66 Probability and Distributions}
}
return(gain)
}
The following R script obtains the average gain for a sample of 10,000 games. For
the example, we set the amount the player pays at $5.
amtpaid <- 5; numtimes <- 10000; gains <- c()
for(i in 1:numtimes){gains <- c(gains,simplegame(amtpaid))}
mean(gains)When we ran this script, we obtained− 3 .5446 as our estimate ofE(G). Exercise
1.8.13 shows thatE(G)=− 3 .54.
EXERCISES1.8.1.Our proof of Theorem 1.8.1 was for the discrete case. The proof for the con-
tinuous case requires some advanced results in in analysis. If, in addition, though,
the functiong(x) is one-to-one, show that the result is true for the continuous case.
Hint:First assume thaty=g(x) is strictly increasing. Then use the change-of-
variable technique with Jacobiandx/dyon the integral∫
x∈SXg(x)fX(x)dx.1.8.2.Consider the random variableXin Example 1.8.5. As in the example, let
Y=1/(1 +X). In the example we found theE(Y) by using Theorem 1.8.1. Verify
this result by finding the pdf ofYand use it to obtain theE(Y).
1.8.3.LetXhave the pdff(x)=(x+2)/ 18 ,− 2 <x<4, zero elsewhere. Find
E(X),E[(X+2)^3 ], andE[6X−2(X+2)^3 ].
1.8.4.Suppose thatp(x)=^15 ,x=1, 2 , 3 , 4 ,5, zero elsewhere, is the pmf of the
discrete-type random variableX. ComputeE(X)andE(X^2 ). Use these two results
to findE[(X+2)^2 ]bywriting(X+2)^2 =X^2 +4X+4.1.8.5.LetXbe a number selected at random from a set of numbers{ 51 , 52 ,..., 100 }.
ApproximateE(1/X).
Hint:Find reasonable upper and lower bounds by finding integrals boundingE(1/X).
1.8.6.Let the pmfp(x) be positive atx=− 1 , 0 ,1 and zero elsewhere.
(a)Ifp(0) =^14 , findE(X^2 ).(b)Ifp(0) =^14 and ifE(X)=^14 , determinep(−1) andp(1).1.8.7.LetXhave the pdff(x)=3x^2 , 0 <x<1, zero elsewhere. Consider a
random rectangle whose sides areXand (1−X). Determine the expected value of
the area of the rectangle.