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Problem:

I have a sum of $N$ random variables $X_i$, where $N$ is distributed according to a binomial distribution and the $X_i$ are independent and identically distributed according to an exponential distribution.
I would like to get the pdf of the sum

$$ Y_N = X_1 + X_2 + dots + X_N. $$

If I write the pmf of $N$ as
$$ p_n = {N choose n} p^n q^{N-n}$$
and the pdf of the $X_i$ as
$$ f(x) = lambda e^{-lambda x},$$
How can I get the pdf $p(y)$ of $Y_N$?

Attempt:

I gather this is called a compound process. I read about it somewhat in Feller: the section is "Sum of a Random Number of Random Variables", but he does not mix discrete and continuous random variables.
I think this is just a discrete time random walk where the steps are exponentially distributed.

I think I can condition something like
$$text{Prob}(Y_N = y) = sum_{n=0}^N text{Prob}(N=n)text{Prob}(X_1 + dots +X_n = y),$$
and this will become

$$p(y) = sum_{n=1}^N p_n f^{n*}(y),$$

but I'm not actually sure how to write the $n$-fold convolution of exponentials (although this should be a Gamma distribution I think).

So, first interjection: is it this?

$$ f^{n*}(y) = int dx_1dots dx_n f(x_1)f(x_2-x_1)f(x_3-x_2-x_1)dots f(y-x_n-dots x_1)?$$

or is it this?

$$ f^{n*}(y) = int dx_1dots dx_n f(y-x_n)f(x_n-x_{n-1})dots f(x_2-x_1)f(x_1)?$$

Then I know I can take a laplace transform to simplify the convolution (although I'm not sure what the convolution should look like exactly yet), to get something like

$$ p(s) = sum_{n=0}^N p_n f(s)^n$$

Does this make sense? Then this is the probability generating function of $p_n$ evaluated at $f(s)$:

$$ G(z) = langle z^n rangle|_{z=f(s)} = sum_{n=0}^N z^n p_n |_{z=f(s)}$$

So the mgf for $ p(y)$ is the pgf for $p_n$ evaluated at the mgf for f(x): Is this correct?

(using this link to keep my words straight)

Following through, the pgf of the binomial distribution $p_n$ is

$$ G(z) = sum_{i=0}^N {N choose i} (zp)^i q^{N-i} = (zp + q)^N,$$

(verified correct here)
and the mgf of the exponential distribution $f(x)$ is

$$ f(s) = int_0^infty dx lambda e^{(s-lambda)x} = frac{lambda}{lambda -s} $$

(verified correct here).
So the mgf for $p(y)$ will be

$$ p(s) = (pfrac{lambda}{lambda-s}+q)^N. $$

So for example the mean value of $Y$ should be

$$mu_Y = frac{d}{ds}(pfrac{lambda}{lambda-s}+q)^N|_{s=0} = N (pfrac{lambda}{lambda-s}+q)^{N-1} pfrac{lambda}{(lambda-s)^2}|_{s=0} = Nplambda^{-1}. $$
(Fixed an error thanks to Clement)

The probability distribution should be the inverse fourier transform of the characteristic function $p(s=ik)$:

$$p(y) = int_{-infty}^{infty} frac{dk}{2pi} e^{-iky}(frac{plambda}{lambda-ik}+p)^N$$

Any thoughts on this integral are appreciated !