2
May 18

Law of total probability - you could have invented this




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.

 

Leave a Reply

You must be logged in to post a comment.