 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/iii/paper-201/2/d/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

Write

p_{m} = P (S_{T} = m)

p_{m + 1} = P (S_{T} = m + 1)

, and

q_{m} = p_{m} + p_{m + 1}

. Part (b) gives

1 = 2^{- m} (1 - q_{m}) + 2^{m} p_{m} + 2^{m + 1} p_{m + 1},

q_{m} \leq 2^{- m}

. The exit-state decomposition also gives

\frac{E S _{T}}{m} = - 1 + 2 q_{m} + \frac{p _{m + 1}}{m} ⟶ - 1.

Combining this with part (c), including the upper overshoot, yields

\frac{E T _{m}}{m} ⟶ \frac{7}{4} .

For example, the same calculation supplies the quantitative bound

0 \leq \frac{7}{4} - \frac{E T _{m}}{m} \leq \frac{7}{4} (2 + \frac{1}{m}) 2^{- m} .

The asymptotic linear growth is determined by the negative mean increment; the exponentially unlikely upper exit gives a vanishing correction.

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