Assessment 1

A solutions sheet for this assessment is available on Minerva. Marks and feedback are available on Minerva and Gradescope respectively.

1. Let \((X_n)\) be a simple random walk that starts from \(X_0 = 0\) and on each step goes up one with probability \(p\) and down one with probability \(q = 1-p\).

Calculate:

(a) \(\mathbb P(X_6 = 0)\), [1 mark]

(b) \(\mathbb EX_6\), [1]

(c) \(\text{Var}(X_6)\), [1]

(d) \(\mathbb E(X_{10} \mid X_4 = 4)\), [1]

(e) \(\mathbb P(X_{10} = 0 \mid X_6 = 2)\), [1]

(f) \(\mathbb P(X_4 = 2 \mid X_{10} = 6)\). [1]

Consider the case \(p = 0.6\), so \(q = 0.4\).

(g) What are \(\mathbb E X_{100}\) and \(\text{Var}(X_{100})\)? [1]

(h) Using a normal approximation, estimate \(\mathbb P(16 \leq X_{100} \leq 26)\). You should use an appropriate “continuity correction”, and explain why you chose it. (Bear in mind the possible values \(X_{100}\) can take.) [3]

2. Consider the gambler’s ruin with draws: Alice starts with £\(a\) and Bob with £\((m-a)\), and at each time step Alice wins £1 off Bob with probability \(p\), loses £1 to Bob with probability \(q\), and no money is exchanged with probability \(s\), where \(p+q+s =1\). We consider the case where Bob and Alice are equally matched, so \(p = q\) and \(s = 1-2p\). (We assume \(0 < p < 1/2\).)

Let \(r_i\) be Alice’s ruin probability from the point she has £\(i\).

(a) By conditioning on the first step, explain why \(pr_{i+1} - (1-s)r_i + pr_{i-1} = 0\), and give appropriate boundary conditions. [2]

(b) Solve this linear difference equation to find an expression for \(r_i\). [2]

Let \(d_i\) be the expected duration of the game from the point Alice has £\(i\).

(c) Explain why \(pd_{i+1} - (1-s)d_i + pd_{i-1} = -1\), and give appropriate boundary conditions. [2]

(d) Solve this linear difference equation to find an expression for \(d_i\). [2]

(e) Compare your answer to parts (b) and (d) with those for the standard gambler’s ruin problem with \(p = 1/2\), and give reasons for the similarities or differences. [2]