To calculate variance, given conditional distribution

Question

Let Y be an exponential random variable with mean $\frac{1}{\theta}$, where $\theta>0$. The conditional distribution of X given Y has Poisson distribution with mean Y. Then, the variance of X is

(A)$\dfrac{1}{\theta^2}$ (B) $\dfrac{\theta+1}{\theta}$ (C) $\dfrac{\theta^2+1}{\theta^2}$ (D) $\dfrac{\theta+1}{\theta^2}$

My Attempt:

$\!f(x; \lambda|Y)= \Pr(X{=}x)= \frac{\lambda^x e^{-\lambda}}{x!}$

$E(X|Y)=\displaystyle\sum xf(x; \lambda|Y)=\lambda=\frac{1}{\theta}$

Variance(X)=$E[X^2]-(E[X])^2$

Now, $E(X^2)=\displaystyle\sum x^2\frac{\lambda^x e^{-\lambda}}{x!}=(1+\lambda)=1+\frac{1}{\theta}$

Because $(1+x)e^x= \sum^\infty_{n=0}{n+1\over n!}x^n$

Hence, Var(X)=$1+\frac{1}{\theta}-\frac{1}{\theta^2}$

Where did I go wrong ? Please advise.

Answer 1

Hints:

We are given that $Y\sim exp(\theta)$ and $X\vert y\sim Po(y)$. Consequently, from the properties of a Poisson distributed random variable with parameter $y$,

$$ \mathbb E[X\vert Y=y] = y\ \ \ \text{ and }\ \ \ \mathbb E[X^2\vert Y=y] = y(y+1)\,. $$

This implies, using the properties of an exponentially distributed random variable with parameter $\theta$, $$ \mathbb E[X] = \mathbb E[\mathbb E[X\vert Y]] = \mathbb E[Y] = \ldots\ \ \ \text{ and }\ \ \ \mathbb E[X^2] = \mathbb E[\mathbb E[X^2\vert Y]] = \mathbb E[Y(Y+1)] = \ldots $$

Since, $$ \mathbb V(X) = \mathbb E[X^2] - (\mathbb E[X])^2 = \mathbb E[ \mathbb E[X^2\vert Y]] - (\mathbb E[ \mathbb E[X\vert Y]])^2, $$

we must have

$$\mathbb V(X) = \frac{2}{\theta^2}+\frac{1}{\theta}-\frac{1}{\theta^2} = \frac{1+\theta}{\theta^2}$$