Math 303
Solutions to Assignment #6
March 4, 2011
Total marks = [25].
1. Suppose that , are distinct SAWs with P, > 0. There could be several ways to get
from to . Specically, suppose that there are m possible pairs (i(j) , g (j) ), j = 1, . . . m,
such t
Math 303
Solutions to Assignment #8
March 20, 2015
Total marks = [25].
1. Given that N (t) = n, the arrival times are independent and uniform on [0, t]. The probability that an arrival time lies in [0, u] is thus u/t; call this event a success. Then N (u)
Math 303
Solutions to Assignment #3
January 30, 2015
Total marks = [25].
P nr r
1. By the ChapmanKolmogorov equation, Pijn = k Pik
Pkj . By assumption, Pr has only
positive entries. Let c > 0 be the value of the smallest one. Then for any states i, j,
X
X
Math 303
Solutions to Assignment #5
February 27, 2015
Total marks = [25].
1. G(s) = 21 + 12 s3 and the extinction probability is the smallest nonnegative root of 21 + 12 s3
s = 0. One rootis s = 1 and long division gives (s 1)(s2 + s 1) = 0, so
the othe
Math 303
Solutions to Assignment #7
March 13, 2015
Total marks = [25].
1. (a) N (5) is Poisson(5) so its mean is 5.
[1]
(b) S3 is Gamma(3, ) so its mean is 3/.
[1]
(c) P (N (5) < 3) = e
5
2
(1 + 5 + 25 /2).
[1]
(d) If you like doing integrals you can inte
Math 303
Solutions to Assignment #6
March 6, 2015.
Total marks = [25].
1. Suppose that , 0 are distinct SAWs with P,0 > 0. There could be several ways to get
from to 0 . Specifically, suppose that there are m possible pairs (i(j) , g (j) ), j = 1, . . . m
Math 303
Solutions to Assignment #1
January 11, 2011
Total marks = [25].
1. (a) By taking into account each of the 8 possibilities for Garys non-glum state on the
following three days, we obtain the result:
[4]
PCC PCC PCC + PCC PCC PCS + PCC PCS PSC + PC
Math 303
Solutions to Assignment #3
January 28, 2011
Total marks = [25].
1. The stationary distribution is the solution = (1 , 2 , 3 ) to P = P with 1 +2 +3 = 1,
where P is the given transition matrix. Thus we solve
1 = .71 + .22 + .13
2 = .21 + .62 + .43
Math 303
Solutions to Assignment #5
February 25, 2011
Total marks = [25].
1. (a) Given Xn , Xn+1 is Bin(m, Xn /m) and hence has expected value Xn (conditional on
Xn ). Thus EXn+1 = EXn for all n, and since X0 = i, EXn = i for all n.
[2]
(b) For i = 1, . .
Math 303
Solutions to Assignment #9
April 1, 2011
Total marks = [25].
1. Suppose the state is (n, m). Male i mates at rate with female j, so mates at rate m.
There are n males, so mating occurs at rate mn. Thus v(n,m) = mn. Each mating is
1
equally likely
Math 303
Solutions to Assignment #8
March 18, 2011
Total marks = [25].
1. This is the gamblers ruin problem, with initial fortune k and total fortune 2k, and with
win probability 1 /(1 + 2 ) (from the perspective of team 1). The parameter a = (1 p)/p
is t
Math 303
Solutions to Test #2
March 23, 2011
Total marks = [30].
1. The ospring distribution has generating function
G(s) =
[2]
1 3 2
+ s
4 4
and the extinction probability is the smallest non-negative root of s = G(s), i.e.,
s=
1 3 2
+ s ,
4 4
or
3s2 4s