= Solution
{c}
Write $A_i=Y_{i-1}Y_i$, so $S_n=\sum_{i=1}^nA_i$. Each product has <mean> one, and its <variance> is
$$
\operatorname{Var}(A_i)=\mathbb E(Y_{i-1}^2)\mathbb E(Y_i^2)-1=2\cdot2-1=3.
$$
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 $n-1$ adjacent pairs. By the <variance of a sum>,
$$
\boxed{\mathbb E S_n=n,\qquad\operatorname{Var}(S_n)=3n+2(n-1)=5n-2,\qquad n\ge1.}
$$
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