I'm working through the problem below:
Q: Let $X∼Poisson(α)$ and $Y∼Poisson(β)$ be two independent random variables. Define a new random variable as $Z=X+Y$. Find the PMF of $Z$.
A: $$ \begin{aligned} P_Z(k) & =P(X+Y=k) \\ & =\sum_{i=0}^k P(X+Y=k \mid X=i) P(X=i) \quad \text { (law of total probability) } \\ & =\sum_{i=0}^k P(Y=k-i \mid X=i) P(X=i) \quad \quad \quad \text { (since } X \text { and } Y \text { are independent) } \\ & =\sum_{i=0}^k P(Y=k-i) P(X=i) \quad \\ & =\sum_{i=0}^k \frac{e^{-\beta} \beta^{k-i}}{(k-i) !} \frac{e^{-\alpha} \alpha^i}{i !} \\ & =e^{-(\alpha+\beta)} \sum_{i=0}^k \frac{\alpha^i \beta^{k-i}}{(k-i) ! i !} \\ & =\frac{e^{-(\alpha+\beta)}}{k !} \sum_{i=0}^k \frac{k !}{(k-i) i !} \alpha^i \beta^{k-i} \\ & =\frac{e^{-(\alpha+\beta)}}{k !} \sum_{i=0}^k\left(\begin{array}{c} k \\ i \end{array}\right) \alpha^i \beta^{k-i} \\ & =\frac{e^{-(\alpha+\beta)}}{k !}(\alpha+\beta)^k \quad \quad \text { (by the binomial theorem). } \end{aligned} $$
I can follow the algbraic manipulations, but my question is how on earth can I come up with the solution in the first place? Seeing the solution makes sense but it's like if you have a hammer everything looks like a nail. What is the intuition here to start with the fact that I have to expand using the LOTP and then express the first expression in terms of Y, and only then substitute with the Poisson PMF.
Or am I just overthinking it and it is just a "trick" that needs to be learned by heart?
Thanks
It's a mix of everything, knowing tricks and knowing where to start. To make it a little more intuitive you can do this \begin{align} P(Z =k) & = \sum_{x} P(Z = k, X = x) \end{align*} true? yes! because we are marginalized $X$, but here the trick is that we know what $Z$ is, so
\begin{align*} P(Z =k) & = \sum_{x} P(Z = k, X = x)\tag{1} \\ & = \sum_{x} P(X+Y = k, X = x) \\ & = \sum_{x} P(Y = k - X, X = x) \\ & = \sum_{x} P(Y = k - x, X = x) \\ & = \sum_{x} P(Y = k - x)P( X = x) \end{align*}
and from now on it is the same as what you suggest.