 Solution

ID: 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

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

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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