hat{t}_{\pi}\right)}=\frac{1}{n}\cdot\frac{n_{2}}{n_{1}}\cdot\frac{\left(n_{1}+\phi n_{2}\right)^{2}}{n_{2}+\phi^{2}\frac{n_{2}^{2}}{n_{1}}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{V_{MESA_{Ney}}\left(\hat{t}_{\pi}\right)}{V_{EST}\left(\hat{t}_{\pi}\right)}=\frac{1}{n}\cdot\frac{n_{2}}{n_{1}}\cdot\frac{\left(n_{1}+\phi n_{2}\right)^{2}}{n_{2}\left(1+\phi^{2}\frac{n_{2}}{n_{1}}\right)}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{V_{MESA_{Ney}}\left(\hat{t}_{\pi}\right)}{V_{EST}\left(\hat{t}_{\pi}\right)}=\frac{1}{n}\cdot\frac{1}{n_{1}}\cdot\frac{\left(n_{1}+\phi n_{2}\right)^{2}}{1+\phi^{2}\frac{n_{2}}{n_{1}}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{V_{MESA_{Ney}}\left(\hat{t}_{\pi}\right)}{V_{EST}\left(\hat{t}_{\pi}\right)}=\frac{1}{n}\cdot\frac{\left(n_{1}+\phi n_{2}\right)^{2}}{n_{1}+\phi^{2}n_{2}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{V_{MESA_{Ney}}\left(\hat{t}_{\pi}\right)}{V_{EST}\left(\hat{t}_{\pi}\right)}=\frac{\left(n_{1}+\phi n_{2}\right)^{2}}{\left(n_{1}+n_{2}\right)\left(n_{1}+\phi^{2}n_{2}\right)}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{V_{MESA_{Ney}}\left(\hat{t}_{\pi}\right)}{V_{EST}\left(\hat{t}_{\pi}\right)}=\frac{\left(n_{1}+\phi n_{2}\right)^{2}}{n_{1}\left(n_{1}+\phi^{2}n_{2}\right)+n_{2}\left(n_{1}+\phi^{2}n_{2}\right)}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{V_{MESA_{Ney}}\left(\hat{t}_{\pi}\right)}{V_{EST}\left(\hat{t}_{\pi}\right)}=\frac{\left(n_{1}+\phi n_{2}\right)^{2}}{n_{1}^{2}+\phi^{2}n_{1}n_{2}+n_{1}n_{2}+\phi^{2}n_{2}^{2}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\left(n_{1}+\phi n_{2}\right)^{2}=\left(n_{1}-\phi n_{2}\right)^{2}+4\phi n_{1}n_{2}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{V_{MESA_{Ney}}\left(\hat{t}_{\pi}\right)}{V_{EST}\left(\hat{t}_{\pi}\right)}=\frac{\left(n_{1}-\phi n_{2}\right)^{2}+4\phi n_{1}n_{2}}{n_{1}^{2}+\phi^{2}n_{1}n_{2}+n_{1}n_{2}+\phi^{2}n_{2}^{2}+2\phi n_{1}n_{2}-2\phi n_{1}n_{2}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{V_{MESA_{Ney}}\left(\hat{t}_{\pi}\right)}{V_{EST}\left(\hat{t}_{\pi}\right)}=\frac{\left(n_{1}-\phi n_{2}\right)^{2}+4\phi n_{1}n_{2}}{n_{1}^{2}-2\phi n_{1}n_{2}+\phi^{2}n_{2}^{2}+\phi^{2}n_{1}n_{2}+n_{1}n_{2}+2\phi n_{1}n_{2}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{V_{MESA_{Ney}}\left(\hat{t}_{\pi}\right)}{V_{EST}\left(\hat{t}_{\pi}\right)}=\frac{\left(n_{1}-\phi n_{2}\right)^{2}+4\phi n_{1}n_{2}}{\left(n_{1}-\phi n_{2}\right)^{2}+\phi^{2}n_{1}n_{2}+n_{1}n_{2}+2\phi n_{1}n_{2}}
\]

\end_inset


\end_layout

\begin_layout Standard
Aplicamos la identidad de la consigna.
 
