20065ee131A_1_EE131AF06HW7Solution

20065ee131A_1_EE131AF06HW7Solution - Solution of Assignment...

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Unformatted text preview: Solution of Assignment 7, EE 131A Due: Monday December 4, 2006 Problem 1. For the two-dimensional random variable (X, Y ) sketch the region of the plane corresponding to the following events. (a) {X − Y ≤ 1}. (b) {max(X, Y ) < 2}. (c) {|X − Y | ≤ 2}. (d) {|X | < |Y |}. (e) {X 2 ≤ Y }. (f) {max(|X |, |Y |) < 2} Solution: y y (2, 2) y 2 −2 1 −1 −2 x x 2 x y y = −x y=x y y= x2 y 2 x x −2 2x −2 Problem 2. The joint density function of X and Y is given by c(y − x)e−y , if −y < x < y and 0 < y < ∞, fX,Y (x, y ) = 0 elsewhere. (a) Find c. (b) Find the marginal pdf of X . (c) Find the marginal pdf of Y . Solution: (a) ∞ ∞ ∞ y fX,Y (x, y ) dx dy = c −∞ −∞ 0 e−y −y ∞ (y − x) dx dy = 2c 0 e−y y 2 dy = 4c. 1 Thus c = 4 . (b) For x ≥ 0, only for the values of y such that y > x the density function fX,Y (x, y ) is non-zero, hence ∞ fX (x) = −∞ fX,Y (x, y ) dy 1∞ = (y − x)e−y dy 4x 1 = − ye−y − e−y + xe−y 4 1 −x = e. 4 ∞ y =x For x ≥ 0, only for the values of y such that y > −x the density function fX,Y (x, y ) is non-zero, hence ∞ fX (x) = −∞ fX,Y (x, y ) dy = 1∞ (y − x)e−y dy 4 −x 1 = − ye−y − e−y + xe−y 4 1 = −2xe−x + ex . 4 ∞ y =−x 2 (c) ∞ fY (y ) = −∞ fX,Y (x, y ) dx 1 −y y e (y − x) dx 4 −y 1 2 −y ye . = 2 = Problem 3. The random vector (X, Y ) has a joint pdf 2e−x e−2y if x > 0 and y > 0, fX,Y (x, y ) = 0 elsewhere. Find the probabilities of the following events: (a) {X + Y ≤ 8}. (b) {X < Y }. (c) {X − Y ≤ 2}. (d) {X 2 < Y }. Solution: (a) The event {X + Y ≤ 8} corresponds to the following region in the plane: A = {(x, y ) : 0 < x ≤ 8, 0 < y ≤ 8 − x}. y 8 y = 8−x 8 x 3 Therefore, P [X + Y ≤ 8] = =2 0 8 fX,Y (x, y ) dx dy A 8 8−x e−x 0 e−2y dy dx −1 −2y e 2 8−x =2 0 8 e−x dx y =0 = 0 e−x − ex−16 dx = 1 + e−16 − 2 e−8 . (b) The event {X < Y } corresponds to the following region in the plane: B = {(x, y ) : 0 < x < ∞, x < y < ∞}. y y=x x Therefore, P [X < Y ] = =2 0 8 fX,Y (x, y ) dx dy B ∞ ∞ e−x x e−2y dy dx −1 −2y e 2 ∞ =2 0 ∞ e −x dx y =x = 0 e−3x dx 1 = . 3 (c) The event {X − Y ≤ 2} corresponds to the region C in the plane where C1 = {(x, y ) : 0 < x ≤ 2, 0 < y < ∞} and C2 = {(x, y ) : 2 < x < ∞, x − 2 ≤ y < ∞}. 4 y C1 C2 y = x−2 x 2 Therefore, P [X − Y ≤ 2] = =2 0 2 fX,Y (x, y ) dx dy + C1 2 C2 ∞ fX,Y (x, y ) dx dy ∞ ∞ e−x 0 e−2y dy dx + 2 2 ∞ e−x x −2 e−2y dy dx =− =1 e−x dx + 2 e−3x+4 dx 0 2 −3 e. −2 (d) The event {X 2 < Y } corresponds to the following region in the plane: D = {(x, y ) : 0 < x < ∞, x2 < y < ∞}. y y = x2 x 5 Therefore, P [X 2 < Y ] = =2 0 ∞ fX,Y (x, y ) dx dy D ∞ ∞ e−x x2 e−2y dy dx −1 −2y e 2 ∞ =2 0 ∞ e −x dx y =x 2 = 0 exp −2x2 − x dx ∞ 0 =e = 1 2 1 8 exp −2 x + ∞ 12 4 dx change of variable t = 2(x + 1 ) 4 e 1 8 1 2 e−t 2 /2 dt To evaluate the last integral, we consider Φ(x) the cdf of the standard normal random variable, where ∞ 1 2 e−t /2 dt. 1 − Φ(x) = √ 2π x Therefore, 1 P [X 2 < Y ] = π e 8 (1 − Φ( 1 )) ≈ 0.438 2 2 6 ...
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