I was wondering if anyone could help me with the steps for this question:
Suppose that the continuous random variable T has the Exponential distribution with expected value 3. Given that T=t, for any t > 0, the discrete random variable S has the Poisson distribution with expected value 2t.
Find E(S) and Var(s).
Any help at all would be much appreciated!
We know that the distribution of $S$ conditioned on the value of $T$ is
$$\mathbb{P}(S=s|T=t)=\frac{e^{-2t}(2t)^s}{s!} \;\; , \;\;\; s \in \mathbb{N}$$
and $T$ is exponentially distributed with parameter 3, i.e. the density of $T$ is
$$f_T(t)=3 e^{-3t} \;\; , \;\;\; t>0 .$$
The expectation of $S$ can be computed as follows
$$\begin{eqnarray}\mathbb{E}(S) & = & \sum_{s=0}^{\infty}s\mathbb{P}(S=s) \\ & = & \sum_{s=0}^{\infty}s \int_0^{\infty}\mathbb{P}(S=s|T=t)f_T(t)dt\end{eqnarray}$$
In the last step, we used the law of total probability. Substituting back our distribution formulas, this becomes
$$\begin{eqnarray}\mathbb{E}(S) & = & \sum_{s=0}^{\infty}s \int_0^{\infty}\frac{e^{-2t}(2t)^s}{s!}3 e^{-3t}dt \\ & = & \sum_{s=0}^{\infty}\frac{3 \cdot 2^s}{(s-1)!} \int_0^{\infty}e^{-5t}t^s dt \\ & = & \sum_{s=0}^{\infty}\frac{3 \cdot 2^s}{(s-1)!} \frac{s!}{5^{s+1}} \\ & = & \frac{3}{5}\sum_{s=0}^{\infty} s \left(\frac{2}{5}\right)^s \\ & = & \frac{3}{5}\frac{\frac{2}{5}}{\left(1-\frac{2}{5}\right)^2}=\frac{2}{3} \end{eqnarray}$$
The computation for the variance can be done in a similar way.