MAT00035I Probability & Statistics: Applied Probability 2020/21
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MAT00035I
BA, BSc and MMath Examinations 2020/21
Probability & Statistics: Applied Probability
1 (of 6).
(All numbers in this question are purely fictional.)
A particular type of tumour cell divides according to the rules of a branching pro- cess: with probability 0.35 the cell dies, and with probability 0.65 the cell divides into two. Let Xn denote the number of cells in generation n, where X0 = 1.
(a) Calculate the extinction probability of this branching process. [5]
Radiotherapy can reduce the chance of a cell dividing to 0.65 - 0.15β, where the value of β e [0, 1] depends upon the strength of the treatment.
(b) Calculate the extinction probability, as a function of β, for a tumour branching
process which is being treated with radiotherapy. [4]
(c) Sketch a graph, as a function of β, of the ratio
P (X never hits zero I radiotherapy)
P (X never hits zero I no radiotherapy) . [5]
(d) Doctor A suggests using radiotherapy strong enough to reduce
P (X never hits zero I radiotherapy) to 70% of the corresponding value with no treatment. Doctor B suggests using a strength which reduces the expected number of cells in generation 3 to 90% of the corresponding value with no treatment. Which of these two suggestions would give the greater probability of the branching process going extinct? [10]
2 (of 6). Two people play a game using a coin satisfying P (Head) = 0.2 = 1 - P (Tail). On
a player’s turn they toss the coin three times, with the following possible outcomes:
• if they score three Heads, they win and the game stops;
• if they score exactly one Head, they get another turn;
• if they score zero Heads or two Heads, play passes to the other player.
(a) Find the probability that the player who starts the game is the winner. [10]
(b) Calculate the expected number of coin tosses before the game ends. [5]
3 (of 6). Let P be a transition matrix for a Markov chain on a state space S with ISI = n.
Let A be an n × n matrix, each of whose entries equals 1/n. For q e [0, 1] let Mq denote the matrix given by
Mq = qP + (1 - q)A .
Carefully explain:
(a) for which values of q e [0, 1] we can be sure that there exists at least one
non-negative vector π satisfying (i πi = 1 and πMq = π; [5]
(b) for which values of q e [0, 1] we can be sure that there exists a unique π
satisfying these properties. [5]
4 (of 6). A Markov chain X has state space S = {1, 2, 3, 4} and transition matrix
1 2 3 4
P = 2(1) 、.
3.. 0 0 2/10 8/10.. .
(a) List the communicating classes, and identify which states are recurrent and
which are transient.
[3]
(b) Explain why limn二3 Pn exists.
(c) Calculate limn二3 Pn .
[2]
[11]
5 (of 6).
Consider the following algorithm for generating a finite sequence of random num- bers from the set A = {1, 3, 6, 8}.
Algorithm 1 1. let X1 , X2 , . . . be i.i.d. random variables, each distributed uniformly on the set A (i.e. P (X1 = a) = 1/4 for all a e A); 2. simulate a value of N ~ Geom(1/4), independently of all the Xi random variables; 3. output the sequence V = (X1 , X2 , . . . , XN ). |
E.g. if (X1 , X2 , X3 , . . . ) = (1, 6, 8, 1, 6, 6, . . . ) and N = 4, then V = (1, 6, 8, 1).
(a) What is the mean length of the output sequence V? [3]
(b) Let SV denote the sum of the entries in V (so in the example above, SV = 16).
Calculate E [SV ], making sure to explain your reasoning carefully. [5]
(c) What is the probability that V contains no 1s? [7]
(Hint: you might find it helpful to condition on the value of N.)
6 (of 6).
Consider an alternative algorithm (different to that in Question 5) for generating a finite sequence of random numbers from the set A = {1, 3, 6, 8}.
Algorithm 2 1. let X1 , X2 , . . . be i.i.d. random variables, each distributed uniformly on the set A (i.e. P (X1 = a) = 1/4 for all a e A); 2. let T = min{n : Xn+1 = Xn }; 3. output the sequence W = (X1 , X2 , . . . , XT ). |
E.g. if (X1 , X2 , X3 , . . . ) = (1, 6, 8, 1, 6, 6, . . . ) then, since X6 = X5 , T = 5 and so the output sequence is W = (1, 6, 8, 1, 6).
