= Solution
Write $s=\sigma_Y\sigma_Z$ and $W=YZ-\mathbb E(YZ)$. The variables $Y$ and $Z$ need not be <independent random variables>. Part (d) and the <Cauchy-Schwarz inequality> imply, for every integer $q\geq2$,
$$
\mathbb E|YZ|^q
\leq\bigl(\mathbb E|Y|^{2q}\mathbb E|Z|^{2q}\bigr)^{1/2}
\leq2^{q+1}q!s^q.
$$
Also $|\mathbb E(YZ)|\leq s$ by Cauchy-Schwarz, because $\mathbb EY^2\leq\sigma_Y^2$ and $\mathbb EZ^2\leq\sigma_Z^2$. Thus $\mathbb EW^2=\operatorname{Var}(YZ)\leq16s^2$, and, for $q\geq3$,
$$
\mathbb EW_+^q
\leq2^{q-1}\left(\mathbb E|YZ|^q+|\mathbb E(YZ)|^q\right)
\leq2^{2q+1}q!s^q
=\frac{q!}{2}(64s^2)(4s)^{q-2}.
$$
Part (c) shows that $W$ is <Sub-Gamma random variable in the right tail>[sub-Gamma in the right tail] with variance parameter $64s^2$ and scale parameter $4s$.
If $\mathbb E(YZ)\geq0$, then $W_+\leq(YZ)_+\leq|YZ|$. Consequently
$$
\mathbb EW_+^q\leq2^{q+1}q!s^q
=\frac{q!}{2}(16s^2)(2s)^{q-2},
$$
while $\mathbb EW^2\leq16s^2$. Another application of part (c) gives the improved variance parameter $16\sigma_Y^2\sigma_Z^2$ and scale parameter $2\sigma_Y\sigma_Z$.
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