Suppose that the conditional distribution of $Y$ given $X=x$ is Poisson with mean $E(Y|x)=x$, $Y|x\sim POI(x)$, and that $X\sim EXP(1)$
Find $V[Y]$
We know that
$$V[Y]=E_x[Var(Y|X)] + Var_x[E(Y|X)]$$
we also know that
$$V(Y|x)=E(Y^2|x)-[E(Y|x)]^2$$
I know we can find $V[Y]$ with those 2 formulas, but I am getting confused with so many expectances and variances around each other. Can anyone explain me how to solve this?
On
We have been given that $Y|x$ is a Poisson Variate with parameter $x$. We know that, in Poisson distribution the mean and variance are equal to its parameter. Hence, the following will hold:
$E[Y|x]=V[Y|x]=x$.
Now, to obtain the variance of $Y$,
$V(Y)=E(V[Y|x])+V(E[Y|x]) = E(x)+V(x)$
Now, $X$ itself is a exponential Variate with parameter $1$. So, it's Expectation and variance both will be $1$. Hence, the required value of Variance of $Y$ after plugging in the appropriate Values will be:
$V(Y)=E(x)+V(x)=1+1=2$
Note: The mean and variance of $X~Exp(\beta)$ is:
$E(X)=\frac{1}{\beta}$, $V(X)=\frac{1}{\beta^2}$
If the conditional distribution of $Y$ given $X$ is Poisson with parameter $X$ then the conditional expectation is also $X$. That is, if $X$ is of exponential distribution with parameter one then
$$E[Y]=\int_0^{\infty}E[Y\mid X=x\,]e^{-x}\ dx=\int_0^{\infty}xe^{-x}\ dx=E[X]=1.$$
Also,
$$E[Y^2]=\int_0^{\infty}E[Y^2\mid X=x\,]e^{-x}\ dx=\int_0^{\infty}(x^2+x)e^{-x}\ dx$$ because the second moment of the Poisson distribution is the square of the parameter plus the parameter.
Now, $$\int_0^{\infty}(x^2+x)e^{-x}\ dx=\int_0^{\infty}x^2e^{-x}\ dx+\int_0^{\infty}xe^{-x}\ dx=E[X^2]+E[X]=2+1=3$$ because the second moment of the exponential distribution of parameter $1$ is $2$.
The variance of $Y$ is then $$E[Y^2]-E^2[Y]=3-1=2.$$
The second moments of the distributions above can be calculated based calculated based on the corresponding MGFs. The MGFs can be found in Wikipedia. If you have the MGF then take the second derivative and substitute $0$.