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{V_{MESA_{Ney}}\left(\hat{t}_{\pi}\right)}{V_{EST}\left(\hat{t}_{\pi}\right)}=\frac{4\phi n_{1}n_{2}}{\phi^{2}n_{1}n_{2}+n_{1}n_{2}+2\phi n_{1}n_{2}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{V_{MESA_{Ney}}\left(\hat{t}_{\pi}\right)}{V_{EST}\left(\hat{t}_{\pi}\right)}=\frac{4\phi}{\phi^{2}+1+2\phi}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{V_{MESA_{Ney}}\left(\hat{t}_{\pi}\right)}{V_{EST}\left(\hat{t}_{\pi}\right)}=\frac{4\phi}{\left(1+\phi\right)^{2}}
\]

\end_inset


\end_layout

\begin_layout Section*
Punto i del Ejercicio de Estratificación
\end_layout

\begin_layout Subsection*
i Construir el estimador de HT 
\begin_inset Formula $\hat{d}_{i\pi}$
\end_inset

 siendo 
\begin_inset Formula $d_{i}=\overline{y}_{U_{i}}-\overline{y}_{U}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
d_{i}=\overline{y}_{U_{i}}-\overline{y}_{U}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $W_{h}=\nicefrac{N_{h}}{N}\Rightarrow\overline{y}_{U}=\sum_{h=1}^{H}W_{h}\overline{y}_{U_{h}}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
d_{i}=\overline{y}_{U_{i}}-\sum_{h=1}W_{h}\overline{y}_{U_{h}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
d_{i}=\left(1-W_{i}\right)\overline{y}_{U_{i}}-\sum_{h\neq i}W_{h}\overline{y}_{U_{h}}
\]

\end_inset


\end_layout

\begin_layout Standard
En Särndall, probamos que
\lang spanish
 siempre se cumple 
\begin_inset Formula $\mathbb{E}\left(\sum_{s}\check{a}_{k}\right)=\sum_{U}a_{k}$
\end_inset

.
 
\end_layout

\begin_layout Standard

\lang spanish
Es decir, el estimador de HT es insertable.
 
\end_layout

\begin_layout Standard

\lang spanish
Sean 
\begin_inset Formula $\hat{\overline{y}}_{s_{h}}=\frac{1}{N_{h}}\sum_{s\cap U_{h}}\frac{y_{h}}{\pi_{h}}=\frac{1}{N_{h}}\sum_{s\cap U_{h}}\check{y}_{h}$
\end_inset


\end_layout

\begin_layout Standard

\lang spanish
\begin_inset Formula $\pi_{h}=\frac{n_{h}}{N_{h}}$
\end_inset


\end_layout

\begin_layout Itemize
\begin_inset Formula 
\[
\hat{d}_{i\pi}=\left(1-W_{i}\right)\hat{\overline{y}}_{s_{i}}-\sum_{h\neq i}W_{h}\hat{\overline{y}}_{s_{h}}
\]

\end_inset


\end_layout

\begin_layout Standard
Probamos que es un estimador de la variable deseada (e insesgado).
 
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\mathbb{E}\left(\hat{d}_{i\pi}\right)=\mathbb{E}\left[\left(1-W_{i}\right)\hat{\overline{y}}_{s_{i}}-\sum_{h\neq i}W_{h}\hat{\overline{y}}_{s_{h}}\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\mathbb{E}\left(\hat{d}_{i\pi}\right)=\left(1-W_{i}\right)\mathbb{E}\left(\hat{\overline{y}}_{s_{i}}\right)-\sum_{h\neq i}W_{h}\mathbb{E}\left(\hat{\overline{y}}_{s_{h}}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
Por cómo están definidos los 
\begin_inset Formula $\hat{\overline{y}}_{s_{h}}$
\end_inset

, el resultado de Särndall aplica inmediatamente.
 
\end_layout

\begin_layout Itemize
\begin_inset Formula 
\[
\mathbb{E}\left(\hat{d}_{i\pi}\right)=\left(1-W_{i}\right)\overline{y}_{U_{i}}-\sum_{h\neq i}W_{h}\overline{y}_{U_{h}}
\]

\end_inset


\end_layout

\begin_layout Subsection*
ii Desarollar la varianza de 
\begin_inset Formula $\hat{d}_{i\pi}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V\left(\hat{d}_{i\pi}\right)=V\left[\left(1-W_{i}\right)\hat{\overline{y}}_{s_{i}}-\sum_{h\neq i}^{H}W_{h}\hat{\overline{y}}_{s_{h}}\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V\left(\hat{d}_{i\pi}\right)=\left(1-W_{i}\right)^{2}V\left(\hat{\overline{y}}_{s_{i}}\right)+\sum_{h\neq i}^{H}W_{h}^{2}V\left(\hat{\overline{y}}_{s_{h}}\right)
\]

\end_inset


\end_layout

\begin_layout Itemize
Este punto está entrocado con el resto del los ejercicios sobre muestreo
 estratificado y por lo tanto asumimos que se extiende a él la elección
 del MESA, siendo 
\begin_inset Quotes fld
\end_inset

cualquier adjudicación
\begin_inset Quotes frd
\end_inset

 referido a variaciones en la cantidad de estratos y métodos para determinar
 los 
\begin_inset Formula $n_{h}$
\end_inset

 siempre dentro de este esquema.
 
