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272 Chapter 4 Special Distributions
Proof
&∞
&∞
1. (1) = 0 y 1−1 e−y dy = 0 e−y dy = 1
2. Integrate the gamma function by parts. Let u = y r −1 and dv = e−y . Then
% ∞
% ∞
'∞
y r −1 e−y dy = −y r −1 e−y '0 +
(r − 1)y r −2 e−y dy
0
0
%
∞
= (r − 1)
y r −2 e−y dy = (r − 1)(r − 1)
0
3. Use part (2) as the basis for an induction argument. The details will be left as
an exercise.
Definition 4.6.2. Given real numbers r > 0 and λ > 0, the random variable Y is
said to have the gamma pdf with parameters r and λ if
f Y (y) =
λr r −1 −λy
y e ,
(r )
y >0
Comment To justify Definition 4.6.2 requires a proof that f Y (y) integrates to 1. Let
u = λy. Then
%
∞
0
λr r −1 −λy
λr
y e dy =
(r )
(r )
1
=
(r )
Theorem
4.6.3
%
∞
0
%
∞
u r −1 −u 1
du
e
λ
λ
u r −1 e−u du =
0
1
(r ) = 1
(r )
Suppose that Y has a gamma pdf with parameters r and λ. Then
1. E(Y ) = r/λ
2. Var(Y ) = r/λ2
Proof
%
% ∞
λr r −1 −λy
λr
y e
dy =
y r e−λy dy
(r )
(r ) 0
0
%
λr (r + 1) ∞ λr +1
y r e−λy dy
=
(r ) λr +1
(r + 1)
0
λr r (r )
=
(1) = r/λ
(r ) λr +1
2. A calculation similar to the integration carried out in part (1) shows that
E(Y 2 ) = r (r + 1)/λ2 . Then
1. E(Y ) =
∞
y
Var(Y ) = E(Y 2 ) − [E(Y )]2
= r (r + 1)/λ2 − (r/λ)2
= r/λ2