Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

188 Some Special Distributions

Upon evaluating these derivatives att= 0, the mean and variance ofZare

E(Z)=0 and Var(Z)=1. (3.4.5)

Next, define the continuous random variableXby

X=bZ+a,

forb>0. This is a one-to-one transformation. To derive the pdf ofX,notethat
the inverse of the transformation and the Jacobian arez=b−^1 (x−a)andJ=b−^1 ,
respectively. Becauseb>0, it follows from (3.4.2) that the pdf ofXis

fX(x)=

1

√

2 πb

exp

{

−

1

2

(

x−a

b

) 2 }

, −∞<x<∞.

By (3.4.5), we immediately haveE(X)=a and Var(X)=b^2. Hence, in the
expression for the pdf ofX, we can replaceabyμ=E(X)andb^2 byσ^2 =Var(X).
We make this formal in the following:

Definition 3.4.1(Normal Distribution).We say a random variableXhas anor-
mal distributionif its pdf is

f(x)=

1

√

2 πσ

exp

{

−

1

2

(

x−μ

σ

) 2 }

, for−∞<x<∞. (3.4.6)

The parametersμandσ^2 are the mean and variance ofX, respectively. We often

write thatXhas aN(μ, σ^2 )distribution.

In this notation, the random variableZwith pdf (3.4.2) has aN(0,1) distribution.
We callZastandard normalrandom variable.
For the mgf ofX, use the relationshipX=σZ+μand the mgf forZ, (3.4.4),
to obtain

E[exp{tX}]=E[exp{t(σZ+μ)}]=exp{μt}E[exp{tσZ}]

=exp{μt}exp

{

1

2

σ^2 t^2

}

=exp

{

μt+

1

2

σ^2 t^2

}

, (3.4.7)

for−∞<t<∞.
We summarize the above discussion, by noting the relationship betweenZand
X:

Xhas aN(μ, σ^2 ) distribution if and only ifZ=Xσ−μhas aN(0,1) distribution.
(3.4.8)
LetX have aN(μ, σ^2 ) distribution. The graph of the pdf ofX is seen in
Figure 3.4.1 to have the following characteristics: (1) symmetry about a vertical
axis throughx=μ; (2) having its maximum of 1/(σ

√
2 π)atx=μ;and(3)having
thex-axis as a horizontal asymptote. It should also be verified that (4) there are