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 Incoming links: Covariance of overlapping independent products

Past exam of the year 1 of the computer science course of the University of Oxford / 2017 / Discrete mathematics, continuous mathematics and probability / 9 / a / iii / Solution 2026-10-31

For the overlapping products, factor using the independence of

T, U, V

:

Cov (T U, U V) = E [T U^{2} V] - E [T U] E [U V] = 1 \cdot 2 \cdot 1 - 1 \cdot 1 = 1.

This is the covariance of overlapping independent products: the shared

U

contributes its second moment. In contrast,

T U

and

VW

use disjoint independent factors and are themselves independent, giving

Cov (T U, U V) = 1, Cov (T U, VW) = 0.

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Past exam of the year 1 of the computer science course of the University of Oxford / 2017 / Discrete mathematics, continuous mathematics and probability / 9 / c / i / Solution 2026-10-31

Put

Z_{i} = Y_{i - 1} Y_{i}

for

1 \leq i \leq n

. Independence of the underlying exponentials gives

E Z_{i} = 1

and

Var (Z_{i}) = E [Y_{i - 1}^{2}] E [Y_{i}^{2}] - 1 = 4 - 1 = 3.

By the covariance of overlapping independent products, adjacent terms have covariance one. Terms separated by at least two indices use disjoint independent factors, so their covariance is zero. There are

n - 1

adjacent pairs. Linearity of expectation and the variance of a sum formula therefore yield

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

In particular, the products themselves are not independent merely because the original

Y_{i}

are.

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