Raising the bar

Law of total probability - you could have invented this

A knight wants to kill (event $latex K$) a dragon. There are two ways to do this: by fighting (event $latex F$) the dragon or by outwitting ($latex O$) it. The choice of the way ($latex F$ or $latex O$) is random, and in each case the outcome ($latex K$ or not $latex K$) is also random. For the probability of killing there is a simple, intuitive formula:

$latex P(K)=P(K|F)P(F)+P(K|O)P(O)$.

Its derivation is straightforward from the definition of conditional probability: since $latex F$ and $latex O$ cover the whole sample space and are disjoint, we have by additivity of probability

$latex P(K)=P(K\cap(F\cup O))=P(K\cap F)+P(K\cap O)=\frac{P(K\cap F)}{P(F)}P(F)+\frac{P(K\cap O)}{P(O)}P(O)$

$latex =P(K|F)P(F)+P(K|O)P(O)$.

This is easy to generalize to the case of many conditioning events. Suppose $latex A_1,...,A_n$ are mutually exclusive (that is, disjoint) and collectively exhaustive (that is, cover the whole sample space). Then for any event $latex B$ one has

$latex P(B)=P(B|A_1)P(A_1)+...+P(B|A_n)p(A_n)$.

This equation is call the law of total probability.

Application to a sum of continuous and discrete random variables

Let $latex X,Y$ be independent random variables. Suppose that $latex X$ is continuous, with a distribution function $latex F_X$, and suppose $latex Y$ is discrete, with values $latex y_1,...,y_n$. Then for the distribution function of the sum $latex F_{X+Y}$ we have

$latex F_{X+Y}(t)=P(X+Y\le t)=\sum_{j=1}^nP(X+Y\le t|Y=y_j)P(Y=y_j)$

(by independence conditioning on $latex Y=y_j$ can be omitted)

$latex =\sum_{j=1}^nP(X\le t-y_j)P(Y=y_j)=\sum_{j=1}^nF_X(t-y_j)P(Y=y_j)$.

Compare this to the much more complex derivation in case of two continuous variables.