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Math 5010-1, Spring 2005 Assignment 8 Problems #2, p. 290. (a) Write p(x, y) = P {X1 = x, X2 = y}, and Ri and Wi for the events that the ith draw is respectively red or white. Note that 8 7 × 12 13 8 5 p(0, 1) = P (R1 )P (W2 | R1 ) = × , 13 12 5 8 × , p(1, 0) = P (W1 )P (R2 | W1 ) = 13 12 5 4 p(1, 1) = P (W1 )P (W2 | W1 ) = × . 13 12 (b) Next, write p(a, b, c) = P {X1 = a, X2 = b, X3 = c} to find that p(0, 0) = P (R2 | R1 )P (R1 ) = 8 7 6 × × , 13 12 11 8 7 5 p(0, 0, 1) = × × , 13 12 11 5 7 8 × × , p(0, 1, 0) = 13 12 11 5 8 7 p(1, 0, 0) = × × , 13 12 11 5 8 4 p(1, 0, 1) = × × , 13 12 11 4 9 5 × × , p(1, 1, 0) = 13 12 11 8 5 4 p(0, 1, 1) = × × , 13 12 11 5 4 3 p(1, 1, 1) = × × . 13 12 11 [These are it, because of the principle of counting.] p(0, 0, 0) = #8, p. 290. (a) Want RR f (x, y) dx dy = 1. In this case, dx dy is easier to compute. So, Z ∞Z ∞ Z ∞Z y f (x, y) dx dy = c (y 2 − x2 )e−y dx dy −∞ −∞ 0 −y Z 2 3 −y ∞ 3 =c 0 2y − y e dy 3 Z ∞ 4c 4c = y 3 e−y dy = Γ(4). 3 0 3 1 Γ(4) = 3! = 6. Therefore, c = 1/8. (b) First fX : Z 1 ∞ 2 y − x2 e−y dy (−∞ < x < ∞) fX (x) = 8 |x| ! Z ∞ Z ∞ 1 e−y dy = y 2 e−y dy − x2 8 |x| |x| ! Z ∞ 1 = y 2 e−y dy − x2 e−|x| . 8 |x| The remaining integral is computed by two applications of integration-by-parts: Z Z 2 −y 2 −y y e dy = −y e + 2 ye−y dy Z 2 −y −y −y = −y e + 2 −ye + e dy = −y 2 e−y − 2ye−y − 2e−y . Therefore, i 1 h 2 −|x| −|x| −|x| fX (x) = x e + 2|x|e + 2e 8 (−∞ < x < ∞). The fY -computation is easier: Z 1 y 2 y − x2 e−y dx fY (y) = 8 −y 1 1 = y 3 e−y − y 3 e−y (y > 0). 4 12 (c) E[X] = R∞ −∞ xfX (x) dx, so E[X] = 0 by symmetry. #14, p. 291. Let X and Y denote the distance to 0 of the accident and the ambulance, respectively. We know that X and Y are independent, both uniformly distributed on [0, 1]. Therefore, 1/L, if 0 < x < L 1/L, if 0 < y < L, fX (x) = , and fY (y) = 0, otherwise 0, otherwise. Thanks to independence, f (x, y) = 1/L2 , if 0 < x < L and 0 < y < L, 0, otherwise. Therefore, for all 0 < a < L, Z Z P {|X − Y | ≤ a} = f (x, y) dx dy, R 2 where R is the region between y = x + a and y = x − a in the square of side L. The only values of interest for a are 0 ≤ a ≤ L. So, Z LZ L Z a Z x+a −2 −2 dy dx dy dx + L P {|X − Y | ≤ a} = L 0 0 = L−2 a Z (x + a) dx + L−2 0 = Z a L x−a (L − x + a) dx a 2a2 + L2 . 2L2 Therefore, the pdf of |X − Y | is f|X−Y | (a) = 2a/L2 , if 0 ≤ a ≤ L, 0, otherwise. In particular, the ambulance is expected to travel Z L 2 2L E |X − Y | = 2 a2 da = units. L 0 3 #15, p. 291. (a) Integrate. (b) In this case, f (x, y) = 1/4 when −2 < x < 2 and −2 < y < 2 (and zero, otherwise). Compute the marginal s: Z 2 1 1 fX (x) = dy = (−2 < x < 2) 2 −2 4 Z 2 1 1 fY (y) = dx = (−2 < y < 2). 2 −2 4 Therefore, in all cases, f (x, y) = fX (x)fY (y). (c) In the general uniform case P {(X, Y ) ∈ A} = area(A)/area(R). Here, area(R) = 4, A = π, so P {X 2 + Y 2 ≤ 1} = π/4. #17, p. 292. Implict here are the assumptions that the Xi ’s are uniform on [0, L], and independent from one another. Note that whenever i, j, and k are all different P {Xi < Xj < Xk } = P {Xj < Xk < Xi } = P {Xk < Xi < Xj } = · · ·. Since there are 3! = 6 such possibilities, and because they are all equi-probable, they must add up to one; i.e., P {X1 < X2 < X3 } = 1/6. #20, p. 292. In the second case, the answer is a resounding, “no.” This is because, there, P {X < Y } = 1. (Work out the details in terms of the pdf’s.) In the first case, compute marginals (do it for the second case, too): Z ∞ −x fX (x) = xe e−y dy = xe−x (x > 0) 0 Z ∞ fY (y) = e−y xe−x dx = e−y Γ(2) = e−y (y > 0). 0 3 Because fX (x)fY (y) = f (x, y), X and Y are independent. Theoretical Exercises #14, p. 297. This was worked out during lecture. 4