790 CHAPTER 38. PROBABILITY

In this case there are no mathematical pathologies and so

M′ (t) = e12 t2σ2+µt (tσ2 +µ

)so letting t = 0 yields E (X) = µ . Then also

M′′ (t) = e12 t2σ2+µt (t2

σ4 +2tσ2

µ +σ2 +µ

2)and so E

(X2)=M′′ (0)=σ2+µ2. Then the variance is E

(X2)−E (X)2 =σ2+µ2−µ2 =

σ2 showing the identification of these parameters.

Example 38.8.6 Let X be a X 2 (r) distribution. Find the moment generating functionvalid for t in some interval containing 0.

By definition, this is∫∞

0

1Γ(r/2)2r/2 x(r/2)−1e−x/2etxdx =

∫∞

0

1Γ(r/2)2r/2 x(r/2)−1e−x( 1

2−t)dx

so change the variable letting u= x( 1

2 − t)

so du=( 1

2 − t)

dx. Let |t|< 12 . Then the integral

is ∫∞

0

1Γ(r/2)2r/2

(u

(1/2− t)

)(r/2)−1

e−u 1(1/2− t)

du

=1

2r/2 (1/2− t)r/2

1Γ(r/2)

∫∞

0ur/2−1e−udu =

1

(1−2t)r/2

Now with this, you can find all the moments desired.

Proposition 38.8.7 Suppose X is normally distributed with mean µ and variance σ2. ThenX−µ

σis normally distributed with mean 0 and variance 1.

Proof: This is real easy to do with the moment generating technique.

E(

exp(

tX−µ

σ

))= E

(exp( t

σX− tµ

σ

))= E

(exp( t

σX)

exp(− tµ

σ

))Now t

σX and − tµ

σare independent. (Check the definition.) Therefore, the above reduces

to

E(

exp( t

σX))

E(

exp(− tµ

σ

))= e

tσ

µ e12 σ2

(t2

σ2

)e−tσ

µ = e12 t2

which is the moment generating function of a random variable which is normally dis-tributed with mean 0 and variance 1. ■

You might call X−µ

σa standard normal deviate.

Corollary 38.8.8 Let X be normally distributed with mean µ and variance σ2. Then(X−µ

σ

)2is distributed as X 2 (1).