(a) Explain how the sequence of random variables in the output sequence W may
be described by a Markov chain with state space {0, 1, 3, 6, 8}, where 0 is an
absorbing state, and write out the transition matrix.
(b) What is the probability that W contains no 1s?
[4]
[5]
(c) Let SW denote the sum of the entries in W. Calculate E [SW ]. [7] (Hint: write mk = E [SW I X1 = k]; use a first-step decomposition argument to express mk as a linear combination of mj for j k.)
(d) Describe some of the differences and similarities between the distributions of the output sequences of Algorithms 1 and 2. (You are not expected to write more than a few sentences.) [4]
SOLUTIONS: MAT00035I
The offspring distribution is given by P (V = 0) = (1 - ε) = 1 - P (V = 2), where we write ε = 3/10.
(a) Using the standard notation from the course, the PGF of the offspring distri-
bution is given by GV (z) = (1 - ε)+ (1+ε)z2 . The extinction probability
is the smallest non-negative root to the equation GV (η) = η . 、(')2 Marks
Solving this quadratic (in the knowledge that one of the two roots is always 1) gives the extinction probability to be
1 - ε
1 + ε
(b) We repeat the calculation from part (a) with the new offspring distribution,
writing ηβ for the extinction probability:
ηβ =
1 - (1 - β)ε
1 + (1 - β)ε .
' e
4 Marks 、 /
(c) The ratio reduces to the following:
P (X never hits zero I radiotherapy) 1 - ηβ (1 + ε)(1 - β) P (X never hits zero I no radiotherapy) = 1 - η = 1 + (1 - β)ε .
When β = 0 there is no effect of radiotherapy, and so the ratio equals 1; when β = 1 the process becomes critical, and the numerator hits zero. The graph should show a continuous, decreasing function, lying above the line y = -x. E.g. when ε = 0.2 the graph looks like this:
1.0
0.8
0.6
0.4
0.2
0.2 0.4 0.6 0.8 1.0
' e 5 Marks 、 /
(d) We calculate the corresponding β values for each suggestion.
Doctor A: write x = 70/100 and solve (1 - ηβ )/(1 - η) = x to obtain
A =
(1 + ε)(1 - x) 1 + ε(1 - x)
= 0.357 .
' e
3 Marks 、 /
Doctor B: The mean of the offspring distribution is µ := 1 + ε with no radiotherapy, and µβ := 1 + (1 - β)ε with treatment. The mean number of cells in generation n is µn and µβ(n), respectively. Writing n = 3 and y = 90/100, we therefore need to solve
(1 + (1 - β)ε)n = y(1 + ε)n
for β, which gives
(1 + ε)(1 - y1/n)
ε
' e
5 Marks
、 /
tion which yields the largest choice of β . 、(')2 Marks
' e
2. (a) Model the game as a four-state Markov chain with state space S = {A, B, WA , WB },
where A and B indicate that it is player A or B’s turn, and WA (WB ) indi- cates that player A (B) has won. Write p = 0.2 for P (Head). The chain has transition matrix
A
B(A) ╱
WB . 0
B
3p2 (1 - p) + (1 - p)3 3p(1 - p)2
0
0
WA
p3
0
1
0
WB
p3(0) 、.
1(0) .. .
' e 5 Marks 、 /
The game is symmetric, so we just need to calculate fA(*),WA , the probability that the chain ever reaches WA if started from A. We could do this via a first-step decomposition, or by calculating the matrix GQ : we do both here.
First-step: By symmetry, fB(*),WA = fA(*),WB = 1 - fA(*),WA . Thus
fA(*),WA = p3 + 3p(1 - p)2 fA(*),WA + (1 - 3p + 6p2 - 4p3 )(1 - fA(*),WA ) . Solving for fA(*),WA we obtain
fA(*) 3p(1 - p)2 = 0.503 .
2022-05-13