Let $Y$ be an exponential random variable with mean $\frac{1}{\theta}$, where $\theta\gt0$. The conditional distribution of $X$ given $Y$ has Poisson distribution with mean $Y$. Then, the variance of $X$ is
(A)$\frac{1}{\theta^2}\quad\quad$(B)$\frac{\theta+1}{\theta}\quad\quad$(C)$\frac{\theta^2+1}{\theta^2}\quad\quad$(D)$\frac{\theta+1}{\theta^2}$
The thing is since $Y$ a continuous random variable , is it possible that the conditional distribution of $X$ given $Y$ follow Poisson distribution (which is discrete in nature) and moreover we don't know whether $X$ is a discrete or a continuous random variable.
How do we find the variance of $X$?
Best Answer
You can use the law of total expectation
$$E[X]=E[E[X|Y]]=E[Y]=\frac{1}{\theta}$$
$$V[X]= E[E[X^2|Y]] -E^2[X]=E[Y^2+Y]-\frac{1}{\theta^2}$$
$$V[X]=\frac{2}{\theta^2}+\frac{1}{\theta}-\frac{1}{\theta^2}=\frac{1+\theta}{\theta^2}$$
Thus the correct answer is D)