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Unformatted text preview: Probability and Stochastic Processes Problem set 7 Solutions March 26, 2010 Problem Solutions : Yates and Goodman, 3.8.1 3.8.3 3.8.4 3.8.5 3.8.6 3.8.8 4.1.1 4.1.3 4.2.2 4.2.6 4.3.2 4.3.4 4.4.1 and 4.4.2 Problem 3.8.1 Solution The PDF of X is f X ( x ) = braceleftbigg 1 / 10 5 ≤ x ≤ 5 otherwise (1) (a) The event B has probability P [ B ] = P [ 3 ≤ X ≤ 3] = integraldisplay 3 3 1 10 dx = 3 5 (2) From Definition 3.15, the conditional PDF of X given B is f X  B ( x ) = braceleftbigg f X ( x ) /P [ B ] x ∈ B otherwise = braceleftbigg 1 / 6  x  ≤ 3 otherwise (3) (b) Given B , we see that X has a uniform PDF over [ a,b ] with a = 3 and b = 3. From Theorem 3.6, the conditional expected value of X is E [ X  B ] = ( a + b ) / 2 = 0. (c) From Theorem 3.6, the conditional variance of X is Var[ X  B ] = ( b a ) 2 / 12 = 3. Problem 3.8.3 Solution The condition right side of the circle is R = [0 , 1 / 2]. Using the PDF in Example 3.5, we have P [ R ] = integraldisplay 1 / 2 f Y ( y ) dy = integraldisplay 1 / 2 3 y 2 dy = 1 / 8 (1) Therefore, the conditional PDF of Y given event R is f Y  R ( y ) = braceleftbigg 24 y 2 ≤ y ≤ 1 / 2 otherwise (2) The conditional expected value and mean square value are E [ Y  R ] = integraldisplay ∞∞ yf Y  R ( y ) dy = integraldisplay 1 / 2 24 y 3 dy = 3 / 8 meter (3) E bracketleftbig Y 2  R bracketrightbig = integraldisplay ∞∞ y 2 f Y  R ( y ) dy = integraldisplay 1 / 2 24 y 4 dy = 3 / 20 m 2 (4) The conditional variance is Var [ Y  R ] = E bracketleftbig Y 2  R bracketrightbig ( E [ Y  R ]) 2 = 3 20 parenleftbigg 3 8 parenrightbigg 2 = 3 / 320 m 2 (5) The conditional standard deviation is σ Y  R = radicalbig Var[ Y  R ] = 0 . 0968 meters. 1 Problem 3.8.4 Solution From Definition 3.8, the PDF of W is f W ( w ) = 1 √ 32 π e w 2 / 32 (1) (a) Since W has expected value μ = 0, f W ( w ) is symmetric about w = 0. Hence P [ C ] = P [ W > 0] = 1 / 2. From Definition 3.15, the conditional PDF of W given C is f W  C ( w ) = braceleftbigg f W ( w ) /P [ C ] w ∈ C otherwise = braceleftbigg 2 e w 2 / 32 / √ 32 π w > otherwise (2) (b) The conditional expected value of W given C is E [ W  C ] = integraldisplay ∞∞ wf W  C ( w ) dw = 2 4 √ 2 π integraldisplay ∞ we w 2 / 32 dw (3) Making the substitution v = w 2 / 32, we obtain E [ W  C ] = 32 √ 32 π integraldisplay ∞ e v dv = 32 √ 32 π (4) (c) The conditional second moment of W is E bracketleftbig W 2  C bracketrightbig = integraldisplay ∞∞ w 2 f W  C ( w ) dw = 2 integraldisplay ∞ w 2 f W ( w ) dw (5) We observe that w 2 f W ( w ) is an even function. Hence E bracketleftbig W 2  C bracketrightbig = 2 integraldisplay ∞ w 2 f W ( w ) dw (6) = integraldisplay ∞∞ w 2 f W ( w ) dw = E bracketleftbig W 2 bracketrightbig = σ 2 = 16 (7) Lastly, the conditional variance of W given C is Var[ W  C ] = E bracketleftbig W 2  C bracketrightbig ( E [ W  C ]) 2 = 16 32 /π = 5 . 81 (8) Problem 3.8.5 SolutionProblem 3....
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This note was uploaded on 02/13/2011 for the course 332 226 taught by Professor Staff during the Spring '08 term at Rutgers.
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