Identify the Distribution of Gamma RV, dependent on Geometric RV Suppose I have $X_1, X_2, ... \overset{iid}{\sim} \Gamma(\alpha = 1, \beta=1)$, and $N\sim \text{Geometric}(p)$, where $X_i \perp N, \forall i$. 
Define $Y = \sum^N_{i=1}X_i$.
I am trying to find the distribution of $Y$. 
My strategy has been to use moment generating functions (MGF), but I got stuck.  
Since MGFs uniquely determine the distribution, MGFs of the convolution of independent RVs is the product of the MGFs, and since the $X$s are independent, I have the following. 
$$M_{Y|N}(t) = \left ( M_{X_1}(t)\right )^N$$
$$= \left ( 1 - t\right )^{-N}$$
Which gets me to the following, 
$$M_Y(t) = E(M_{Y|N}(t))$$
$$ = E(\left ( 1 - t\right )^{-N})$$
$$ = E\left ( \text{e}^{Nlog\left(\frac{1}{1-t}\right )}\right )$$
$$ = \frac{p \text{e}^{Nlog\left(\frac{1}{1-t}\right )}}{1-(1-p)\text{e}^{Nlog\left(\frac{1}{1-t}\right )}}$$
Which simplifies to 
$$ = \frac{p}{p-t}$$
I have two questions: (1) I did not find a recognizable MGF, so I did not find the distribution of $Y$.  Is there another way to go about this?  Is there something else to be said from this?  (2) From my understanding MGFs are defined for $t$ where $t$ can be any real number. Does that mean what I did in the third to last step is not allowed, and that I should have used characteristic functions instead?
Help appreciated, thanks in advance!
UPDATE:
I am now convinced I did something wrong because of the following check $$M^{'}_Y(0) =0\neq E(Y) =E(E(Y|N)) = E(N) = \frac{1}{p}$$
Help still appreciated.
 A: There is a solution using the principles of superposition and thinning in Poisson processes (refer to here for a definition), which state:


*

*The combined arrivals process of two independent Poisson processes with intensity parameters $\lambda$ and $\mu$ is a Poisson process with intensity $\lambda + \mu$. [Superposition]

*If we have a Poisson process of intensity $\lambda$ where each arrival independently with probability $p$ is of type $1$ and with probability $1 - p$ is of type $2$, the two can be modeled as two simultaneous independent Poisson processes: those of type $1$, with intensity $p\lambda$, and those of type $2$, with probability $(1 - p)\lambda$. [Thinning]
Notice that $\Gamma(\alpha = 1, \beta = 1)$ has the same distribution as $\text{Expon}(1)$. Consider a Poisson process with intensity $\lambda = 1$, whose arrivals are either type $1$ with probability $p$ or type $2$ with probability $1-p$. It follows that the number of type $2$ arrivals before the first type $1$ arrival (let's call this $N$) has $N \sim \text{Geo}(p)$. So what does the random variable 
$$T = \sum_{i = 1}^N X_i, \text{ } X_i \sim \text{Expon}(1)$$
represent? Recalling that the Poisson arrivals process has the property of independent increments (which says that the number of arrivals in two non-overlapping time intervals are independent), we can model this sum as the total time that passes before the first type $1$ arrival (equivalently, the first type $2$ arrival time). Why? Because we know that the intensity parameter $\lambda$ is constant (think of this as the rate of arrivals), the wait time for the first arrival of type $2$ is exponentially distributed with intensity $\lambda$. Similarly, the increment (i.e. the difference between the second type $2$ arrival time and the first type $2$ arrival time) also follows an exponential distribution with intensity $\lambda$, and so on. When we sum these random variables together, we get the total wait time until the first type $1$ arrival.
But by the principle of thinning outlined above, we know that the process of type two arrivals is a Poisson process of intensity $p \lambda$. But we know well that the first arrival time of any Poisson process is exponentially distributed with the intensity of the Poisson process, so we have $T \sim \text{Expon}(p \lambda) = \text{Expon}(p)$, as $\lambda = 1$ in your case. If you wish to express this as a Gamma random variable, then it is equivalent to the distribution $\Gamma(\alpha = 1, \beta = p)$.
Hope this helps!
