Estimador

Suficiencia

Si $\hat{\theta }$ del parámetro $\theta$ emplea efectivamente la información de una m.a. de tamaño $n$ decimos que es un estimador suficiente, tal que cumple.

$$\begin{array}{r}\text{Condicion de suficiencia}\\ f(x_1,x_2,\dots ,x_n;\theta )=g(\hat{\theta };\theta )h(x_1,x_2,\dots ,x_n)\\ \text{Donde}\{\begin{array}{l}f(x_1,x_2,\dots x_n;\theta )={\prod }_{i=1}^{n}f(x_i;\theta )\text{por ser una m.a.}\\ g(\hat{\theta };\theta ):\text{Funcion que depende de }\hat{\theta },\theta \\ h(x_1,x_2,\dots ,x_n):\text{Funcion que solo depende de la m.a.}\end{array}\end{array}$$

Ejemplos

Sea $\bar{X}$ La media muestral verificar que es un estimador suficiente.

Sea $X$ una v.a. poblacional con distribución normal con media ${\mu }_x$ (Desconocida) y ${\sigma }_X^2$ (Conocida). Sol. $$ \begin{align} x \to N(\mu_{x},\sigma_{o}^{2}) \quad E(\bar{X})=\mu_{x} \quad \text{Insesgado} \ f(x;\theta=\mu_{x})=\frac{1}{\sqrt{ 2 \pi }\sigma_{o}}e^{- \frac{1}{2}\left( \frac{x-\mu_{x}}{\sigma_{o}} \right)^{2}}\quad-\infty<x<\infty \ \text{m.a: }X_{1},X_{2},\dots,X_{n} \ f(x_{1},x_{2},\dots,x_{n};\mu_{x})=\prod_{i=1}^{n}{f(x_{i};\mu_{x})}=\prod_{i=1}^{n}{\frac{1}{\sqrt{ 2 \pi }\sigma_{o}}e^{- \frac{1}{2}\left( \frac{x_{i}-\mu_{x}}{\sigma_{o}} \right)^{2}}} \ =\frac{1}{(\sqrt{ 2 \pi }\sigma_{o})^{2}}e^{- \frac{1}{2} \sum_{i=1}^{n}{ \left( \frac{x_{i}-\mu_{x}}{\sigma_{o}} \right)^{2}}} \ \text{primero} \sum_{i=i}^{n}{(X_{i}-\mu_{x})^{2}}=\sum_{i=1}^{n}{[(x_{i}-\bar{x})-(\mu_{x}-\bar{x})]^{2}}=\sum_{i=1}^{n}{[(x_{i}-\bar{x})^{2}-2(\mu_{x}-\bar{x})(x_{i}-\bar{x})+(\mu_{x}-\bar{x})^{2}]} \ =\sum_{i=1}^{n} (x_{i}-\bar{x})^{2}-2(\mu_{x}-\bar{x})\sum_{i=1}^{n}(x_{i}-\bar{x})+n(\mu_{x}-\bar{x})^{2}=\sum_{i=1}^{n}(X_{i}-\bar{x})^{2}+n(\mu_{x}-\bar{x})^{2} \ \text{aplicando en la original} \ =ke^{-1/2 \left[ n(\mu_{x}-\bar{x})^{2}+\sum_{i=1}^{n}(X_{i}-\bar{x})^{2} \right]}=ke^{-\frac{n}{2\sigma_{o}^{2}}(\mu_{x}-\bar{x})^{2}}e^{-\frac{1}{2\sigma_{o}^{2}}\sum_{i=1}^{n}(x_{i}-\bar{x})^{2} } \ \text{Que es de la forma } ke^{g(\bar{x};\mu_{x})}e^{h(x_{1},x_{2},\dots,x_{n})} \end{align}