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Lecture

# ch2_1.pdf

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School
University of Waterloo
Department
Statistics
Course
STAT 330
Professor
Christine Dupont
Semester
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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