 Solution

ID: past-exam-of-the-probability-course-of-the-university-of-oxford/2023/1/c/iii/solution

For

α > 0

E U_{1}^{α} = \frac{1}{b} \int_{0}^{b} u^{α} d u = \frac{b ^{α}}{1 + α}, E W_{n}^{α} = (\frac{b ^{α}}{1 + α})^{n} .

P (W_{n} > ε) \leq ε^{- α} (\frac{b ^{α}}{1 + α})^{n} .

The parenthesized factor is less than one precisely when

α lo g b - lo g (1 + α) < 0

. This expression vanishes at

α = 0

lo g b - 1

there. If

b < e

, the derivative is negative, so some sufficiently small

α > 0

makes the factor strictly less than one. The bound then tends to zero geometrically, proving

W_{n} ⟶ 0 in probability for every 0 < b < e .

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