Download MTH5121 Probability Models Exercise Sheet 2: Solutions

MTH5121 Probability Models
Exercise Sheet 2: Solutions
1. Suppose that the sequence xn satisfies the difference equation xn+1 =
xn + xn−1 .
a) Using results from the course find the general solution for xn .
b) Find the particular solution when x0 = 0 and x1 = 1.
Solution:
a) We rearrange the equation to xn+1 − xn − xn−1 = 0. Now we solve
the auxiliary equation√θ2 − θ −√1 = 0. This has solutions (by the
quadratic formula) 1+2 5 and 1−2 5 . Hence by Lemma 6 (see Notes)
the general solution is
√ !n
√ !n
1+ 5
1− 5
xn = A
+B
.
2
2
b) To find the particular solution we solve for x0 and x1 . From x0 we
have
√ !0
√ !0
1− 5
1+ 5
+B
= A + B.
0 = x0 = A
2
2
so B = −A. From x1 we have
√ !1
√ !1
1+ 5
1− 5
1 = x1 = A
+B
2
2
!
√
√
1+ 5 1− 5
=A
−
2
2
√
= A 5.
√
√
Hence A = 1/ 5 and B = −1 5 so
√ !n
1+ 5
1
1
xn = √
−√
2
5
5
since B = −A
√ !n
1− 5
.
2
2. The roulette wheel at a casino has integers from 1 to 36, together with 0.
Half of the non-zero numbers are red, the other half are black, and 0 is
green. Any of the numbers between 0 and 36 is equally likely to occur
each time the wheel is spun.
Fred has £100 to gamble on roulette at the casino. Each time he plays if
red comes up he wins £1 and otherwise he loses £1. He decides to play
until he loses all his money or has tripled his money to £300.
a) Find the probability that Fred triples his money.
b) Suppose that instead Fred decided to play for bigger stakes: on each
game he either wins £10 or loses £10 (but still starting with £100).
What is the probability of him tripling his money this time?
c) Now, suppose that instead Fred went to a different casino with a
‘fair’ roulette wheel: i.e., a roulette wheel without a 0 (so it has 36
numbers 18 of which are red and 18 of which are black.) What is
his chance of tripling his money when playing for the original £1
stake in this casino?
d) Finally what is his chance of tripling his money if he plays for the
larger £10 stake in this fair casino?
Hint. Read the part of the notes which explains the relation between the
Gambler’s ruin problem and the random walk on an interval.
Solution:
a) The amount of money Fred has follows a random walk with probabilities q = 19/37 and p = 18/37. We want to know the probability
that the walk reaches 300 before zero starting from 100. The formula
from the notes is
n
q
−1
p
q
rn (M, N, ) = N
p
q
−1
p
where M = 0, N = 300 and n = 100 . Substituting in the values
shows that the probability Fred wins is approximately 0.00002 or 1
in 50, 000.
b) When Fred plays for bigger stakes it is as if his money takes a random
walk in units of £10, so it is as if he starts at 10 and we want to
know the probability it reaches 30 before 0. Since he is playing the
same game the chance the walk moves in each direction is the same.
Thus the probability he wins this time is r10 (0, 30, 18
)
19
10
1 − pq
18
r10 (0, 30, ) =
30 ≈ 0.17
19
1 − pq
so this time he has more than a one in six chance of winning.
c) This is a random walk between 0 and 300 as in part a, but this
time the probabilities are balanced: p = q = 1/2. Hence we use the
formula for this case: the probability he wins is
n
100
1
=
= .
N
300
3
d) As in part b this is a random walk between 0 and 30, but with equal
probabilities. Thus the probability he wins is:
n
10
1
=
= .
N
30
3
3. Consider the following random walk on the integers. (Note this is similar
to but not the same as the walk defined in lectures).
The RW is always at an integer and it moves one step to the left or right
or stays still at each time step. The probability it moves left is q, the
probability it moves right is p and the probability it stays still is s and
p + q + s = 1. (You may assume that all three of these probabilities are
non-zero.)
a) Let rn be the probability that a particle starting at position n
reaches N before it reaches 0. Express rn in terms of rn−1 and
rn+1 .
b) What is the probability that the particle moves to the left conditioned on it moving? What is the probability it moves to the right
conditioned on it moving?
Solution:
a) We condition on the first step: let B1 be the event the first step is
to the right, B2 the event that the first step is to the left, and B3
the event that at the first step the particle stays still. Obviously
these are a partition and P(B1 ) = p, P(B2 ) = q and P(B3 ) = s.
What is P(An | B1 ). If B1 occurs then the first step is to the right,
and after this first step the random walk is starting at n + 1 so the
probability that we reach N before 0 is exactly P(An+1 ) = rn+1 .
Similarly if B2 occurs then after the first step the RW follows a
random walk starting at n−1. Hence P(An | B2 ) = P(An−1 ) = rn−1 .
Finally, if B3 occurs then at the first time step the RW does not
move, so after the first time step the RW is starting at n. Hence
P(An | B3 ) = P(An ) = rn .
Now we apply the Theorem of Total Probability:
rn = P(An ) = P(An | B1 )P(B1 ) + P(An | B2 )P(B2 ) + P(An | B3 )P(B3 )
= P(An+1 )p + P(An−1 )q + P(An )s
= prn+1 + qrn−1 + srn
which (since 1 − s = p + q) rearranges to
rn =
p
q
rn+1 +
rn−1
p+q
p+q
b) Let Al be the event the particle moves to the left, Ar that the
particle moves to the right and B be the probability the particle
moves. Then P(Al ∩ B) = P(Al ) = q, P(Ar ∩ B) = P(Ar ) = p and
P(B) = p + q. Thus,
P(Al | B) =
P(Al ∩ B)
q
=
,
P(B)
p+q
P(Ar | B) =
p
P(Ar ∩ B)
=
.
P(B)
p+q
and
4. In this question we revise some standard distributions (ie random variables) from Introduction to Probability. For each of the following distributions write down its probability mass function, its mean and its
variance. In each case I give a broad description of the distribution as
well as its name.
a) The Bernoulli distribution with parameter p — i.e., a single trial
with probability p of success.
b) The Binomial distribution with parameters N and p — i.e, N independent trials each of which has probability p of success.
c) Geometric distribution with parameter p — i.e., a sequence of independent trials each with probability of success p: the random
variable is the number of trials until the first success. In my definition this number includes the first successful trial; other people
sometimes just count the number of failures.
d) Poisson distribution with parameter λ. (The real world interpretation here is less obvious but we will cover it later in the course.)
Solution: This is just revision.