Proove that Var Y=$n\alpha\beta /(\alpha+\beta)^2$

Let $(X_i, P_i)$, with $_i$ being an integer be an independent random variable s.t. $X_i|P_i$ ~ $Bernoulli(P_i)$ and $P_i$ ~ $Beta(\alpha, \beta$.

Here's what I have so far: Var Y=Var(E(Y|P)+E(Var(Y|P) is what we know as our formula;

If we break down the two components on the right side we get: Var(E(Y|P))=$\alpha\beta/(\alpha+\beta)^2 (\alpha+\beta+1)$ and E(Var(Y|P)=$E(P_i(1-P_i))=\alpha/(\alpha+\beta) (1-\alpha/(\alpha+\beta))$.

Somehow I'm not getting the right result.

Is this correct and if not, could you please correct my understanding?

• How is $Y$ related to the $X_i$'s? – Gregory Oct 15 '15 at 3:57
• Y is the summation of $X_i$ from 1 to n – cambelot Oct 15 '15 at 3:57
• The first paragraph also is cut off... I'm not sure what you mean with $P_iBeta$? – Gregory Oct 15 '15 at 3:58

First, we recall the conditional mean and variance of $X_i \mid P_i$: $$\operatorname{E}[X_i \mid P_i] = P_i,$$ and $$\operatorname{Var}[X_i \mid P_i] = P_i(1-P_i).$$ Consequently, the unconditional variance of $X_i$ is, by the law of total variance, \begin{align*} \operatorname{Var}[X_i] &= \operatorname{Var}[\operatorname{E}[X_i \mid P_i]] + \operatorname{E}[\operatorname{Var}[X_i \mid P_i]] \\ &= \operatorname{Var}[P_i] + \operatorname{E}[P_i(1-P_i)] \\ &= \operatorname{E}[P_i^2] - \operatorname{E}[P_i]^2 + \operatorname{E}[P_i] - \operatorname{E}[P_i^2] \\ &= \operatorname{E}[P_i](1 - \operatorname{E}[P_i]) \\ &= \frac{\alpha\beta}{(\alpha+\beta)^2},\end{align*} since $$\operatorname{E}[P_i] = \frac{\alpha}{\alpha+\beta}.$$ Hence
$$\operatorname{Var}[Y] = \operatorname{Var}\left[\sum_{i=1}^n X_i \right] \overset{\text{ind}}{=} \sum_{i=1}^n \operatorname{Var}[X_i] = \frac{n \alpha \beta}{(\alpha+\beta)^2}.$$
Hint: The formula is correct, but it is the result for each $X_i$ not for $Y$. The $var(Y) = var \sum_i X_i = \sum_i var(X_i)$ (since the \$X_i are i.i.d.).