\end_layout

\begin_layout Standard
\begin_inset Formula $V_{MAS}\left(\hat{\overline{y}}_{s_{h}}\right)=\frac{1-f_{h}}{n_{h}}\cdot S_{yU_{h}}^{2}=$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V\left(\hat{d}_{i\pi}\right)=\left(1-W_{i}\right)^{2}\frac{1-f_{i}}{n_{i}}\cdot S_{yU_{i}}^{2}+\sum_{h\neq i}^{H}W_{h}^{2}\frac{1-f_{h}}{n_{h}}\cdot S_{yU_{h}}^{2}
\]

\end_inset


\end_layout

\begin_layout Subsection*
iii ¿Cuál es la adjudicación de Neyman para minimizar la varianza de 
\begin_inset Formula $\hat{d}_{i\pi}$
\end_inset

 para un tamaño de muestra fijo?
\end_layout

\begin_layout Standard
Para no tener que optimizar desde cero la adjudicación sobre los mismos
 supuestos, intentamos lograr amoldar el problema a las hipótesis del caso
 conocido con un cambio de variable.
 Necesitamos una variable cuya varianza vamos a evaluar estrato por estrato.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\hat{d}_{i\pi}=\left(1-W_{i}\right)\hat{\overline{y}}_{s_{i}}-\sum_{h\neq i}W_{h}\hat{\overline{y}}_{s_{h}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\hat{d}_{i\pi}=\left(1-W_{i}\right)\left(\frac{1}{N_{i}}\sum_{s\cap U_{i}}\check{y}_{k}\right)-\sum_{h\neq i}W_{h}\left(\frac{1}{N_{h}}\sum_{s\cap U_{h}}\check{y}_{k}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
Por ser un MESA: 
\begin_inset Formula $\forall y_{h}\in U_{h}:\pi_{k}=\frac{n_{h}}{N_{h}}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\hat{d}_{i\pi}=\left(1-W_{i}\right)\left(\frac{1}{N_{i}}\sum_{s\cap U_{i}}\frac{y_{k}}{\nicefrac{n_{i}}{N_{i}}}\right)-\sum_{h\neq i}W_{h}\left(\frac{1}{N_{h}}\sum_{s\cap U_{h}}\frac{y_{k}}{\nicefrac{n_{h}}{N_{h}}}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\hat{d}_{i\pi}=\left(1-W_{i}\right)\left(\frac{1}{n_{i}}\sum_{s\cap U_{i}}y_{k}\right)-\sum_{h\neq i}W_{h}\left(\frac{1}{n_{h}}\sum_{s\cap U_{h}}y_{k}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\hat{d}_{i\pi}=W_{i}\sum_{s\cap U_{i}}\frac{\left(\nicefrac{1}{W_{i}}-1\right)}{n_{i}}y_{k}+\sum_{h\neq i}\left(\sum_{s\cap U_{h}}\left(-\frac{W_{h}}{n_{h}}\right)y_{k}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\hat{d}_{i\pi}=\sum_{s\cap U_{i}}\frac{W_{i}}{n_{i}}\left[\left(\nicefrac{1}{W_{i}}-1\right)y_{k}\right]+\sum_{h\neq i}\left(\sum_{s\cap U_{h}}\left(-\frac{W_{h}}{n_{h}}\right)y_{k}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $z_{k}=\begin{cases}
\left(\nicefrac{1}{W_{i}}-1\right)y_{k} & k\in h_{i}\\
-y_{k} & k\notin h_{i}
\end{cases}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\hat{d}_{i\pi}=\sum_{s\cap U_{i}}z_{k}
\]

\end_inset


\end_layout

\begin_layout Standard
Ahora la ecuación tiene una forma que nos deja aplicar la fórmula ya conocida
 para la asignación óptima de Neyman:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
n_{h}=n\cdot\frac{N_{h}S_{zU_{h}}}{\sum_{k}N_{k}S_{zU_{k}}}
\]

