How to find a variance of sample sequence
$\sigma_r^2=\sigma_y^2\sigma_v^2$ iff

1) $\mathbb{E}\{y\} = \mathbb{E}\{v\}=0$.

AND

2) $y$ and $v$ are independent, OR uncorrelated and their squares ($y^2$ and $v^2$) are also uncorrelated.



Proof: assuming independence,

$\sigma_r^2=\mathbb{E}\{(yv-\mathbb{E}\{yv\})^2\}$ (by definition)

$~~~~=\mathbb{E}\{(yv-\mathbb{E}\{y\}\mathbb{E}\{v\})^2\}$ (by independence)

$~~~~=\mathbb{E}\{y^2v^2+\mathbb{E}^2\{y\}\mathbb{E}^2\{v\}-2yv\mathbb{E}\{y\}\mathbb{E}\{v\}\}$

$~~~~=\mathbb{E}\{y^2v^2\}-\mathbb{E}^2\{y\}\mathbb{E}^2\{v\}$

$~~~~=\mathbb{E}\{y^2\}\mathbb{E}\{v^2\}-\mathbb{E}^2\{y\}\mathbb{E}^2\{v\}$ (by independence, since the square function is a measurable function).\

But, $\mathbb E\{y^2\} = \sigma_y^2 + \mathbb E^2\{y\}$
and $\mathbb E\{v^2\} = \sigma_v^2 + \mathbb E^2\{v\}$ and so,
\begin{align}\sigma_r^2&amp;=(\sigma_y^2 + \mathbb E^2\{y\})(\sigma_v^2 + \mathbb E^2\{v\}) - \mathbb E^2\{y\}\mathbb E^2\{v\}\\
&amp;= \sigma_y^2\sigma_v^2 + \sigma_y^2\mathbb E^2\{v\} +
\sigma_v^2\mathbb E^2\{y\}
\end{align}

which does not equal $\sigma_y^2\sigma_v^2$ unless both $y$ and $v$ have zero mean as the OP has stated: having at least one mean be zero (which does not preclude the other mean from being nonzero) is not enough (as was claimed in an earlier version of this answer).



Proof: assuming $y,v$ uncorrelated and $y^2,v^2$ also uncorrelated,

\begin{align}\sigma_r^2 &amp;=\mathbb{E}\left[(yv-\mathbb{E}[yv])^2\right] &amp;{\scriptstyle{\text{(by definition)}}}\\
&amp;=\mathbb{E}\left[(yv-\mathbb{E}[y]\mathbb{E}[v])^2\right]  &amp; {\scriptstyle{E[yv]=E[y]E[v] ~\text{since}~ y ~\text{and}~v~\text{are uncorrelated}}}\\
&amp;=\mathbb{E}\left[y^2v^2\right] &amp;{\scriptstyle{\mathbb E[y] = \mathbb E[v]  = 0~\text{by assumption}}}\\
&amp;=\mathbb{E}[y^2]\mathbb{E}[v^2] &amp;{\scriptstyle{E[y^2v^2]=E[y^2]E[v^2] ~\text{since}~ y^2 ~\text{and}~v^2~\text{are uncorrelated}}}\\
&amp;= \sigma_y^2\sigma_v^2 &amp;{\scriptstyle{\text{since}~\mathbb E[y]= \mathbb E[v] = 0~\text{by assumption}}}
\end{align}

Note that $y$ and $v$ being independent is a subcase of the above assumptions about $y$ and $v$ being uncorrelated and $y^2$ and $v^2$ also being uncorrelated.  When $y$ and $v$ are independent (and hence uncorrelated), so are $y^2$ and $v^2$ independent (which implies that $y^2$ and $v^2$ are uncorrelated).