Let X and y be independent random variables taking only integer values. Let Z=X+Y, which also takes only integer values. Its probability mass function can be computed by the convolution formula: for any integer z,
$P_Z(z)=P(Z=z)=P(X+Y=z)$
$=\sum_{x=-\infty}^{\infty} P(X=x, X+Y=z) \tag1$ $=\sum_{x=-\infty}^{\infty} P(P(X=x,Y=z-x) \tag2$ $=\sum_{x=-\infty}^{\infty} P(P(X=x)*P(Y=z-x) \tag3$ $=\sum_{x=-\infty}^{\infty} p_X(x)P_Y(z-x) \tag4$
Question 1: Explain why steps 1,2 and 3 hold.
Is it because X and Y are independent and so there's no overlap?
Best Answer
Your expressions $\mathbb P(\mathbb P(\cdot)$ are meaningless because a probability measure assigns a real number between $0$ and $1$ to an event, and hence cannot be iterated (as you cannot assign a probability to a number). The expression for $\mathbb P(Z=n)$ should instead be something like \begin{align} \mathbb P(Z=n) &= \mathbb P(X+Y=n)\\ &=\sum_{k=-\infty}^\infty \mathbb P(X=k, X+Y=n)\\ &=\sum_{k=-\infty}^\infty \mathbb P(X=k, Y = n-k)\\ &=\sum_{k=-\infty}^\infty \mathbb P(X=k)\mathbb P(Y=n-k). \end{align} These equalities are valid because of independence of $X$ and $Y$, hence $$ \{X=k\}\cap\{X+Y=n\} = \{X=k\}\cap\{Y=n-k\}.\tag1 $$ The events in $(1)$ are disjoint and so summing their probabilities over the support of $X$ yields the correct result.