Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 216 / 3 / d

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- Solution d

Solution

Subtracting the expected value of

f

does not alter either side, so suppose

π (f) = 0

. Since reversibility makes

K

a self-adjoint operator and stationarity makes it a contraction,

E_{K^{2}} (f) = ∥ f ∥_{L^{2} (π)}^{2} - ∥ K f ∥_{L^{2} (π)}^{2} .

The Discrete-time Poincaré inequality for a Markov kernel is therefore equivalent to

∥ K f ∥_{2}^{2} \leq (1 - \frac{1}{C}) ∥ f ∥_{2}^{2} .

Applying this inequality successively to

f, K f, \dots, K^{t - 1} f

yields

Var_{π} (K^{t} f) = ∥ K^{t} f ∥_{2}^{2} \leq (1 - \frac{1}{C})^{t} ∥ f ∥_{2}^{2} = (1 - \frac{1}{C})^{t} Var_{π} (f) .

Conversely, the asserted variance contraction with

t = 1

rearranges to the Poincaré inequality. Hence the two statements are equivalent.

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