174 Some Special Distributions

Ifα>1, an integration by parts shows that

Γ(α)=(α−1)

∫∞

0

yα−^2 e−ydy=(α−1)Γ(α−1). (3.3.1)

Accordingly, ifαis a positive integer greater than 1,

Γ(α)=(α−1)(α−2)···(3)(2)(1)Γ(1) = (α−1)!.

Since Γ(1) = 1, this suggests we take 0! = 1, as we have done. The Γ function is

sometimes called the factorial function.

We say that the continuous random variableXhas a Γ-distribution with pa-

rametersα>0andβ>0, if its pdf is

f(x)=

{ 1

Γ(α)βαx

α− (^1) e−x/β 0 <x<∞
0elsewhere.
(3.3.2)
In which case, we often write thatXhas Γ(α, β) distribution.
To verify thatf(x) is a pdf, note first thatf(x)>0, for allx>0. To show
that it integrates to 1 over its support, we use the change-of-variablez=x/β,
dz=(1/β)dxin the following derivation:
∫∞
0
1
Γ(α)βα
xα−^1 e−x/βdx =
1
Γ(α)βα
∫∞
0
(βz)α−^1 e−zβdz

1
Γ(α)βα
βαΓ(α)=1;
hence,f(x) is a pdf. This change-of-variable used is worth remembering. We use a
similar change-of-variable in the following derivation ofX’s mgf:
M(t)=
∫∞
0
etx
1
Γ(α)βα
xα−^1 e−x/βdx

∫∞
0
1
Γ(α)βα
xα−^1 e−x(1−βt)/βdx.
Next, use the change-of-variabley=x(1−βt)/β, t < 1 /β,orx=βy/(1−βt), to
obtain
M(t)=
∫∞
0
β/(1−βt)
Γ(α)βα
(
βy
1 −βt
)α− 1
e−ydy.
That is,
M(t)=
(
1
1 −βt
)α∫∞
0
1
Γ(α)
yα−^1 e−ydy

1
(1−βt)α
,t<
1
β
.
Now
M′(t)=(−α)(1−βt)−α−^1 (−β)