{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

20065ee131A_1_EE131AF06HW7Solution

# 20065ee131A_1_EE131AF06HW7Solution - Solution of Assignment...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}