 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/discrete-mathematics-continuous-mathematics-and-probability/7/b/ii/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability / 7 / b / ii / Solution by

Codex  0 2026-10-31

The absorbing states give

e_{0} = e_{n} = 0

. Multiplying the difference equation in (i) by

3

gives

e_{k + 1} - 2 e_{k} + e_{k - 1} = - 3.

The second finite difference of

k^{2}

is

2

, so a particular solution is

- 3 k^{2} /2

; the homogeneous solutions are

A + B k

. Imposing the two boundary conditions gives

A = 0, B = 3 n /2

. Therefore

e_{k} = \frac{3}{2} k (n - k), 0 \leq k \leq n .

Uniqueness follows because the difference of any two solutions has zero second finite difference, hence is affine function, and vanishes at both endpoints. This is the mean absorption time of a lazy symmetric random walk with holding probability

1/3

.

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