final_solutions

final_solutions - EE 505 B Fall 2011 Final Exam Solutions...

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Unformatted text preview: EE 505 B Fall 2011 Final Exam Solutions 1. [35 points] (a) [10 points] We use the formula f X (x) = f U (u) d du g(u) u = u i ∑ i x There is only one solution u1 that sat1 isfies the equation x = − λ ln(u): x u1 = e − λ x e−λx Substituting u1 into the pdf: f U ( u1 ) = = = 0 u1 < 0 or u1 > 1 1 0 ≤ u1 ≤ 1 0 e−λx < 0 or e−λx > 1 1 0 ≤ e−λx ≤ 1 0 x<0 1 x≥0 . The denominator is d g (u) du = u = u1 1 λu = u=e−λ x 1 λx e λ Finally, f X (x) = 0 x<0 λe − λ x x≥0 X is an exponentially distributed random variable. 1 . 1 u (b) [10 points] There are two ways to solve this problem. Your choice depends on how confident you are about your answer to part (a). The first method uses the probability density function from part (a). For λ ≥ 0, Pr − 1 ln(U ) ≤ λ λ = Pr(X ≤ λ) λ = 0 λe−λx dx 1 = λ − e−λx λ = 1−e λ 0 − λ2 For λ < 0, Pr − 1 ln(U ) ≤ λ λ =0 Therefore, Pr − 1 ln(U ) ≤ λ λ λ<0 0 = 1−e − λ2 λ≥0 The second method uses the probability density function for U . Pr − 1 ln(U ) ≤ λ λ = Pr − ln(U ) ≤ λ2 = Pr ln(U ) ≥ −λ2 = Pr U ≥ e−λ 2 2 For e−λ > 1 or λ < 0, Pr U ≥ e−λ 2 =0 2 For 0 < e−λ ≤ 1 or λ ≥ 0, Pr U ≥ e−λ 2 = 1 2 e−λ du = 1 − e−λ 2 2 (e−λ can never be less than 0.) Therefore, 1 Pr − ln(U ) ≤ λ λ 2 = λ<0 0 1−e − λ2 λ≥0 (c) [15 points] We can write down an expression for the conditional probability density function f X |{Λ=λ} ( x ): 0 x<0 λe − λ x f X |{Λ=λ} ( x ) = x≥0 We then find the probability density function of X using the principle of total probability: ∞ f X (x) = −∞ f X | Λ= λ ( x ) f Λ (λ) dλ For x < 0, f X (x) = 0 For x ≥ 0, 1 f X (x) = 0 λe − λ x dλ Integrating by parts with 1 v = − e−λx x dv = e−λx dλ u=λ du = dλ λ f X ( x ) = − e−λx x = = = = 1 + 0 1 0 1 −λx e dλ x 1 1 1 −x − e−λx − e −0 + x x x 1 1 − e− x − 2 e− x − 1 x x − xe− x − e− x + 1 x2 − x ( x + 1) 1−e x2 Therefore, 0 3 x<0 1− e − x ( x + 1) x2 f X (x) = x≥0 1 0 2. [40 points] (a) [15 points] We start by finding the margin probability density functions. For x ≤ 0, f X (x) = 0 y For x > 0, ∞ f X (x) = = −∞ ∞ f XY ( x, y) dy x 2e− x e−y dy x = 2e − x e − y x x x ∞ = 2e −2 x f X (x) = 0 x≤0 2e −2 x x>0 For y ≤ 0, f Y (y) = 0 y For y > 0, ∞ f Y (y) = = −∞ y 0 f XY ( x, y) dx y 2e− x e−y dx = 2e − y e − y y 0 y −y x = 2e − y 1 − e f Y (y) = 0 y≤0 2e − y ( 1 − e − y ) y>0 Therefore X and Y are not independent, because for x > 0 and y > 0, f X ( x ) f Y (y) = 4e−2x−y 1 − e−y = f XY ( x, y) 4 (b) [10 points] The