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(MATH246)[2003](f)final~PPSpider^sol_10491.pdf
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MATH 246 Probability and Random Processes
Solution to Final Examination
Fall 2003 Course Instructor: Prof. Y. K. Kwok

Time allowed: 100 minutes
1. (a) Stationary increments
The increments of the Poisson process N(t) over two time intervals of equal length have the same probability distribution, independent of the starting time of the interval. That is,
P [N(t1 + ) . N(t1)] = P [N(t2 + ) . N(t2)]
for any t1,t2 and .
Independent increments
The increments of the Poisson process over any two non-overlapping time intervals are independent.
(b) (i) Assume t1 <t2,
CN (t1,t2)= E[(N(t1) . t1)(N(t2) . t2)] = E[(N(t1) . t1){[N(t2) . N(t1) . (t2 . t1)] + (N(t1) . t1)}] = E[N(t1) . t1]E[N(t2) . N(t1) . (t2 . t1)] + E[{N(t1) . (t1)}2] = var(N(t1)) = min(t1,t2).
(ii) P (N(t1)=1|N(t2) = 1]
P [N(t1)=1,N(t2) . N(t1) = 0]

=
P [N(t2) = 1]
P [N(t1) = 1] P [N(t2 . t1) = 0]

= P [N(t2) = 1]
.t1 e.(t2.t1)
t1et1
== .
.t2
t2et2
.Yn + Yn.1 .
2. (a) E[Xn]= E =+= .
2 22
(b) Sum of two independent Poisson random variables remain to be Poisson so that Yn + Yn.1 is a Poisson random variable with parameter 2. Now
(2)2k 13
.2
P [Xn = k]= P [Yn + Yn.1 =2k]= e ,k =0,, 1,,
(2k)! 22
1
(c) RX (i, j)= E[(Yi + Yi.1)(Yj + Yj.1)]
4
1
= {E[YiYj ]+ E[YiYj.1]+ E[Yi.1Yj]+ E[Yi.1Yj.1]}
4
1
. var(Yi)+ E[Yi]2 if i = jNote that E[YiYj ]= E[Yi]E[Yj ] if i =.j
. + 2 if i = j
= 2 .
if i .= j
(i) When i = j
1

RX (i, i)= {E[Y 2]+2E[YiYi.1]+ E[Y 2 ]}
ii.1
4
1 (2 + 1)
=[ + 2 +22 + + 2)= .
42
(ii) When i = j +1
1

RX (i, i . 1) = {E[YiYi.1]+ E[YiYi.2]+ E[Y 2 ]+ E[Yi.1Yi.2]}
4 i.1
1 (4 + 1)
= (32 + + 2)= .
44 (4 + 1)
(iii) When i = j . 1, we also obtain RX (i, i +1)= .
4
(iv) When i ...
= j and |i . j| = 1, we have RX (i, j)= 2
3. (a) A random process is stationary if X(t) and X(t + ) have the same statistics for any .
A random process is wide sense stationary if
mX (t)= m for all t
CX (t1,t2)= CX (t1 . t2) for all t1 and t2.

(b) mX (t)= E[U sin 0t] = sin 0tE[U]=0
CX (t1,t2)= E[U2 sin 0t1 sin 0t2] . mX (t)2 = sin 0t1 sin 0t2(Var(U)+ E[U]2) = sin 0t1 sin 0t2.
X(t) is not wide sense stationary since CX (t1,t2) is not a function of t2 . t1.
4. (a)
.. ..
(1 . )2 2(1 . ) 2
(1 . ) + (1 . )(1 . ) (1 . )
2 2(1 . ) (1 . )2

P =
P [X1 =0,X0 = 1] = P [X1 =0|X0 = 1]P [X0 = 1] = (1 . )/2.
(b)
is obtained by solving
=
P.
2 2 2
To verify that
, consider
=
( + )2 ( + )2 ( + )2
..
(1 . )2 2(1 . ) 2
(1 . ) + (1 . )(1 . ) (1 . )
2 2(1 . ) (1 . )2

2 2 2
LHS =
( + )2 ( + )2 ( + )2
=
. .
. . .
.
T
2(1.)2+22(1.)+22 (+)2
22(1.)+2[+(1.)(1.)]+22(1.)
(+