Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 6 / c

Write

A_{i} = Y_{i - 1} Y_{i}

, so

S_{n} = \sum_{i = 1}^{n} A_{i}

. Each product has mean one, and its variance is

Var (A_{i}) = E (Y_{i - 1}^{2}) E (Y_{i}^{2}) - 1 = 2 \cdot 2 - 1 = 3.

Adjacent products share one factor, giving covariance one by part (a)(iii). Products whose indices differ by at least two involve disjoint independent random variables, so their covariance is zero. There are

n - 1

E S_{n} = n, Var (S_{n}) = 3 n + 2 (n - 1) = 5 n - 2, n \geq 1.

ii

0 \leq P (\frac{S _{n}}{n} > 1 + ϵ) \leq P (\frac{S _{n}}{n} - 1 \geq ϵ) \leq \frac{Var ( S _{n} )}{n ^{2} ϵ ^{2}} = \frac{5 n - 2}{n ^{2} ϵ ^{2}} ⟶ 0.

This actually proves the stronger statement

S_{n} / n P 1

, namely convergence in probability, despite the statistical dependence between adjacent products.

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