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Unformatted text preview: MATH 471 HW 7 Solutions
Pengsheng Ji October 31, 2006 4.1.16 By deﬁnition, we have EX =
∞ 1 x · (ρ − 1)x−ρ dx = ρ − 1 2− ρ ∞ ρ − 1 x 1 = ρ−2 ρ−2 Note that the last calculation equality is valid only for ρ > 2. From the antiderivative we see that Ex = ∞ when ρ ≤ 2. 3.5.8 Let X and Y be their arrival times in hour after 5:00PM and 0 ≤ X, Y ≤ 1. Since X and Y are independent and uniformly distributed on the unit square, the probability ¨hey will meet¨ the area of {(x, y ) : x−y  ≤ 1 }, which is 1−2·(3/4)2 /2 = t ıs 4 7/16. 3.5.12 (a) FY = P (X1 ≤ y, . . . , Xn ≤ y ) = F (y )n The last equality is due to the independence of X1 , . . . , Xn . (b) Similarly, P (Z > z ) = P (X1 > z, . . . , Xn > z ) = (1 − F (z ))n . So FZ (z ) = 1 − (1 − F (z ))n 3.7.8 Note fx (x) = 1, if 0 ≤ x ≤ 1, 0 otherwise, and fY (y ) = 1/2, if 0 ≤ y ≤ 2, 0 otherwise. Using formula (7.2), we have fX +Y (z ) = . If 0 < z < 1, fX + Y (z ) = If 1 ≤ z < 2, fX +Y (z ) = If 2 ≤ z < 3, fX +Y (z ) =
1 0 z 0 1 1 · 2 dx = z . 2 1 2 fX (x)fY (z − x)dx 1· = 1. 2 =
3−z 2 1 1 z −2 · 2 3.7.12 Using Formula (7.2), we have fX +Y (z ) = = λµe−µx
z 0 λe−λx µe−µ(z−x) dx = λµe−µx z 0 e(µ−λ)x dx 1 λµ (e(µ−λ)x − 1) = (e−λx − e−µx ) µ−λ µ−λ ...
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 Derivative

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