definition of the conditional probability density function is f X |Y ( x ) = f XY ( x, y) f Y (y) For y > 0, f X |Y ( x ) = = 0 x < 0 or x ≥ y 2e − x − y 2e − y (1− e − y ) 0<x<y 0 x < 0 or x ≥ y e− x 1− e − y 0<x<y For y ≤ 0, f X |Y ( x ) is undefined f X |Y ( x ) 1 1− e − y e−y 1− e − y y x Figure 1: The conditional probability density function f X |Y ( x ) for 0 < x < y. 5 (c) [15 points] Cov{ X , Y } = E{ XY } − E{ X } E{Y } ∞ ∞ E{ XY } = = −∞ −∞ ∞ ∞ x 0 ∞ = 0 ∞ = = 0 0 ∞ xy f XY ( x, y) dy dx y 2 xye− x e−y dy dx 2 xe− x ∞ ye−y dy x x dx x 2 xe− x e− x ( x + 1) d x x2e−2x dx + ∞ 0 x x2 2e−2x dx The first term is simply the expected value of an exponentially distributed random variables with λ = 2, and the second term is the second moment of and exponentially random variable with λ = 2. Therefore, E{ XY } = E{ X } λ=2 + E{ X 2 } λ=2 12 =+ 24 =1 We need values for E{ X } and E{Y }. X is a exponentially distributed random variable with λ = 2, so E{ X } = 1 2 f Y ( y ) = 2 e − y 1 − e − y = 2 e − y − 2 e − 2y y E {Y } = 0 =2 = y2e−y − y2e−2y dy y 0 = 2− y>0 ye−y dy − y 0 y2e−2y dy 1 2 3 2 Therefore, Cov{ X , Y } = 1 − 6 1 2 3 2 = 1 4 3. [25 points] We use f XY ( x, y) | J ( x, y)| f WZ (w, z) = x = h1 ( w , z ) y = h2 ( w , z ) w = g1 ( x , y ) = 2 x − y z = g2 ( x , y ) = x − 2 y 2 w − z = 2 (2 x − y ) − ( x − 2 y ) = 3 x w − 2 z = (2 x − y ) − 2 ( x − 2 y ) = 3 y 2w − z 3 w − 2z y = h2 ( w, z ) = 3 → → ∂ g1 ∂x ∂ g2 ∂x J ( x, y) = det = det x = h1 ( w, z ) = ∂ g1 ∂y ∂ g2 ∂y 2 −1 1 −2 = −3 therefore | J ( x, y)| = 3. Therefore, f WZ (w, z) = 2 e − x − 2y 3 − x = 2w3 z w −2z y= 3 2w −4z 2w − z 2 = e−( 3 )−( 3 ) 3 2w − z +2w −4z 2 ) 3 = e−( 3 4w −5z 2 = e−( 3 ) 3 For this expression to be complete, we need the ranges for w and z. The joint probability distribution function for X and Y is non zero for { ( x , y ) : { x ≥ 0} ∩ { y ≥ 0} Substituting the expressions for x = w+z and y = w−z , we can find the ranges of w 2 2 and z for the non zero part of the joint probability distribution for W and Z. w − 2z ≥0 3 w = ( w, z ) : { z ≤ 2w } ∩ z ≤ 2 {( x, y) : { x ≥ 0} ∩ {y ≥ 0}} = ( w, z ) : 2w − z ≥0 ∩ 3 This set of (w, z) values is plotted in the figure. 7 z w w 2 z= 2w z= Putting everything together, 2 − ( 4w −5z ) e 3 f WZ (w, z) = 3 0 8 w < 0 and z ≤ 2w or w w ≥ 0 and z ≤ 2 otherwise ...
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This note was uploaded on 03/01/2012 for the course EE 101 taught by Professor Munk during the Spring '12 term at Alaska Anch.

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