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Demostar que el valor esperado de la varianza muestral es igual al valor esperado poblacional

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Es decir E\left [ s^{2} \right ]=\sigma ^{2}.

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Demostracíón: 

Recordemos que la varianza muestra es

 

S^{2}=\frac{1}{n-1}\sum_{1}^{n} \left ( x_{i}-\bar{x_{n}} \right )^{2 }.

También que var(X)=E\left [ \left ( X-E\left [ X \right ] \right ) ^{2 }\right ] . 

 

Primero se mostrara la siguiente igualdad 

\sum_{1}^{n} \left ( X_{i}-E\left [ X \right ] \right ) ^{2 }=\sum_{1}^{n} \left ( X_{i}-\bar{X_{n}}\right )^{2} +n \left (\bar{X_{n}}\right - E\left [ X \right ] )^{2} \, \, \bigstar

puesto que,

\sum_{1}^{n} \left ( X_{i}-E\left [ X \right ] \right ) ^{2 }\\ = \sum_{1}^{n} \left ( X_{i}-\bar{X}+\bar{X}-E\left [ X \right ] \right ) ^{2 } \\ =\sum_{1}^{n} \left ( (X_{i}-\bar{X})+(\bar{X}-E\left [ X \right ] \right) ) ^{2 } \\ =\sum_{1}^{n}(X_{i}-\bar{X})^{2} +2(X_{i}-\bar{X})(\bar{X}-E\left [ X \right ] )+ (\bar{X}-E\left [ X \right ] ) ^{2}\\ =\sum_{1}^{n}(X_{i}-\bar{X})^{2} +2(\bar{X}-E\left [ X \right ] )\sum_{1}^{n}(X_{i}-\bar{X})+n(\bar{X}-E\left [ X \right ] ) ^{2}\\ =\sum_{1}^{n} \left ( X_{i}-\bar{X_{n}}\right )^{2} +n \left (\bar{X_{n}}\right - E\left [ X \right ] )^{2}

 

Observemos que 

2(\bar{X}-E\left [ X \right ] )\sum_{1}^{n}(X_{i}-\bar{X})=2(\bar{X}-E\left [ X \right ] )(0)=0

ya que 

 

\sum_{1}^{n}(X_{i}-\bar{X})=\sum_{1}^{n}(X_{i})-\sum_{1}^{n}(\bar{X})\\ = \sum_{1}^{n}(X_{i})-n(\bar{X})\\ = \sum_{1}^{n}(X_{i})- \sum_{1}^{n}(X_{i}) =0.

 

Usando la ecuación \bigstar, se obtiene 

 E\left [ S^{2} \right ]=E [(1/n-1)\sum_{1}^{n} ( X_{i}-\bar{X} ) ^{2 } ] \\ =\frac{1}{n-1}\, E [ \sum_{1}^{n} ( X_{i}-E[X]) ^{2 }-n( \bar{X}-E[X] ) ^{2 } ] \\ =\frac{1}{n-1} \sum_{1}^{n} E [ ( X_{i}-E[X]) ^{2 }-n( \bar{X}-E[X] ) ^{2 } ]\\ = \frac{1}{n-1} \left \{ \sum_{1}^{n} var(X)-n \, var(\bar{X}) \right \}\\ =\frac{1}{n-1} \left \{ n \, var(X)-n \, var(\bar{X}) \right \}\\ =\frac{1}{n-1} \left \{ n \, var(X)-n \, \frac{var(X)}{n} \right \} =var(X).

 

 

respondido por anónimo Oct 23
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