S OME S PECIAL D ISTRIBUTIONS
Gamma Function
• Gamma function:
Z
∞
Γ(α) =
y α−1 e−y dy
0
• If α = 1,
Z
∞
Γ(1) =
e−y dy = 1.
0
If α
> 1, an integration by part shows that
Z ∞
y α−2 e−y dy = (α − 1)Γ(α − 1).
Γ(α) = (α − 1)
0
If α is a positive integer greater than 1,
Γ(α) = (α − 1)(α − 2) · · · (3)(2)(1) = (α − 1)!
S OME S PECIAL D ISTRIBUTIONS
• Let y = x/β , where β > 0. Then
Z ∞
x α−1 −x/β 1
Γ(α) =
( ) e
( )dx
β
β
0
or, equivalently,
Z
∞
1=
0
• We see that
f (x) =


1
α−1 −x/β
x
e
dx.
α
Γ(α)β
1
α−1 −x/β
x
e
α
Γ(α)β
 0
0<x<∞
elsewhere,
is a pdf of a random variable of the continuous type.
S OME S PECIAL D ISTRIBUTIONS
Gamma Distribution
• Gamma distribution with parameters α and β . Write X has a Γ(α, β)
distribution.
• mgf:
Z
∞
M (t) =
0
1
tx
α−1 −x/β
e
x
e
dx =
α
Γ(α)β
1
=
,
α
(1 − βt)
Z
∞
0
1
t< .
β
• Mean and variance: µ = αβ and σ 2 = αβ 2 .
1
α−1 −x(1−βt)/β
x
e
dx
α
Γ(α)β
S OME S PECIAL D ISTRIBUTIONS
Chi-square Distribution
• Consider a special case of gamma distribution in which α = r/2
where r is a positive integer, and β = 2. A random variable X of the
continuous type that has the pdf


f (x) =
1
r/2−1 −x/2
x
e
,
Γ(r/2)2r/2
 0
0<x<∞
elsewhere
and the mgf
−r/2
M (t) = (1 − 2t)
1
, t<
2
is said to be have a chi-square distribution.
S OME S PECIAL D ISTRIBUTIONS
• Mean and variance: µ = r and σ 2 = 2r.
• For no obvious reason, we call parameter r the number of degrees of
freedom of the chi-square distribution.
• We write χ2 (r), which means the random variable X has a
chi-square distribution with r degrees of freedom.
S OME S PECIAL D ISTRIBUTIONS
Examples
• Let X has the pdf f (x) = 41 xe−x/2 for x > 0. Then X is χ2 (4).
Hence µ = 4 and σ 2 = 8, and M (t) = (1 − 2t)−2 , t < 1/2.
• If X has the mgf M (t) = (1 − 2t)−8 , t < 1/2, then X is χ2 (16).
S OME S PECIAL D ISTRIBUTIONS
Theorem 3.3.1
• Let X have a χ2 (r) distribution. If k > −r/2, then EX k exists and
it is given by
r
k
2
Γ(
+ k)
k
2
EX =
.
r
Γ( 2 )
S OME S PECIAL D ISTRIBUTIONS
Examples
• Let X be χ2 (10), then by Table II of Appendix C with r = 10,
P (3.25 ≤ X ≤ 20.5) = P (X ≤ 20.5) − P (X ≤ 3.5)
= 0.975 − 0.025 = 0.95.
Again, as an example, if P (a
< X) = 0.05, the
P (X ≤ a) = 0.95, and a = 18.3 from Table II.
• Let X have a gamma distribution with α = r/2 and β > 0. Define
Y = 2X/β . Then Y is χ2 (r).
S OME S PECIAL D ISTRIBUTIONS
Theorem 3.3.2
• Let X1 , . . . , Xn be independent random variables. Suppose that Xi
Pn
has Γ(αi , β) distribution. Let Y =
i=1 Xi . Then Y has
Pn
Γ( i=1 αi , β) distribution.
• Corollary. Let X1 , . . . , Xn be independent χ2 (ri ) distribution,
Pn
Pn
2
respectively. Let Y =
i=1 Xi . Then Y has χ (
i=1 ri )
distribution.
S OME S PECIAL D ISTRIBUTIONS
beta distribution
• Let X1 and X2 be two independent random variables that have Γ
distribution and the joint pdf
1
β−1 −x1 −x2
xα−1
x
e
, x1 , x2 > 0.
h(x1 , x2 ) =
2
1
Γ(α)Γ(β)
• Let Y1 = X1 + X2 and Y2 = X1 /(X1 + X2 ).
g(y1 , y2 ) =
for 0
y1α+β−1 e−y1
< y1 < ∞, 0 < y2 < 1.
Γ(α + β) α−1
y2 (1 − y2 )β−1 ,
Γ(α)Γ(β)
S OME S PECIAL D ISTRIBUTIONS
• We shall show that Y1 and Y2 are independent and Y2 has the beta
distribution
g(y2 ) =


Γ(α+β) α−1
y (1
Γ(α)Γ(β) 2
− y2 )β−1 0 < y2 < 1
 0
elsewhere
• beta(α, β) with parameters α and β .
α
αβ
2
µ=
, σ =
α+β
(α + β + 1)(α + β)2