\end_inset


\end_layout

\begin_layout Standard
Buscamos la varianza de esta nueva variable:
\end_layout

\begin_layout Standard
\begin_inset Formula $V\left(z_{kh}\right)=\begin{cases}
V\left[\left(\nicefrac{1}{W_{i}}-1\right)y_{k}\right] & k\in h_{i}\\
V\left(-y_{k}\right) & k\notin h_{i}
\end{cases}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $S_{zU_{h}}=\begin{cases}
\left(\nicefrac{1}{W_{i}}-1\right)^{2}S_{yU_{i}} & h=i\\
S_{yU_{h}} & h\neq i
\end{cases}$
\end_inset


\end_layout

\begin_layout Standard
Especializamos con la varianza obtenida
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
n_{h}=\begin{cases}
n\cdot\frac{N_{i}\cdot\left(\nicefrac{1}{W_{i}}-1\right)^{2}S_{yU_{i}}}{\sum_{k\neq i}N_{k}S_{yU_{k}}+N_{i}\cdot\left(\nicefrac{1}{W_{i}}-1\right)^{2}S_{yU_{i}}} & h=i\\
n\cdot\frac{N_{h}S_{yU_{h}}}{\sum_{k\neq i}N_{k}S_{yU_{k}}+N_{i}\cdot\left(\nicefrac{1}{W_{i}}-1\right)^{2}S_{yU_{i}}} & h\neq i
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Subsection*
iv ¿Cuándo existirán entre esta adjudicación y la de Neyman considerando
 la variable de estratificación?
\end_layout

\begin_layout Standard
Recordamos la adjudicación de Neyman:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
n_{h_{Ney}}=n\cdot\frac{N_{h}\cdot S_{yU_{h}}}{\sum_{k=1}^{H}N_{k}S_{yU_{k}}}
\]

\end_inset


\end_layout

\begin_layout Itemize
Efecto sobre 
\begin_inset Formula $n_{i}$
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
n_{i}=n_{i_{Ney}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
n\cdot\frac{N_{i}\cdot\left(\nicefrac{1}{W_{i}}-1\right)^{2}S_{yU_{i}}}{\sum_{k\neq i}N_{k}S_{yU_{k}}+N_{i}\cdot\left(\nicefrac{1}{W_{i}}-1\right)^{2}S_{yU_{i}}}=n\cdot\frac{N_{i}\cdot S_{yU_{i}}}{\sum_{k=1}^{H}N_{k}S_{yU_{k}}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{1\cdot\left(\nicefrac{1}{W_{i}}-1\right)^{2}}{\sum_{k\neq i}N_{k}S_{yU_{k}}+N_{i}\cdot\left(\nicefrac{1}{W_{i}}-1\right)^{2}S_{yU_{i}}}=\frac{1}{\sum_{k=1}^{H}N_{k}S_{yU_{k}}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{1}{\frac{\sum_{k\neq i}N_{k}S_{yU_{k}}}{\left(\nicefrac{1}{W_{i}}-1\right)^{2}}+N_{i}\cdot S_{yU_{i}}}=\frac{1}{\sum_{k\neq i}N_{k}S_{yU_{k}}+N_{i}\cdot S_{yU_{i}}}
\]

\end_inset


\end_layout

\begin_layout Standard
Como la función 
\begin_inset Formula $\frac{1}{x}$
\end_inset

 es biyectiva cuando restringimos los valores de 
\begin_inset Formula $x$
\end_inset

 a 
\begin_inset Formula $\mathbb{R}^{+}$
\end_inset

 y este es el caso de nuestro denominador, podemos asegurar que esta igualdad
 sólo se da cuando los denominadores son iguales:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{1}{\frac{\sum_{k\neq i}N_{k}S_{yU_{k}}}{\left(\nicefrac{1}{W_{i}}-1\right)^{2}}+N_{i}\cdot S_{yU_{i}}}=\frac{1}{\sum_{k\neq i}N_{k}S_{yU_{k}}+N_{i}\cdot S_{yU_{i}}}\text{\ensuremath{\Rightarrow}}\frac{\sum_{k\neq i}N_{k}S_{yU_{k}}}{\left(\nicefrac{1}{W_{i}}-1\right)^{2}}+N_{i}\cdot S_{yU_{i}}=\sum_{k\neq i}N_{k}S_{yU_{k}}+N_{i}\cdot S_{yU_{i}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\sum_{k\neq i}N_{k}S_{yU_{k}}}{\left(\nicefrac{1}{W_{i}}-1\right)^{2}}+N_{i}\cdot S_{yU_{i}}=\sum_{k\neq i}N_{k}S_{yU_{k}}+N_{i}\cdot S_{yU_{i}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\sum_{k\neq i}N_{k}S_{yU_{k}}}{\left(\nicefrac{1}{W_{i}}-1\right)^{2}}=\sum_{k\neq i}N_{k}S_{yU_{k}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\sum_{k\neq i}N_{k}S_{yU_{k}}}{\sum_{k\neq i}N_{k}S_{yU_{k}}}=\left(\nicefrac{1}{W_{i}}-1\right)^{2}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\left(\nicefrac{1}{W_{i}}-1\right)^{2}=1
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $W_{i}\leq1\Rightarrow\left(\nicefrac{1}{W_{i}}-1\right)>0$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{1}{W_{i}}-1=1
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\nicefrac{1}{W_{i}}=2
\]

