Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

1.9. Some Special Expectations 71

Theorem 1.9.2.LetXandY be random variables with moment generating func-
tionsMXandMY, respectively, existing in open intervals about 0. ThenFX(z)=
FY(z)for allz∈Rif and only ifMX(t)=MY(t)for allt∈(−h, h)for some
h> 0.

Because of the importance of this theorem, it does seem desirable to try to make
the assertion plausible. This can be done if the random variable is of the discrete
type. For example, let it be given that

M(t)= 101 et+ 102 e^2 t+ 103 e^3 t+ 104 e^4 t

is, for all real values oft, the mgf of a random variableXof the discrete type. If

we letp(x) be the pmf ofXwith support{a 1 ,a 2 ,a 3 ,...}, then because

M(t)=

∑

x

etxp(x),

we have

1

10 e

t+^2

10 e

2 t+^3

10 e

3 t+^4

10 e

4 t=p(a 1 )ea 1 t+p(a 2 )ea 2 t+···.

Because this is an identity for all real values oft, it seems that the right-hand
member should consist of but four terms and that each of the four should be equal,
respectively, to one of those in the left-hand member; hence we may takea 1 =1,
p(a 1 )= 101 ;a 2 =2,p(a 2 )= 102 ;a 3 =3,p(a 3 )= 103 ;a 4 =4,p(a 4 )= 104. Or, more
simply, the pmf ofXis

p(x)=

{ x

10 x=1,^2 ,^3 ,^4

0elsewhere.

On the other hand, supposeXis a random variable of the continuous type. Let

it be given that

M(t)=

1

1 −t

,t< 1 ,

is the mgf ofX.Thatis,wearegiven

1

1 −t

=

∫∞

−∞

etxf(x)dx, t < 1.

It is not at all obvious howf(x) is found. However, it is easy to see that a distri-
bution with pdf

f(x)=

{

e−x 0 <x<∞

0elsewhere

has the mgfM(t)=(1−t)−^1 ,t<1. Thus the random variableXhas a distribution
with this pdf in accordance with the assertion of the uniqueness of the mgf.
Since a distribution that has an mgfM(t) is completely determined byM(t),
it would not be surprising if we could obtain some properties of the distribution
directly fromM(t). For example, the existence ofM(t)for−h<t<himplies that
derivatives ofM(t) of all orders exist att= 0. Also, a theorem in analysis allows