HW10sol - 1 ,..,X n )]= P[min 1 (X 1 , .., X n )...

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EE/NIS 605-CS505 Probability and Stochastic Processes I Homework 10 Solutions Professor Heffes Textbook: S. Ross, A First Course in Probability, Seventh Edition , Prentice Hall, 2006 Chapter 7 : Problems: 26,39,58,64; Theoretical Exercises: 1,22 P39 : For j=0; cov(Y n , Y n )=Var(Y n )=3 σ 2 For j=1; cov(Y n , Y n+1 )=cov(X n +X n+1 +X n+2 , X n+1 + X n+2 +X n+3 ) = cov(X n+1 +X n+2 , X n+1 + X n+2 )= Var(X n+1 +X n+2 ) = 2 σ 2 . For j=2; cov(Y n , Y n+2 )=cov(X n+2 , X n+2 ) = σ 2 . For j> 3; cov(Y n , Y n+j )=0 P58 : Let X= # flips required. Condition on first flip. a) E[X]=E[X|heads first]p + E[X|tails first](1-p) = [1+1/(1-p)]p + [1+1/p](1-p) = p/(1-p) + 1/p b) P[heads last] = P[tails first] = 1-p P64 : a) E[X] = E[X|type 1]p + E[X|type 2](1-p) = p μ 1 + (1-p) μ 2 . b) E[X 2 ] = E[X 2 |type 1]p + E[X 2 |type 2](1-p) = {Var[X 2 |type 1]+ E 2 [X|type 1]}p + {Var[X 2 |type 2]+ E 2 [X|type 2]}(1-p) = σ 1 2 p + μ 1 2 p + σ 2 2 (1-p) + μ 2 2 (1-p). Var(X) = E[X 2 ] – E 2 [X]= σ 1 2 p + μ 1 2 p + σ 2 2 (1-p) + μ 2 2 (1-p) – [p μ 1 + (1-p) μ 2 ] 2 . (b) E[min (X
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Unformatted text preview: 1 ,..,X n )]= P[min 1 (X 1 , .., X n ) > t]dt = = P[ 1 X 1 > t] ... P[X n > t]dt = (1 t) n dt = 1 n + 1 1 P 26 : (a) E[max (X 1 ,..,X n )]= P[max 1 (X 1 ,..,X n ) > t]dt= {1 P[max 1 (X 1 ,...,X n ) t]}dt = {1 P[ 1 X 1 t] ... P[X n t]}dt = {1 t n }dt = n n + 1 1 TE1 : Let = E[X]. Then E[(X-a) 2 ]= E[(X- + -a) 2 ]= E[(X- ) 2 ]+ (-a) 2 + 2E[(X-29 ( -a)] = E[(X- ) 2 ]+ (-a) 2 + 2E[(X-29] ( -a) = E[(X- ) 2 ]+ (-a) 2 ; minimized when a= TE22 : cov(X,Y) = cov(X, a + bX)= cov(X, bX) + cov(X,a) = cov(X, bX) + 0 = bVar(X) Var(Y) = b 2 Var(X) Thus = +1 ; if b>0, and = -1; if b< 0. Professor Heffes Also from d da E[(X a) 2 ] = 2E[X a] = 0 a = E[X] = (X,Y) = cov(X,Y) X Y = bVar(X) Var(X) b 2 Var(x) = b b...
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HW10sol - 1 ,..,X n )]= P[min 1 (X 1 , .., X n )...

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