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May 18

## Law of total probability - you could have invented this

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

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

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

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

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

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

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

This equation is call the law of total probability.

## Application to a sum of continuous and discrete random variables

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

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

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

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

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