# Thus we have m x s m n s e s m t i s parenleftbigg 1

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Thus we have, M X ( s ) = M N ( s ) | e s = M T i ( s ) = parenleftBigg 1 3 + 2 3 3 e s 6 ( s 3) parenrightBigg 12 (c) By the law of iterated expectations, E [ N ] = E [ E [ N | P ]]. We can compute E [ N | P ] from the fact that N is a binomial with parameter P , where P is a random variable uniformly distributed between 0 and 1. Thus E [ N ] = E [12 P ] = 12 E [ P ] = 12 1 2 = 6 We compute the var ( N ) using the law of conditional variance: var ( N ) = E [ var ( N | P )]+ var ( E [ N | P ]) = E [12 P (1 P )] + var (12 P ) Page 6 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) = 12 E [ P (1 P )] + 144 var ( P ) = 12(1 / 2 1 / 3) + 12 = 14. (d) M N ( s ) = E [ E [ e sN | P ]] = E [ M P ( s )] = E [1 P + Pe s ] = 1 E [ P ]+ e s E [ P ] = 1 1 2 + 1 2 e s = 1 2 + 1 2 e s 10. Let A t (respectively, B t ) be a Bernoulli random variable which is equal to 1 if and only if the t th toss resulted in 1 (respectively, 2). We have E [ A t B t ] = 0 and E [ A t B s ] = E [ A t ] E [ B s ] = p 1 p 2 for s negationslash = t . We have E [ X 1 X 2 ] = E [( A 1 + · · · + A n )( B 1 + · · · + B n )] = n E [ A 1 ( B 1 + · · · + B n )] = n ( n 1) p 1 p 2 , and cov( X 1 ,X 2 ) = E [ X 1 X 2 ] E [ X 1 ] E [ X 2 ] = n ( n 1) p 1 p 2 np 1 np 2 = np 1 p 2 . 11. (a) Here it is easier to find the PDF of Y . Since Y is the sum of independent Gaussian random variables, Y is Gaussian with mean 2 μ and variance 2 σ 2 X + σ 2 Z . (b) i. The transform of N is M N ( s ) = 1 11 (1 + e s + e 2 s + · · · + e 10 s ) = 1 11 10 summationdisplay k =0 e ks Since Y is the sum of a random sum of Gaussian random variables an independent Gaussian random variable, M Y ( s ) = parenleftbigg M N ( s ) | e s = M X ( s ) parenrightbigg M Z ( s ) = parenleftbigg 1 11 10 summationdisplay k =0 ( e + s 2 σ 2 X 2 ) k parenrightbigg e s 2 σ 2 Z 2 = parenleftbigg 1 11 10 summationdisplay k =0 e skμ + s 2 2 X 2 parenrightbigg e s 2 σ 2 Z 2 = 1 11 10 summationdisplay k =0 e skμ + s 2 ( 2 X + σ 2 Z ) 2 In general, this is not the transform of a Gaussian random variable. ii. One can differentiate the transform to get the moments, but it is easier to use the laws of iterated expectation and conditional variance: E Y = E X E N + E Z = 5 μ var( Y ) = E N var( X ) + ( E X 2 )var( N ) + var( Z ) = 5 σ 2 X + 10 μ 2 + σ 2 Z iii. Now, the new transform for N is M N ( s ) = 1 9 ( e 2 s + · · · + e 10 s ) = 1 9 10 summationdisplay k =2 e ks Page 7 of 8
Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006) Therefore, M Y ( s ) = parenleftbigg M N ( s ) | e s = M X ( s ) parenrightbigg M Z ( s ) = parenleftbigg 1 9 10 summationdisplay k =2 ( e + s 2 σ 2 X 2 ) k parenrightbigg e s 2 σ 2 Z 2 = parenleftbigg 1 9 10 summationdisplay k =2 e skμ + s 2 2 X 2 parenrightbigg e s 2 σ 2 Z 2 = 1 9 10 summationdisplay k =2 e skμ + s 2 ( 2 X + σ 2 Z ) 2 (c) Given Y , the linear least-squared estimator of X k is given by ˆ X k = E X k + cov( X k ,Y ) var( Y ) ( Y E Y ) = μ + cov( X k ,Y ) var( Y ) ( Y E Y ) . To determine the mean and variance of Y we first determine those of N : E N = parenleftbigg 1 4 10 + 3 4 5 parenrightbigg = 25 4 var( N ) = E var( N | timeofday ) + var( E N | timeofday ) = 10 + 75 16 = 235 16 Now E Y = EE Y | N = E N E X + E Z = E N E X = 25 4 μ var( Y ) = E N var( X ) + ( E X 2 )var( N ) + var( Z ) = 25 4 σ 2 X + 235 16 μ 2 + σ 2 Z cov( X k ,Y ) = E ( X k μ )( Y 25 μ/ 4) = EE ( X k μ )( Y 25 μ/ 4) | N Since E ( X k μ )( Y 25 μ/ 4) | N = braceleftBigg σ 2 X if N k, 0 otherwise then cov( X k ,Y ) = σ 2 X P ( N k ) = σ 2 X braceleftBigg 0 . 25 k 10 k e - 10 k !
• 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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