= Solution
Put $Z_i=Y_{i-1}Y_i$ for $1\leq i\leq n$. Independence of the underlying exponentials gives $\mathbb EZ_i=1$ and
$$
\operatorname{Var}(Z_i)=\mathbb E[Y_{i-1}^2]\mathbb E[Y_i^2]-1=4-1=3.
$$
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
$$
\boxed{\mathbb ES_n=n,\qquad\operatorname{Var}(S_n)=3n+2(n-1)=5n-2,\qquad n\geq1.}
$$
In particular, the products themselves are not independent merely because the original $Y_i$ are.
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