#### Read ST3236: Stochastic Process Tutorial - solution 10 text version

`solution 10ST3236: Stochastic Process Tutorialsolution 10 Pang ZhenDepartment of Statistics and Applied Probability[email protected]solution 10 Question 1The conditional distribution of W5 is P(W5  s|X (t) = 4) = P(X (s)  5|X (t) = 4) = P(X (s) - X (t)  1|X (t) = 4) = P(X (s) - X (t)  1) = 1 - P(X (s) - X (t) = 0) = 1 - e-(s-t) The density function of W5 is fW5 (s) = e-(s-t)solution 10 Question 1ThenE(W5 |X (t) = 4) =t sfW5 (s)ds se-(s-t) ds (s + t)e-s ds=t =0= t+1 solution 10 Question 1The conditional distribution of W1 , W2 , W3 , W4 is 4!/t 4 , therefore w3 4!/t 4 dw1 dw2 dw3 dw40&lt;w1 &lt;w2 &lt;w3 &lt;w4 &lt;t w4 w3 w2E(W3 |X (t) = 4) = =0···t(0(0(0w3 4!/t 4 dw1 )dw2 )dw3 )dw4=3 t 5solution 10 Question 2The probability we are interested isnP{Xi (t)  1, for i = 1, . . . , n} =i=1 nP{Xi (t)  1} (1 - P{Xi (t) = 0})i=1 n==i=1(1 - e-t )= (1 - e-t )n Therefore, F (t) = 1 - (1 - e-t )n .solution 10 Question 3Let Y1 , Y2 , . . . , Yn be IID uniformly distributed on [0,t]. P{W1  s} = P{min(Y1 , Y2 , . . . , Yn )  s} = 1 - P{min(Y1 , Y2 , . . . , Yn )  s} = 1 - P{Y1  s}P{Y2  s} . . . P{Yn  s} s = 1 - (1 - )n t Therefore, fW1 (s) = n (1 - s )n-1 , for 0  s  t. t t ThentE(W1 ) =0s n s (1 - )n-1 ds t t1= t0sn(1 - s)n-1 ds=t n+1solution 10 Question 3Similarly, P{Wn  s} = P{max(Y1 , Y2 , . . . , Yn )  s} = P{Y1  s}P{Y2  s} . . . P{Yn  s} s = ( )n t Therefore, fW1 (s) = n ( s )n-1 , for 0  s  t. t t Then n s s ( )n-1 ds t t 0 t n n s ds n 0 t n t n+1tE(Wn ) = = =solution 10 Question 4Let Y1 , Y2 be IID uniformly distributed on [0,t], then t E(Y1 ) = E(Y2 ) = 2 . 2 Therefore, E(W1 W2 ) = E(Y(1) Y(2) ) = E(Y1 Y2 ) = t4solution 10 Question 5 5aLet Y1 , Y2 , . . . , Y5 be IID uniformly distributed on [0,t], then t E(Y1 ) = E(Y2 ) = · · · = E(Y5 ) = 2 . Then E{W1 + W2 + · · · + W5 |X (t) = 5} = E{Y(1) + Y(2) + · · · + Y(5) } = E{Y1 + Y2 + · · · + Y5 } 5t = 2solution 10 Question 5 5bWe can get E(W6 |X (t) = 5) = t + Question 1. Then1 using the same method asE{W1 + W2 + · · · + W6 |X (t) = 5} = E{W1 + W2 + · · · + W5 |X (t) = 5} + E(W6 |X (t) = 5) 5t 1 = +t + 2  1 7t + = 2 `

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