Introduction to Probability and Statistics for Engineers and Scientists

108 Chapter 4:Random Variables and Expectation

then

E[X]= 0

( 1

3

)

+ 1

( 2

3

)

=^23

is a weighted average of the two possible values 0 and 1 where the value 1 is given twice as
much weight as the value 0 sincep(1)= 2 p(0).
Another motivation of the definition of expectation is provided by the frequency
interpretation of probabilities. This interpretation assumes that if an infinite sequence
of independent replications of an experiment is performed, then for any eventE, the pro-
portion of time thatEoccurs will beP(E). Now, consider a random variableXthat must
take on one of the valuesx 1 ,x 2 ,...,xnwith respective probabilitiesp(x 1 ),p(x 2 ),...,p(xn);
and think ofXas representing our winnings in a single game of chance. That is, with
probabilityp(xi) we shall winxiunitsi=1, 2,...,n. Now by the frequency interpreta-
tion, it follows that if we continually play this game, then the proportion of time that we
winxiwill bep(xi). Since this is true for alli,i=1, 2,...,n, it follows that our average
winnings per game will be

∑n

i= 1

xip(xi)=E[X]

To see this argument more clearly, suppose that we playNgames whereNis very large.
Then in approximatelyNp(xi) of these games, we shall winxi, and thus our total winnings
in theNgames will be

∑n

i= 1

xiNp(xi)

implying that our average winnings per game are

∑n

i= 1

xiNp(xi)

N

=

∑n

i= 1

xip(xi)=E[X]

EXAMPLE 4.4a FindE[X]whereXis the outcome when we roll a fair die.

SOLUTION Sincep(1)=p(2)=p(3)=p(4)=p(5)=p(6)=^16 , we obtain that

E[X]= 1

( 1

6

)

+ 2

( 1

6

)

+ 3

( 1

6

)

+ 4

( 1

6

)

+ 5

( 1

6

)

+ 6

( 1

6

)

=^72

The reader should note that, for this example, the expected value ofXis not a value thatX
could possibly assume. (That is, rolling a die cannot possibly lead to an outcome of 7/2.)
Thus, even though we callE[X]theexpectationofX, it should not be interpreted as the
value that weexpect Xto have but rather as the average value ofXin a large number of