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Consider the sum

$$ y = \sum_{i=0}^n x_i, $$ where $n$ is a discrete random variable with pmf $q_n$, normalized via $\sum_{n=0}^\infty q_n = 1$, and $x_i$ is a continuous random variable with pdf $p(x)$ where $x \in \mathbb{R}$.

$y$ is therefore a continuous random variable. How can I derive its pdf?

Attempt: I figure if $n$ were fixed, the pdf of $y$ would be $$\text{pdf}(y) =\int_{\mathbb{R}^n} p(y-x_1)p(x_2-x_3)\dots p(x_{n-1}-x_n)dx_1 dx_2\dots dx_n$$

Since $n$ is random, this has to be weighted over all possibilities somehow. Is it $$ \text{pdf}(y) = \sum_{i=0}^\infty q_i \int_{\mathbb{R}^i} p(y-x_1)\dots p(x_{i-1}-x_i)dx_1 \dots dx_i ? $$

In particular I'd like to consider when $p(x)$ is an exponential distribution and $q_n$ is a binomial distribution. Is there anywhere I can read about this problem?

How do things change if $x$ and $n$ are not independent? Thanks for any guidance.

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