Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

62 Probability and Distributions

Remark 1.8.1.The terminology of expectation or expected value has its origin

in games of chance. For example, consider a game involving a spinner with the

numbers 1, 2, 3 and 4 on it. Suppose the corresponding probabilities of spinning

these numbers are 0.20, 0.30, 0.35, and 0.15. To begin a game, a player pays $5 to

the “house” to play. The spinner is then spun and the player “wins” the amount in

the second line of the table:

Number spunx 1 2 3 4

”W i n s ” $2 $3 $4 $12

G=Gain −$3 −$2 −$1 $7

pG(x) 0.20 0.30 0.35 0.15

”Wins” is in quotes, since the player must pay $5 to play. Of course, the random

variable of interest is the gain to the player; i.e.,Gwith the range as given in the

third row of the table. Notice that 20% of the time the player gains−$3; 30% of

the time the player gains−$2; 35% of the time the player gains−$1; and 15% of

the time the player gains $7. In mathematics this sentence is expressed as

(−3)× 0 .20 + (−2)× 0 .30 + (−1)× 0 .35 + 7× 0 .15 =− 0. 50 ,

which, of course, isE(G). That is, the expected gain to the player in this game is

−$0.50. So the player expects to lose 50 cents per play. We say a game is afair

game, if the expected gain is 0. So this spinner game is not a fair game.

Let us consider a function of a random variableX. Call this functionY=g(X).
BecauseY is a random variable, we could obtain its expectation by first finding
the distribution ofY. However, as the following theorem states, we can use the
distribution ofXto determine the expectation ofY.

Theorem 1.8.1. LetXbe a random variable and letY=g(X)for some function
g.

(a) SupposeXis continuous with pdffX(x).If

∫∞

−∞|g(x)|fX(x)dx <∞, then the

expectation ofY exists and it is given by

E(Y)=

∫∞

−∞

g(x)fX(x)dx. (1.8.2)

(b) SupposeXis discrete with pmfpX(x). Suppose the support ofXis denoted
bySX.If

∑

x∈SX|g(x)|pX(x)<∞, then the expectation ofY exists and it is

given by

E(Y)=

∑

x∈SX

g(x)pX(x). (1.8.3)

Proof: We give the proof in the discrete case. The proof for the continuous case
requires some advanced results in analysis; see, also, Exercise 1.8.1.