Below is a copy and paste of Dartmouth's explanation of the sum of independent random variables:
- Suppose $X$ and $Y$ are two independent discrete random variables with distribution functions $m_1(x)$ and $m_2(x)$. Let $Z = X + Y$. We would like to determine the distribution function $m_3(x)$ of $Z$. To do this, it is enough to determine the probability that $Z$ takes on the value $z$, where $z$ is an arbitrary integer. Suppose that $X = k$, where $k$ is some integer.Then $Z = z $ if and only if $Y = z−k$. So the event $Z=z$ is the union of the pairwise disjoint events $(X=k)$ and $(Y =z−k)$
My question is:
- Why are the events $(X = k)$ and $(Y = z-k)$ disjoint? Doesn't being disjoint mean that the events cannot occur simultaneously? How can these events be disjoint if what we are trying to find is EXACTLY when $k+Y = z$?
In other words, wouldn't we want to find $X=k$ and $Y=z-k$ such that their sum $k + z - k = z$ implying their intersection is not empty implying they are not disjoint?
The two events $X{=}k$ and $Y{=}z-k$ are not disjoint, what are disjoint are the pairs of events $\{(X{=}k, Y{=}z-k) \mid \forall k\in\Bbb N\}$
That is $(X{=}0, Y{=}z)$ is disjoint from $(X{=}1, Y{=}z-1)$, and so forth.
Measuring the probability of all such events is measuring that of the Union of the pairs.
$$\begin{align}\mathsf P(X{+}Y{=}z) \; & = \; \mathsf P\Big(\bigcup_{k\in \Bbb N} \big(X{=}k, Y{=}z-k\big)\Big)\\ ~\; & = \; \sum_{k\in\Bbb N} \mathsf P\big(X{=}k, Y{=}z-k\big) & \text{since disjoint}\end{align}$$