\end_inset


\end_layout

\begin_layout Itemize
\begin_inset Formula 
\[
W_{i}=\frac{1}{2}
\]

\end_inset


\end_layout

\begin_layout Standard
El caso de igualdad es cuando el estrato a considerar representa la mitad
 de la muestra.
 
\end_layout

\begin_layout Standard
Observando la igualdad 
\begin_inset Formula $\frac{1}{\frac{\sum_{k\neq i}N_{k}S_{yU_{k}}}{\left(\nicefrac{1}{W_{i}}-1\right)^{2}}+N_{i}\cdot S_{yU_{i}}}=\frac{1}{\sum_{k\neq i}N_{k}S_{yU_{k}}+N_{i}\cdot S_{yU_{i}}}$
\end_inset

 vemos que estaremos cerca de ella cuando el factor 
\begin_inset Formula $\left(\nicefrac{1}{W_{i}}-1\right)^{2}$
\end_inset

 cumple 
\begin_inset Formula $\left(\nicefrac{1}{W_{i}}-1\right)^{2}\approx1$
\end_inset

.
\end_layout

\begin_layout Standard
Si el estrato es pequeño, su peso 
\begin_inset Formula $W_{i}$
\end_inset

 lo es también y la expresión 
\begin_inset Formula $\nicefrac{1}{W_{i}}$
\end_inset

 tiende a crecer así como lo hace su cuadrado.
 El numerador entonces disminuye y obtenemos un 
\begin_inset Formula $n_{i}$
\end_inset

 mayor.
 
\end_layout

\begin_layout Itemize
Para los estratos 
\begin_inset Formula $h\neq i$
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
n_{h}=n_{h_{Ney}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
n\cdot\frac{N_{h}\cdot S_{yU_{h}}}{\sum_{k\neq i}N_{k}S_{yU_{k}}+N_{i}\cdot\left(\nicefrac{1}{W_{i}}-1\right)^{2}S_{yU_{i}}}=n\cdot\frac{N_{h}\cdot S_{yU_{h}}}{\sum_{k=1}^{H}N_{k}S_{yU_{k}}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{1}{\sum_{k\neq i}N_{k}S_{yU_{k}}+N_{i}\cdot\left(\nicefrac{1}{W_{i}}-1\right)^{2}S_{yU_{i}}}=\frac{1}{\sum_{k=1}^{H}N_{k}S_{yU_{k}}}
\]

\end_inset


\end_layout

\begin_layout Standard
Otra vez, tenemos valores similares cuando 
\begin_inset Formula $\left(\nicefrac{1}{W_{i}}-1\right)^{2}\approx1$
\end_inset

.
\end_layout

\begin_layout Standard
En este caso, en la medida en que el estrato 
\begin_inset Formula $i$
\end_inset

 tenga un peso 
\begin_inset Formula $W_{i}$
\end_inset

 pequeño, el factor 
\begin_inset Formula $\left(\nicefrac{1}{W_{i}}-1\right)^{2}$
\end_inset

 se va a ver incrementado y por lo tanto vamos a asignarle menos muestras
 a los demás estratos.
\end_layout

\begin_layout Itemize
Como era de esperar, conocer más información sobre una de las variables
 siempre nos obliga a asignarle más muestras, en este caso hallamos la ley
 de acuerdo a la cual eso sucede.
 
\end_layout

\begin_layout Itemize
Los valores que adopta la variable en el universo de estudio no son drásticament
e importantes, el efecto se da de acuerdo al peso 
\begin_inset Formula $W_{i}$
\end_inset

 del estrato cuya diferencia respecto a la media nos importa.
\end_layout

\begin_layout Itemize
En nuestro marco el marco construido con nuestra variable auxiliar, las
 empresas clasificadas como pequeñas representan un estrato muy poco significati
vo y por lo tanto la diferencia en la asignación va a ser apreciable.
\end_layout

\end_body
\end_document