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ID: past-exam-of-the-algorithmic-foundations-of-learning-course-of-the-university-of-oxford/2025/1/b/solution

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Past exam of the algorithmic foundations of learning course of the University of Oxford / 2025 / 1 / b / Solution by

Codex  0 2026-10-25

Let

M_{n} = max_{1 \leq i \leq n} X_{i}

. For

0 < s < 1/ c

, the exponential of a maximum is bounded by a sum, giving

e^{s M_{n}} \leq \sum_{i = 1}^{n} e^{s X_{i}}, E e^{s M_{n}} \leq n exp (\frac{ν s ^{2}}{2 ( 1 - cs )}) .

Applying Jensen's inequality to the logarithm yields

E M_{n} \leq \frac{1}{s} lo g E e^{s M_{n}} \leq \frac{l o g n}{s} + \frac{ν s}{2 ( 1 - cs )} .

For

n > 1

, put

a = 2 lo g n / ν

and choose

s = a / (1 + c a)

. This gives

1 - cs = 1/ (1 + c a)

and therefore

\frac{l o g n}{s} + \frac{ν s}{2 ( 1 - cs )} = \frac{l o g n}{a} + c lo g n + \frac{ν a}{2} = 2 ν lo g n + c lo g n .

For

n = 1

, let

s ↓ 0

in the preceding upper bound, obtaining

E X_{1} \leq 0

. Consequently

E 1 \leq i \leq n max X_{i} \leq 2 ν lo g n + c lo g n (n \geq 1) .

The proof uses only the individual moment-generating function bounds for Sub-Gamma random variables in the right tail; it requires no independence. Under the paper's literal one-sided definition, the negative part need not be integrable. The positive part of

M_{n}

is integrable by the displayed exponential bound; if

E M_{n} = - \infty

, the conclusion holds in the extended sense. Otherwise the application of Jensen's inequality is ordinary. Equivalently, truncate

M_{n}

below and pass to the limit.

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