Stat 5101 Lecture Notes - School of Statistics

92 Stat 5101 (Geyer) Course NotesExample 3.3.1 (Random Sum of Random Variables).Suppose X 0 , X 1 , ... is an infinite sequence of identically distributed randomvariables, having mean E(X i )=µ X , and suppose N is a nonnegative integervaluedrandom variable independent of the X i and having mean E(N) =µ N .It is getting a bit ahead of ourselves, but we shall see in the next section thatthis impliesE(X i | N) =E(X i )=µ X . (3.7)Question: What is the expectation ofS N = X 1 + ···+X N(a sum with a random number N of terms and each term X i a random variable)where the sum with zero terms when N = 0 is defined to be zero?Linearity of expectation, which applies to conditional as well as unconditionalprobability, impliesE(S N | N) =E(X 1 +···X n |N)=E(X 1 |N)+···+E(X n |N)=E(X 1 )+···+E(X N )=Nµ Xthe next to last equality being (3.7). Hence by the iterated expectation axiomE(S N )=E{E(S N |N)}=E(Nµ X )=E(N)µ X =µ N µ X .Note that this example is impossible to do any other way than using the iteratedexpectation formula. Since no formulas were given for any of the densities,you can’t use any formula involving explicit integrals.If we combine the two conditional probability axioms, we get the following.Theorem 3.1. If X and Y are random variables and g and h are functionssuch that g(X) and h(Y ) are in L 1 , thenE{g(X)E[h(Y ) | X]} = E{g(X)h(Y )}. (3.8)Proof. Replace Y by g(X)h(Y ) in Axiom CE2 obtainingE{E[g(X)h(Y ) | X]} = E{g(X)h(Y )}.then apply Axiom CE1 to pull g(X) out of the inner conditional expectationobtaining (3.8).The reader should be advised that our treatment of conditional expectationis a bit unusual. Rather than state two axioms for conditional expectation,standard treatments in advanced probability textbooks give just one, whichis essentially the statement of this theorem. As we have just seen, our twoaxioms imply this one, and conversely our two axioms are special cases of this