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Dr. Hackney STA Solutions pg 67

Dr. Hackney STA Solutions pg 67 - Z ∞-∞ B 1 ω σ 2 dω...

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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 x 1 /n = nf Y ( nx ). 5.6 a. For Z = X - Y , set W = X . Then Y = W - Z , X = W , and | J | = 0 1 - 1 1 = 1 . Then f Z,W ( z, w ) = f X ( w ) f Y ( w - z ) · 1, thus f Z ( z ) = -∞ f X ( w ) f Y ( w - z ) dw . b. For Z = XY , set W = X . Then Y = Z/W and | J | = 0 1 1 /w - z/w 2 = - 1 /w . Then f Z,W ( z, w ) = f X ( w ) f Y ( z/w ) · |- 1 /w | , thus f Z ( z ) = -∞ |- 1 /w | f X ( w ) f Y ( z/w ) dw . c. For Z = X/Y , set W = X . Then Y=W/Z and | J | = 0 1 - w/z 2 1 /z = w/z 2 . Then f Z,W ( z, w ) = f X ( w ) f Y ( w/z ) · | w/z 2 | , thus f Z ( z ) = -∞ | 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
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Unformatted text preview: Z ∞-∞ B 1+ ( ω σ ) 2 dω = σπB, Z ∞-∞ D 1+ ( ω-z τ ) 2 dω = τπD and Z ∞-∞ " Aω 1+ ( ω σ ) 2-Cω 1+ ( ω-z τ ) 2 # dω = Z ∞-∞ " Aω 1+ ( ω σ ) 2-C ( ω-z ) 1+ ( ω-z τ ) 2 # dω-Cz Z ∞-∞ 1 1+ ( ω-z τ ) 2 dω = A σ 2 2 log ´ 1+ ± ω σ ² 2 µ-Cτ 2 2 log " 1+ ¶ ω-z τ · 2 #³ ³ ³ ³ ³ ∞-∞-τπCz. The integral is finite and equal to zero if A = M 2 σ 2 , C = M 2 τ 2 for some constant M . Hence f Z ( z ) = 1 π 2 στ ´ σπB-τπD-2 πMz τ µ = 1 π ( σ + τ ) 1 1+ ( z/ ( σ + τ )) 2 , if B = τ σ + τ , D = σ σ + τ ) , M =-στ 2 2 z ( σ + τ ) 1 1+ ( z σ + τ ) 2 ....
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