Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 208 / 1 / c

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- Solution c

Solution

Put

q_{i} = min (p_{i}, 1 - p_{i})

. Applying part (b) after possibly replacing

X_{i}

by

1 - X_{i}

gives

H (X_{i}) \leq 2 q_{i} + q_{i}

. If

q_{i} < δ_{i}

, then, because

0 < δ_{i} < 1

, this is less than

3 δ_{i}

, contradicting the assumption. Thus

p_{i}, 1 - p_{i} \geq δ_{i}

.

The centered variables

Y_{i} = lo g P_{i} (X_{i}) + H (X_{i})

are independent and have mean zero. Their two possible values differ by

lo g \frac{p _{i}}{1 - p _{i}} \leq ∣ lo g δ_{i} ∣.

Applying Hoeffding inequality to

\sum_{i} Y_{i}

gives

P (lo g P (X_{1}, \dots, X_{n}) + H (X_{1}, \dots, X_{n}) \leq - t) \leq exp (- \frac{2 t ^{2}}{\sum _{i} ( l o g δ _{i} ) ^{2}}) .

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