# probability distribution of discrete time continuous state markov process

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.