I am given two Poisson distributions, $X_1$ and $X_2$ with parameters $\lambda_1=1$ and $\lambda_2=2$ respectively.
Also given that $Y = -3X_1$ + $2X_2$, how do I represent $Y$ as a compound Poisson distribution?
I know that I have to use the methods of convolution. However, I am entirely lost on this one.
I assume $X_1$ and $X_2$ are independent.
First, here's some intuition: whenever you get an arrival in process $X_1$ (rate 1), your value of $Y$ goes down by 3. Whenever you get an arrival in process $X_2$ (rate 2), your value goes up by 2. Therefore, we could build a merged poisson process with rate 3, and for each arrival we obtain value -3 with probability 1/3 and value 2 with probability 2/3. Our total value $Y$ is the sum of all the rewards collected at each poisson arrival.
Formalizing, if we define $N\sim \text{Pois}(\lambda_1 + \lambda_2)$, $M_n\sim\left\{\begin{array}{ll} -3 & w.p.~\lambda_1/(\lambda_1+\lambda_2) \\ 2 & w.p.~\lambda_2/(\lambda_1+\lambda_2)\end{array}\right.$, then $Y\sim\sum_{n=1}^N M_n$, a compound poisson distribution.
To prove the result, note that the number of times -3 is added to $Y$ in $Y = -3X_1 + 2X_2$ is poisson distributed with rate $\lambda_1$, and the number of times 2 is added is poisson distributed with rate $\lambda_2$, and the counts are independent. By the properties of merged poisson processes, this is exactly the case with our merged poisson process.
I assume that $X_1$ and $X_2$ are independent. The trick here is to fix either $X_1$ or $X_2$ and calculate the distribution of $-3X_1 + 2X_2$ depending on that value first. This is exactly what a convolution does, but here it looks a bit simpler as we can replace integrals by sums:
$$\eqalign{P[Y=y] &= P[-3X_1 + 2X_2 = y] \\ &= \sum_{x \in \mathbb N_0} P[X_1 = x] P[-3X_1+2X_2 = y | X_1 = x] \\ &= \sum_{x \in \mathbb N_0} P[X_1 = x] P[-3x+2X_2 = y]}$$
where you can now calculate both factors and hopefully simplify the sum afterwards.
