Read ST3236: Stochastic Process Tutorial - solution 10 text version

solution 10

ST3236: Stochastic Process Tutorial

solution 10 Pang Zhen

Department of Statistics and Applied Probability

[email protected]

solution 10 Question 1

The 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 1

Then

E(W5 |X (t) = 4) =

t

sfW5 (s)ds se-(s-t) ds (s + t)e-s ds

=

t

=

0

= t+

1

solution 10 Question 1

The conditional distribution of W1 , W2 , W3 , W4 is 4!/t 4 , therefore w3 4!/t 4 dw1 dw2 dw3 dw4

0<w1 <w2 <w3 <w4 <t w4 w3 w2

E(W3 |X (t) = 4) = =

0

···

t

(

0

(

0

(

0

w3 4!/t 4 dw1 )dw2 )dw3 )dw4

=

3 t 5

solution 10 Question 2

The probability we are interested is

n

P{Xi (t) 1, for i = 1, . . . , n} =

i=1 n

P{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 3

Let 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 Then

t

E(W1 ) =

0

s n s (1 - )n-1 ds t t

1

= t

0

sn(1 - s)n-1 ds

=

t n+1

solution 10 Question 3

Similarly, 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+1

t

E(Wn ) = = =

solution 10 Question 4

Let 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 5a

Let 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 5b

We can get E(W6 |X (t) = 5) = t + Question 1. Then

1

using the same method as

E{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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ST3236: Stochastic Process Tutorial - solution 10

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