Mean of a function of a Stochastic Process Let $X(t)$ a Poisson Process with parameter $\lambda$, Consider the compound poisson process:
$$A(t)=\sum_{i=1}^{X(t)}Y_i$$
If $\{Y_i\}_{i\ge 1}$ are a i.i.d. random variables $exp(\theta$.Define:
$$T=\inf\{t\ge 0: A(t)>a\}$$
for $a>0$.
Find $E[T]$.
${\bf idea}$
I Think that a way to compute this mean is find the moment generating function of T, but I couldn't find it. There is a simplest way of compute it?
 A: $
\begin{align*}
P(T\leq t)&=P(A(t)>a) \\
&=\sum_{n=1}^{\infty}P(A(t)>a\mid X(t)=n)P(X(t)=n) \\
&=\sum_{n=1}^{\infty}P\big(\sum_{i=1}^n Y_i>a\big)\frac{(\lambda t)^n e^{-\lambda t}}{n!} \\
&=\sum_{n=1}^{\infty}\frac{\Gamma(n,\theta a)}{(n-1)!}\frac{(\lambda t)^ne^{-\lambda t}}{n!}.
\end{align*}
$
The last line is due to the sum of $n$ i.i.d. $\text{Exp}(\theta)$ random variables having the distribution of a $\text{Gamma}(n,\theta)$ random variable.
Differentiate with respect to $t$ to get the pdf as 
$
\begin{align*}
p(t)&=\sum_{n=1}^{\infty}\frac{\Gamma(n,\theta a)}{(n-1)!}\left(\frac{\lambda^n t^{n-1}}{(n-1)!}-\frac{\lambda^{n+1}t^n}{n!}\right)e^{-\lambda t} \\
&=\sum_{n=1}^{\infty}\left(\frac{\Gamma(n+1,\theta a)}{n!}-\frac{\Gamma(n,\theta a)}{(n-1)!}\right)\frac{\lambda^{n+1}t^n}{n!}e^{-\lambda t}+\Gamma(1,\theta a)\lambda e^{-\lambda t}.
\end{align*}
$
The expectation is
$
\begin{align*}
E(T)&=\int_0^{\infty}tp(t)\,dt \\
&=\sum_{n=1}^{\infty}\left(\frac{\Gamma(n+1,\theta a)}{n!}-\frac{\Gamma(n,\theta a)}{(n-1)!}\right)\frac{1}{n!}\int_0^{\infty}(\lambda t)^{n+1}e^{-\lambda t}\,dt+\Gamma(1,\theta a)\int_0^{\infty}\lambda te^{-\lambda t}\,dt \\
&=\sum_{n=1}^{\infty}\left(\frac{\Gamma(n+1,\theta a)}{n!}-\frac{\Gamma(n,\theta a)}{(n-1)!}\right)\frac{(n+1)!}{n!\lambda}+\frac{\Gamma(1,\theta a)}{\lambda} \\
&=\sum_{n=1}^{\infty}\left(\Gamma(n+1,\theta a)-n\Gamma(n,\theta a)\right)\frac{n+1}{n!\lambda}+\frac{\Gamma(1,\theta a)}{\lambda} \\
&=\frac{1}{\lambda}\sum_{n=0}^{\infty}(\Gamma(n+1,\theta a)-n\Gamma(n,\theta a))\frac{n+1}{n!} \\
&=\frac{1}{\lambda}\sum_{n=0}^{\infty}\frac{(\theta a)^ne^{-\theta a}}{n!}(n+1) \\
&=\frac{1}{\lambda}(\theta a+1).
\end{align*}
$
