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solutions_ch5_exercises

# solutions_ch5_exercises - Chapter 5 Properties of a Random...

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Chapter 5 Properties of a Random Sample 5.1 Let X = # color blind people in a sample of size n . Then X binomial( n, p ), where p = . 01. The probability that a sample contains a color blind person is P ( X > 0) = 1 - P ( X = 0), where P ( X = 0) = ( n 0 ) ( . 01) 0 ( . 99) n = . 99 n . Thus, P ( X > 0) = 1 - . 99 n > . 95 n > log( . 05) / log( . 99) 299 . 5.3 Note that Y i Bernoulli with p i = P ( X i μ ) = 1 - F ( μ ) for each i . Since the Y i ’s are iid Bernoulli, n i =1 Y i binomial( n, p = 1 - F ( μ )). 5.5 Let Y = X 1 + · · · + X n . Then ¯ X = (1 /n ) Y , a scale transformation. Therefore the pdf of ¯ X is f ¯ X ( x ) = 1 1 /n f Y parenleftBig x 1 /n parenrightBig = nf Y ( nx ). 5.6 a. For Z = X - Y , set W = X . Then Y = W - Z , X = W , and | J | = vextendsingle vextendsingle vextendsingle vextendsingle 0 1 - 1 1 vextendsingle vextendsingle vextendsingle vextendsingle = 1 . Then f Z,W ( z, w ) = f X ( w ) f Y ( w - z ) · 1, thus f Z ( z ) = integraltext -∞ f X ( w ) f Y ( w - z ) dw . b. For Z = XY , set W = X . Then Y = Z/W and | J | = vextendsingle vextendsingle vextendsingle vextendsingle 0 1 1 /w - z/w 2 vextendsingle vextendsingle vextendsingle vextendsingle = - 1 /w . Then f Z,W ( z, w ) = f X ( w ) f Y ( z/w ) · |- 1 /w | , thus f Z ( z ) = integraltext -∞ |- 1 /w | f X ( w ) f Y ( z/w ) dw . c. For Z = X/Y , set W = X . Then Y=W/Z and | J | = vextendsingle vextendsingle vextendsingle vextendsingle 0 1 - w/z 2 1 /z vextendsingle vextendsingle vextendsingle vextendsingle = w/z 2 . Then f Z,W ( z, w ) = f X ( w ) f Y ( w/z ) · | w/z 2 | , thus f Z ( z ) = integraltext -∞ | w/z 2 | f X ( w ) f Y ( w/z ) dw . 5.7 It is, perhaps, easiest to recover the constants by doing the integrations. We have integraldisplay -∞ B 1+ ( ω σ ) 2 = σπB, integraldisplay -∞ D 1+ ( ω - z τ ) 2 = τπD and integraldisplay -∞ bracketleftBigg 1+ ( ω σ ) 2 - 1+ ( ω - z τ ) 2 bracketrightBigg = integraldisplay -∞ bracketleftBigg 1+ ( ω σ ) 2 - C ( ω - z ) 1+ ( ω - z τ ) 2 bracketrightBigg - Cz integraldisplay -∞ 1 1+ ( ω - z τ ) 2 = A σ 2 2 log bracketleftbigg 1+ parenleftBig ω σ parenrightBig 2 bracketrightbigg - 2 2 log bracketleftBigg 1+ parenleftbigg ω - z τ parenrightbigg 2

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