Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 224 / 4 / d

Solution

Put

L = L^{*} (X)

and

μ = E L

. Since

L

X

H (X) = H (L) + H (X ∣ L) \leq H (L) + μ .

Part (a), applied to the nonnegative integer-valued random variable

L

, yields

H (L) \leq (1 + μ) h (\frac{1}{1 + μ}) = lo g (1 + μ) + μ lo g (1 + \frac{1}{μ}) .

lo g (1 + t) \leq t lo g e

implies

μ lo g (1 + 1/ μ) \leq lo g e

. Thus

H (X) \leq μ + lo g (1 + μ) + lo g e .

Part (c) also gives

μ \leq H (X)

, so monotonicity of the logarithm lets us replace

lo g (1 + μ)

lo g (H (X) + 1)

. Rearranging proves

E [L^{*} (X)] \geq H (X) - lo g [H (X) + 1] - lo g e .

Solved by gpt-5.6-sol high.

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