GO Classes GATE DA | Probability | Weekly Quiz 5 | Question: 14

Answer 1

Since the conditional distribution of $X$ given $Y$ has Poisson distribution with mean $Y$, it follows that $E(X \mid Y)=\operatorname{Var}(X \mid Y)=Y$. Now, the variance of $X$ is given by

$$
\begin{aligned}
\operatorname{Var}(X) & =E(\operatorname{Var}(X \mid Y))+\operatorname{Var}(E(X \mid Y)) \\
& =E(Y)+\operatorname{Var}(Y) \\
& =\frac{1}{\theta}+\frac{1}{\theta^{2}} \\
& =\frac{\theta+1}{\theta^{2}}
\end{aligned}
$$

Hence option (D) is the correct choice.

Comments

How did you get this equation??  . 

Please give some explanation

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Variance of X comes from:

Variation within each conditional distribution

Variation between conditional distributions

Add them → total variance.

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This property always hold for every conditional distribution like $X|Y \sim Exponential(Y), X \sim Poison(Y)$, etc.

 We can't directly use $Var(X) = E[X] = E[E[X|Y]] = E[Y]$ to find the $Var(X)$ as here $X|Y \sim Poison(Y)$ not $X|Y \sim Poison(\lambda)$