sum of $N$ gamma distributions with $N$ being a poisson distribution I have an event having poisson distribution with time intervals of one minute. Every event has accomplishment time with gamma distribution.
I $N$ number of events start in $t$ minutes, the what will be the distribution of total time of all the $N$ events?
Let $X_i$ be the time for every event $i$
the total time is $N$ summations of gamma distribution.
But I am not able to solve ahead.
Note: all the gamma distributions have same parameters
the time $t$ is the time in which $N$ events started. It is not necessary that these events/tasks would end in time $t$ too.
 A: In addition to the other excellent answer, I will try another approach. Such problems are often attacked with moment generating functions. 
Let $X_1, \dotsc, X_n$ be iid and $N \sim \text{Po}(\lambda)$. We are interested in the sum of a random number of terms, $X_1+\dotsm+X_N$. Let the moment generating function (mgf) of $X_1$ be $M_X(t)$ and then define $M_n(t)=M_{X_1+\dotsm+X_n}(t)= M_X(t)^n$. Then we can calculate the mgf of the sum of a random number of terms as 
$$\DeclareMathOperator{\E}{\mathbb{E}}
   M(t)=M_{X_1+\dotsm+X_N}(t)=\E \E \left\{ e^{t(X_1+\dotsm+X_N} \mid N=n\right\} \\
= \sum_{n=0}^\infty M(t)^n e^{-\lambda} \frac{\lambda^n}{n!}=\sum_{n=0}^\infty e^{-\lambda}\frac{(\lambda M(t))^n}{n!} = e^{-\lambda} e^{\lambda M(t)}
$$
and then checking that the mgf for the gamma distribution is 
$$
  M_X(t)= (1- t/\beta)^{-\alpha}
$$
(for $t< \beta$.)  Inserting this we finally find that the mgf of the random sum is 
$$
   M(t)= e^{-\lambda} e^{\lambda (1-t/\beta)^{-\alpha}}
$$
I cannot recognize that as the mgf of a known distribution, but one can get good approximations starting with the mgf. One possibility is the saddlepoint approximation, see How does saddlepoint approximation work? (and search this site for examples).
A: So $N$, the total number of events follows a Poisson distribution, which means
$$
p(n) = P(N=n) = \frac{e^{-\lambda}\lambda^n}{n!}.
$$
Also, they tell you the conditional distribution of the sum $T = \sum_{i=1}^N X_i$:
$$
f_{T\mid N=n}(t\mid n) = \frac{1}{\Gamma(n \alpha )\beta^{n\alpha} }\exp\left[ - \frac{t}{\beta}\right]t^{n\alpha-1} .
$$
I assume they're all iid, here.
If you want the joint distribution of both random variables, just multiply together. If you want the marginal of $T$, you have to sum out the unwanted $n$ from the joint:
$$
f_T(t) = \sum_{n=0}^{\infty}f_{T\mid N=n}(t\mid n)p(n) = \sum_{n=0}^{\infty} \underbrace{\frac{e^{-\lambda}\lambda^n}{n!}}_{P(N=n)}\underbrace{\frac{1}{\Gamma(n \alpha )\beta^{n\alpha} }\exp\left[ - \frac{t}{\beta}\right]t^{n\alpha-1}}_{f_{T\vert N=n}(t\vert n)}.
$$
Does that help?
