Valor esperado y varianza de relación de dos cantidades de los dos conjuntos de variables aleatorias

Question

Que $X_1,X_2,\ldots,X_n$ ser iid $\operatorname{Gamma}(\alpha,\beta)$ variables aleatorias. Supongamos que, condicionalmente $X_1,X_2,\ldots,X_n$, el $Y_1,Y_2,\ldots,Y_n$ de variables aleatorias son independientes y $Y_{i}\mid X_{i}\sim \operatorname{Gamma}(\alpha,\beta X_i)$. Mostrar que $$E\left(\frac{\bar{Y}}{\bar{X}}\right)=\alpha\beta$$ and $% $ $\operatorname{Var}\left(\frac{\bar{Y}}{\bar{X}}\right)=\alpha\beta^2E\left(\frac{\sum X_i^2}{(\sum X_i)^2}\right).$

Se trata de un examen completo pasado. La pista dice "usar las fórmulas de esperanza y varianza iteradas". Pero no veo en ningún lugar puede utilizar esta sugerencia.

Answer 1

Si $X_i \backsim \operatorname{Gamma}(\alpha,\beta)$ donde $\alpha$ es la forma y el $\beta$ es el parámetro de escala $$ \mathbb{E}\left[ X_i \right] = \alpha \beta \quad \quad \mbox{and} \quad \quad \mathbb{V}\mbox{ar}\left[ X_i \right] = \alpha \beta^2 $ $

De las propiedades de la gamma distribución $$ \overline{X} \backsim \operatorname{Gamma}\left(n \alpha, \beta/n \right) $ $ que significa $$ \mathbb{E}\left[ \bar{X}\right] = \alpha\beta \quad \quad \mbox{and} \quad \quad \mathbb{V}\mbox{ar}\left[\bar{X}\right] = \alpha \beta^2/n $ $ entonces para $$ Y_i | X_i \backsim \operatorname{Gamma}\left(\alpha, \beta X_i \right) $ $ $$ \mathbb{E}\left[ Y_i | X_i \right] =\alpha \beta X_i \quad \quad \mbox{and} \quad \quad \mathbb{V}\mbox{ar}\left[Y_i | X_i \right] = \alpha (\beta X_i )^2 $ $

De la ley de la expectativa total tenemos\begin{equation} \begin{split} \mathbb{E}\left[\frac{\bar{Y}}{\bar{X}}\right]&= \left. \mathbb{E}\left[ \mathbb{E}\left[ \frac{\bar{Y}}{\bar{X}} \right| X_1, \ldots, X_n \right] \right] \\ &= \mathbb{E}\left[ \frac{1}{\bar{X}} \frac{1}{n} \sum_{i=1}^n\mathbb{E} [ Y_i \big| X_1, \ldots, X_n ] \right] \\ &= \mathbb{E}\left[ \frac{1}{\bar{X}} \frac{1}{n} \sum_{i=1}^n\mathbb{E}[ Y_i \big| X_i ] \right] \\ & = \mathbb{E}\left[ \frac{1}{\bar{X}} \frac{1}{n} \sum_{i=1}^n \alpha \beta X_i \right] \\ & = \alpha \beta \mathbb{E}\left[ \frac{1}{\bar{X}} \frac{1}{n} \sum_{i=1}^n X_i \right] \\ & = \alpha \beta \mathbb{E}\left[ \frac{1}{\bar{X}} \bar{X} \right] \\ & = \alpha \beta \mathbb{E}\left[ 1 \right] \\ & = \alpha \beta \\ \end{dividido} \end{equation}

De la ley de la varianza total que tenemos\begin{equation*} \begin{split} \mathbb{V}\mbox{ar}\left[\frac{\bar{Y}}{\bar{X}}\right] &= \left. \mathbb{V}\mbox{ar}\left[ \mathbb{E}\left[ \frac{\bar{Y}}{\bar{X}} \right| X_1, \ldots, X_n \right] \right] + \left. \mathbb{E}\left[ \mathbb{V}\mbox{ar} \left[ \frac{\bar{Y}}{\bar{X}} \right| X_1, \ldots, X_n \right] \right] \\ &= \left. \mathbb{V}\mbox{ar}\left[ \frac{1}{\bar{X} } \frac{1}{n} \mathbb{E}\left[ \sum_{i=1}^nY_i\right| X_1, \ldots, X_n \right] \right] + \left. \mathbb{E}\left[ \frac{1}{\bar{X}^2 } \frac{1}{n^2} \mathbb{V}\mbox{ar} \left[ \sum_{i=1}^nY_i \right| X_1, \ldots, X_n \right] \right] \\ &= \mathbb{V}\mbox{ar}\left[ \frac{1}{\bar{X} } \frac{1}{n} \sum_{i=1}^n \mathbb{E}\left[ Y_i\big| X_i\right] \right] + \mathbb{E}\left[ \frac{1}{\bar{X}^2 } \frac{1}{n^2} \sum_{i=1}^n \mathbb{V}\mbox{ar} [ Y_i \big| X_i] \right] \\ &= \mathbb{V}\mbox{ar}\left[ \frac{1}{\bar{X} } \frac{1}{n} \sum_{i=1}^n \alpha \beta X_i \right] +\mathbb{E}\left[ \frac{1}{\bar{X}^2 } \frac{1}{n^2} \sum_{i=1}^n \alpha (\beta X_i )^2 \right] \\ &= \alpha^2 \beta^2 \mathbb{V}\mbox{ar}\left[ \frac{1}{\bar{X} } \bar{X} \right] +\mathbb{E}\left[ \frac{n^2}{ (\sum_{i=1}^n X_i)^2 } \frac{\alpha \beta^2}{n^2} \sum_{i=1}^n X_i^2 \right] \\ &= \alpha^2 \beta^2 \mathbb{V}\mbox{ar}\left[ 1 \right] + \alpha \beta^2 \mathbb{E}\left[ \frac{1}{ (\sum_{i=1}^n X_i)^2 } \sum_{i=1}^n X_i^2 \right] \\ &= \alpha \beta^2 \mathbb{E}\left[ \frac{ \sum_{i=1}^n X_i^2 }{ (\sum_{i=1}^n X_i)^2 } \right] \\ \end{dividido} \end{equation*}