Find $\text{E}(X2^X)$ for $X\sim\text{Po}(3)$

Question

Let $X\sim\text{Po}(3)$. Find $\text{E}(Y), \; Y:=X2^X$. HINT: First consider the random variable $Z:=Xs^{X-1}.$

My attempt:

$X$ is poisson distributed with $\lambda=3$. This means that $\text{P}(X=k)=e^{-3}\frac{3^k}{k!}$.

Hence the generating function of $X$ is $$\phi_X(s):=\sum_{k=0}^\infty \text{P}(X=k)s^k = \sum_{k=0}^\infty e^{-3}\frac{3^k}{k!} s^k$$

(note that $\phi_X(s) = \text{E}(s^X)$)

Hence $\phi_X'(s)=\text{E}(Xs^{X-1})=\text{E}(Z)$

Let $T:=X-1$ then $X=T+1$ and $$\text{E}(Z)=\text{E}((T+1)s^T)=\text{E}(Ts^T)+\text{E}(s^T)$$ $$= \text{E}(Ts^T)+\phi_T(s)$$

But I do not know how to go on from here. I think I have a more complicated expression that the one I started with, how should I proceed?

Answer 1

I think I got it now, thanks to @madprob

$$\phi'(2)=E(X2^{X-1})=E(X2^X2^{-1})=1/2E(Y)$$ thus $$E(Y)=2\phi'(s)$$

$$2\phi'(s)=2\sum_{k=0}^\infty e^{-3}\frac{3^k}{k!}k2^{k-1} = ... =6e^{3}$$

Answer 2

You are almost done. What is the relation between $\phi'_{X}(2)$ and $Y$?

Also, \begin{align*} \phi_{X}(s) &= \ldots = \exp(3s-3)\sum_{i=0}^{\infty} \frac{\exp(-3s)(3s)^k}{k!} = \exp(3s-3) \end{align*}