3.4. The Normal Distribution 187

This integral exists because the integrand is a positive continuous function that is

bounded by an integrable function; that is,

0 <exp

(

−z^2

2

)

<exp(−|z|+1), −∞<z<∞,

and ∫∞

−∞

exp(−|z|+1)dz=2e.

To evaluate the integralI,wenotethatI>0andthatI^2 may be written

I^2 =

1

2 π

∫∞

−∞

∫∞

−∞

exp

(

−

z^2 +w^2

2

)

dzdw.

This iterated integral can be evaluated by changing to polar coordinates. If we set

z=rcosθandw=rsinθ,wehave

I^2 =

1

2 π

∫ 2 π

0

∫∞

0

e−r

(^2) / 2
rdrdθ

1
2 π
∫ 2 π
0
dθ=1.
Because the integrand of display (3.4.1) is positive onRand integrates to 1 over
R, it is a pdf of a continuous random variable with supportR. Wedenotethis
random variable byZ. In summary,Zhas the pdf
f(z)=
1
√
2 π
exp
(
−z^2
2
)
, −∞<z<∞. (3.4.2)
Fort∈R,themgfofZcan be derived by a completion of a square as follows:
E[exp{tZ}]=
∫∞
−∞
exp{tz}
1
√
2 π
exp
{
−
1
2
z^2
}
dz
=exp
{
1
2
t^2
}∫∞
−∞
1
√
2 π
exp
{
−
1
2
(z−t)^2
}
dz
=exp
{
1
2
t^2
}∫∞
−∞
1
√
2 π
exp
{
−
1
2
w^2
}
dw, (3.4.3)
where for the last integral we made the one-to-one change of variablew=z−t.By
the identity (3.4.2), the integral in expression (3.4.3) has value 1. Thus the mgf of
Zis
MZ(t)=exp
{
1
2
t^2
}
, for−∞<t<∞. (3.4.4)
The first two derivatives ofMZ(t) are easily shown to be
MZ′(t)=texp
{
1
2
t^2
}
MZ′′(t)=exp
{
1
2
t^2
}
+t^2 exp
{
1
2
t^2
}
.