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Lecture

# ch2_1.pdf

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University of Waterloo

Statistics

STAT 330

Christine Dupont

Fall

Description

Exercise 2.2.6 (Page 8) and Properties of Gamma
function (Page 9)
Exercise 2.2.2 (Page 8):
If X is a random variable with p.f.
x
▯(1 ▯ p)
f(x) = xlogp ; x = 1;2;:::; 0 < p < 1;
show that 1
X
f(x) = 1:
x=1
Solution 1: You can answer this question by using the following power series result about
log(1 ▯ y), i.e.,
X yx
log(1 ▯ y) = ▯ :
x=1x
The above result can be obtained from Page 26, 2.10.9 (Logarithmic series) or from wikipedia:
http://en.wikipedia.org/wiki/Taylor series.
Then 1 1
X X ▯(1 ▯ p) 1
f(x) = = log(1 ▯ (1 ▯ p)) = 1:
x=1 x=1 xlogp logp
Solution 2: If you do not remember the Logarithmic series and you still know the summation
of geometric series, then you may consider the following method.
Note that 1 1 x 1 x
X X ▯(1 ▯ p) 1 X ▯(1 ▯ p)
f(x) = = :
x=1 x=1 xlogp logpx=1 x
P 1 ▯(1▯p)
Let g(p) = x=1 x . Then g(1) = 0 and
▯ X ▯(1 ▯ p)▯0 X1 df▯(1 ▯ p) g=dp X1 1
g (p) = = = (1 ▯ p)▯1= = 1=p:
x x 1 ▯ (1 ▯ p)
x=1 x=1 x=1
0
Note that the boundary condition g(1) = 0 and g (p) = 1=p implies that g(p) = logp.
Therefore
X 1
f(x) = g(p) = 1:
x=1 logp
Note: derivative helps us to cancel the x on the denominator, which simpli▯es the
summation. This is a commonly used trick in the power series.
Properties of Gamma Function
The Gamma function is de▯ned as
Z
1 ▯▯1 ▯y
▯(▯) = y e dy; ▯ > 0:
0
It has the following three useful properties:
1 (1) ▯(▯) = (▯ ▯ 1)▯(▯ ▯ 1), ▯ > 1.
(2) ▯(n) = (n ▯ 1)!.
p
(3) ▯(1=2) = ▯.
Note: the proof itself is less important, but the two techniques that we used to ▯nd the
integral: integration by substitution (changing the variable) and integration by parts, are
very important.
Proof:
(1) We notice that

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