1.6. Discrete Random Variables 47correct amperage, the lot is accepted. If, in fact, there are 20 defective fuses in the
lot, the probability of accepting the lot is, under appropriate assumptions,
( 80
5)
( 100
5)=0. 31931.More generally, let the random variableXbe the number of defective fuses among
the five that are inspected. The pmf ofXis given by
pX(x)={ ( 20
x)(
80
5 −x)
(^1005 )
forx=0, 1 , 2 , 3 , 4 , 5
0elsewhere.(1.6.4)Clearly, the space ofXisD={ 0 , 1 , 2 , 3 , 4 , 5 }, which is also its support. This is an
example of a random variable of the discrete type whose distribution is an illustra-
tion of ahypergeometric distribution, which is formally defined in Chapter 3.
Based on the above discussion, it is easy to graph the cdf ofX; see Exercise 1.6.5.1.6.1 Transformations
A problem often encountered in statistics is the following. We have a random
variableXand we know its distribution. We are interested, though, in a random
variableY which is sometransformationofX,say,Y =g(X). In particular,
we want to determine the distribution ofY. AssumeXis discrete with spaceDX.
Then the space ofYisDY={g(x):x∈DX}. We consider two cases.
In the first case,gis one-to-one. Then, clearly, the pmf ofY is obtained as
pY(y)=P[Y=y]=P[g(X)=y]=P[X=g−^1 (y)] =pX(g−^1 (y)). (1.6.5)Example 1.6.3.Consider the random variableXof Example 1.6.1. Recall thatX
was the flip number on which the first head appeared. LetYbe the number of flips
before the first head. ThenY=X−1. In this case, the functiongisg(x)=x−1,
whose inverse is given byg−^1 (y)=y+1. The space ofYisDY={ 0 , 1 , 2 ,...}.The
pmf ofXis given by (1.6.1); hence, based on expression (1.6.5), the pmf ofY is
pY(y)=pX(y+1)=(
1
2)y+1
, fory=0, 1 , 2 ,....Example 1.6.4.LetXhave the pmfpX(x)={
3!
x!(3−x)!( 2
3)x( 1
3) 3 −x
x=0, 1 , 2 , 3
0elsewhere.We seek the pmfpY(y) of the random variableY =X^2. The transformation
y=g(x)=x^2 mapsDX={x:x=0, 1 , 2 , 3 }ontoDY ={y:y=0, 1 , 4 , 9 }.In
general,y=x^2 does not define a one-to-one transformation; here, however, it does,
