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Solution STAT/MATH 8
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Solutions for HW8
fX (x) = y 2ey x4 ex , x > 0, fY (y) = ,y > 0 (5) (3) x4 y 2 exy , x > 0, y > 0 (5)(3)
1: (a)
Scince X and Y are independent fX,Y (x, y) = fX (x)fY (y) =
(b)W = X + Y, Z = X/(X + Y ), so W > 0, 0 < Z < 1 (c) X=WZ, Y=W(1-Z). So
y 1 (x+y)2 (w, z) 1 J(x, y) = = = x 1 (x+y)2 (x, y) x+y
So the joint density of W and Z is fX,Y (h1 (w, z)) fW,Z (w, z) = |J(h1 (w, z))| (wz)4 w 2 (1 z)2 ew /((5)(3)) = 1/w 7 4 2 w w z (1 z) e = (5)(3) (d) fW (w) = w 7 z 4 (1 z)2 ew dz (5)(3) 0 w 7 ew w 7 ew B(5, 3) = = (5)(3) (8)
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So W (8, 1) fZ (z) = w 7z 4 (1 z)2 ew dw (5)(3) 0 z 4 (1 z)2 ) z 4 (1 z)2 (8) = = (3)(5) B(5, 3)
So Z B(5, 3)
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(e) It is easy to check fW,Z (w, z) = fW (w)fZ (z). So W and Z are independent. 2: (a) for x > 0, FX (x) = P (X x) = P ( U x) (1 U) x x = P (U )= 1 + x 1 + x
fX (x) =
(1 + x) 2 x dFX (x) = dx (1 + x)2 ) = (1 + x)2
u x (b) x = = h(u) (1u)2 , so h1 (x) = 1+x 1 1u+u h (u) = (1u)2 = (1u)2 so fU (h1 (x)) fX (x) = 1 = |h (h (x))| (1 + x)2
3 (a) FX (x) =
x
1 tan1 (x) ds = + 1/2 (1 + s2 )
(b)
1 FX (x) = tan(x /2)
(c)
1 g(u) = FX (u) = tan(u /2)
4 (a) X Bin(4, 2/5). so EX = 4 2/5 = 1.6 (b) X HypeGem(M = 10, N = 25, n = 4), EX = nM = 8/5 = 1.6 N (c) X NegBin(r = 4, = 3/5), so EX = r(1) = 8/3 5: P (Y = k) = (1 )k if k 49 (1 )50 if k = 50
So EY = 49 k(1 )k + 50(1 )5 0 k=1 Let S = 49 k(1 )k , then k=1
49 50
(1 )S =
k=1 50
k(1 )k+1 =
j=2 50 j
(j 1)(1 )j
=
j=2
j(1...