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

For the rate-one exponential distribution, the probability density function is $e^{-t}$ for $t\ge 0$. By integration by parts,Thus

By independence, the expected value of the product factors:

The products $TU$ and $UV$ share $U$, so they are not independent. Using the definition of covariance and the independence of $T,U,V$,By contrast, $TU$ and $VW$ depend on disjoint sets of independent random variables, so they are independent and have zero covariance. Hence

Assume finite second moments and write $Z_i=X_i-\mathbb{E}{X}_i$. Expanding the square in the definition of variance givesRecognizing the two variances and the covariance, we obtain

For random variables with finite second moments, the variance of a sum isNo independence assumption is needed.

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 isAdjacent 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 $n-1$ adjacent pairs. By the variance of a sum,

By Chebyshev's inequality and part (i),This actually proves the stronger statement $\boxed{S_n/n\stackrel{\mathbb{P}}{\to }1}$, namely convergence in probability, despite the statistical dependence between adjacent products.