Probability distribution of binomial variable multiplied by a uniform variable

Question

Given is a binomial variable with success rate $p$, and on success, you receive anywhere between $X_1$ and $X_2$ dollars, uniformly distributed. On failure, you receive nothing.

From what I've gathered, the sum of multiple uniform variables would be an Irwin-Hall distribution which approximates a normal distribution in the limit, but this case is a bit different.

Question: What would be the Probability Mass Function (PMF) of the total amount of money you receive after $N$ trials?

Approximations could be acceptable here (e.g. maybe it's similar to summing $N\cdot p$ normal variables which results in a normal distribution; or maybe a binomial multiplied by the average of $X_1$ and $X_2$.)

P.S.: I should clarify that I plan to use an existing Java library or code my own distribution myself to calculate the PMF, so if the resulting distribution is very complicated but can be approximated well by a simpler distribution, that would be preferred.

Answer 1

Edit: Added an exact solution at the end.

For $N=1$ we want to know something about $Y=U*X$ where $U$ has a uniform distribution on $(x_1, x_2)$ and $X$ has a Bernoulli distribution with probability $p$.

We have a mixture distribution in that there is a probability mass of $1-p$ at zero and a "uniform" distribution on $(x_1,x_2)$. The $k$-th moment of that distribution is

$$m_k=\int_{x_1}^{x_2} \frac{p x^k}{x_2-x_1} \, dx+0^k p = \frac{p \left(x_2^{k+1}-x_1^{k+1}\right)}{(k+1) (x_2-x_1)}$$

The moment generating function for a single observation is

$$\text{mgf}=1+\sum _{m=1}^{\infty } \frac{t^m p \left(x_2^{m+1}-x_1^{m+1}\right)}{m! (m+1) (x_2-x_1)}$$

So for the sum of $n$ observations the moment generating function is $\text{mgf}^{~n}$. The mean and variance of the sum of these random variables are found from the moment generating function:

$$\mu=\frac{1}{2} n p (x_1+x_2)$$ $$\sigma^2=\frac{1}{12} n p \left((4-3 p) x_1^2+(4-6 p) x_1 x_2+(4-3 p) x_2^2\right)$$

Depending on the values of $n$ and $p$, one might be able to use the above mean and variance with a normal approximation.

Exact solution

If you are able to switch to Mathematica, then for samples sizes at least up to $n=20$ finding the exact distribution is doable. One just sums the products of the probability of obtaining $k$ successes and the density for the generalized Irwin-Hall distribution with sample size $k$.

There is a probability of $(1-p)^n$ at zero. The "density" part of the distribution is found with the following code (using a specific example):

n = 10;
p = 1/4;
x1 = 10;
x2 = 25;
pdf = Sum[Binomial[n, k] p^k (1 - p)^(n - k) PDF[UniformSumDistribution[k, {x1, x2}], y],
  {k, 1, n}];

Plot[pdf, {y, 0, 250}, Exclusions -> None]

Alternatively, we can plot the cdf which does exist.

Plot[((1 - p)^n) + Integrate[pdf, {y, 0, yy}], {yy, 0, 250}, Frame -> True]