# G x e y x and e 1 z 1 so gx e 1 z e g x xz e e y x xz

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g ( X ) = E [ Y | X ] and E [1 | Z ] = 1 So, g(X) E [1 | Z ] = E [ g ( x ) | X,Z ] = E [ E [ Y | X ] | X,Z ] The Pull-Through and Tower Properties are not limited to discrete random variables. These properties are also true in the continuous case. We can prove this by using the same approach we used for the discrete case. 5. (a) Since E [ X ] = 0, We have E [ E [ X | Y ]] = E [ X ] = 0. Hence cov( X,E [ X | Y ]) = E [ XE [ X | Y ]] = E [ E [ XE [ X | Y ] | Y ]] = E [( E [ X | Y ]) 2 ] 0 . Page 2 of 8

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Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006) (b) We can actually prove a stronger statement than what is asked for in the problem, namely that both pairs of random variables have the same covariance (from which it immediately follows that their correlation coefficients have the same sign. We have cov( Y,E [ X | Y ]) = E [ Y E [ X | Y ]] = E [ E [ XY | Y ]] = E [ XY ] = cov( X,Y ) . 6. (a) The transform M J ( s ) given is a transform of a binomial random variable with parameters n = 10 and p = 2 3 . Thus the PMF for J is: p J ( j ) = parenleftBigg n j parenrightBigg ( 1 3 ) n - j ( 2 3 ) j for j = 0 , 1 , 2 ,... 10 (b) Again by inspection, K is a geometric random variable shifted to the right by 3 with parameter p = 1 5 . This is because we can rewrite M K ( s ) = e 3 s 1 5 e s 1 - 4 5 e s . Thus, p K ( k ) = ( 4 5 ) k - 4 1 5 for k = 4 , 5 , 6 , ... E [ K ] = 3 + 1 p = 3 + 5 = 8 Var( K ) = 1 p p 2 = 4 5 1 25 = 20 (c) Note that L = K 1 + K 2 + ...K J , thus L is a random sum of random variables. So, determining the transform of L is easier than determining the PMF for L . M L ( s ) = M J ( s ) | e s = M K ( s ) = ( 1 3 + 2 3 ( 1 5 e 4 s 1 4 5 e s )) 10 The expectation of L is E [ L ] = E [ K ] E [ J ] = 8 20 3 = 160 3 The variance of L is Var( L ) = Var( K ) E [ J ] + Var( J )( E [ K ]) 2 = (20)(10 2 3 ) + (10 2 3 1 3 )(64) = 2480 9 (d) P (person donates) = 1 4 . Let M = total # of donors from all living groups, and define X i = braceleftBigg 1 if ith person donates 0 otherwise. The PMF for X is just p X ( x ) = braceleftBigg 1 4 if x = 1 3 4 if x = 0. Then, M = X 1 + X 2 + ...X L . Therefore the transform of M is: Page 3 of 8
Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006) M M ( s ) = M L ( s ) | e s = M X ( s ) The transform of X is (by inspection) M X ( s ) = ( 3 4 + 1 4 e s ) Therefore, M M ( s ) = ( 1 3 + 2 3 ( 1 5 ( 3 4 + 1 4 e s ) 4 1 4 5 ( 3 4 + 1 4 e s ) )) 10 To obtain P ( M = 0), we simply evaluate the transform of M at e s = 0. p M (0) = M M ( s ) | e s =0 = ( 1 3 + 2 3 ( 1 5 ( 3 4 ) 4 1 4 5 ( 3 4 ) )) 10 . The expectation of M is E [ M ] = E [ X ] E [ L ] = 40 3 The variance of M is Var( M ) = Var( X ) E [ L ] + Var( L )( E [ X ]) 2 = 27 . 22 7. (a) Let the random variable T represent the number of widgets in 1 crate and let K i represent the number of widgets in the i th carton. T = K 1 + K 2 + ... + K N The transform of T is obtained by substituting the transform of N for every value of e s in the transform of K . M T ( s ) = M N ( s ) | e s = M K ( s ) M T ( s ) = (1 p ) e μ ( e s - 1) 1 pe μ ( e s - 1) . Since T is a non-negative discrete random variable, P ( T = 1) = d de s M T ( s ) | e s =0 = μ (1 p ) e - μ (1 pe - μ ) + μp (1 p ) e - 2 μ (1 pe - μ ) 2 .

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• Spring '06
• Munther Dahleh
• Computer Science, Electrical Engineering, Probability theory, Massachusetts Institute of Technology, Probabilistic Systems Analysis, Department of Electrical Engineering & Computer Science

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