I read the following statement:
If X and Y are independent random variables, the distribution of their sum W = X + Y can be obtained by computing and then inverting the transform $M_W (s) = M_X (s)M_Y (s)$.
Where $M_X(s)$ is the familiar transform:
$$M_X(s) = E[e^{sX}]$$
I was trying to find an explicit formula, but embarrassingly, yielded things that didn't make sense. I was trying to do this first for discrete r.v. and then for continuous. I am aware that there is a different way of doing this, namely, the convolution but I wanted to see a different approach.
So this is what I have so far:
Our goal is to find a formula for $p_W(w)$ (and we know it should be $\sum_x{P_X(x)P_Y(w-x)}$ .) So let start with the obvious:
$M_{W}(s) = E[e^{kW}] = \sum_w{P_W(w)e^{sw}}$
$M_{X+Y}(s) = E[e^{s(X+Y)}] = E[e^{sX}]E[e^{sY}] = \sum_x p_X(x)e^{sx} \sum_y p_Y(y) e^{sy}$
And then I guess I am unsure how to procede.
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The suggested method was from Tsitsiklis probability book, page 163 (or section 4.2 from page 13 from that section), where it talks about convolutions.
Let's start with your $$\sum_x p_X(x)e^{sx} \sum_y p_Y(y) e^{sy}$$ and let $w=y+x$, i.e. $y=w-x$, so this becomes $$\sum_x p_X(x)e^{sx} \sum_w p_Y(w-x) e^{s(w-x)}$$ and then rearrange this to $$\sum_w \sum_x p_X(x) p_Y(w-x) e^{sw}$$ which is the moment generating function of $W$ $$\sum_w p_W(w)e^{sw}$$ where $p_W(w)=\sum_x p_X(x) p_Y(w-x)$.