I have a sequence such as

$$r[n] = y[n]v[n]$$

$y[n]$ and $v[n]$ are zero-mean and statistically independent.

I need to find a variance of $r[n]$ and show that it is white and equal to $\sigma ^2_y\sigma^2_v$.

I know that $y[n]$ has variance of $\sigma ^2_y$ and I also know that $$Cor(v_n, v_{n+m}) = \sigma ^2_v\delta[m]$$

This is what I have so far :

Finding the variance of $r[n]$: $$\sigma_r^2 = \mathbb{E}([r[n]-m_r)^2]=\mathbb{E}[r^2[n]] - 2\mathbb{E}[r[n]]m_r+m_r^2$$


$$\sigma_r^2 = \mathbb{E}[r^2[n]] - 2m_r m_r+m_r^2 = \mathbb{E}[r^2[n]]-m_r^2 $$

I am stuck at this point. Could you suggest how do I proceed from here?

The process is WSS random process.

  • 1
    $\begingroup$ Welcome to DSP.SE! Are $y[n]$ and $v[n]$ uncorrelated? Do you know the means of these processes? $\endgroup$
    – Tendero
    Feb 16, 2019 at 23:33
  • 4
    $\begingroup$ I'm voting to close this question as off-topic because it's only very peripherally a signal processing question – especially, there's a post on cross-validated, the stat/stoch sister SE site, that answers this: stats.stackexchange.com/questions/52646/… $\endgroup$ Feb 17, 2019 at 12:27
  • $\begingroup$ Hi @Tendero , yes y[n] and v[n] are zero-mean and statistically independent $\endgroup$
    – ViniLL
    Feb 17, 2019 at 20:47
  • $\begingroup$ @ViniLL. How about taking the time to edit your question so as to put the important information that $y[n]$ and $v[n]$ are zero-mean and independent right up there as part of the question instead of just being tucked away in a comment that a reader might well miss? -1 pending the edit requested. $\endgroup$ Feb 17, 2019 at 21:07
  • $\begingroup$ @DilipSarwate Question is edited. $\endgroup$
    – ViniLL
    Feb 17, 2019 at 21:19

2 Answers 2


$\sigma_r^2=\sigma_y^2\sigma_v^2$ iff

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


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)



$~~~~=\mathbb{E}\{y^2\}\mathbb{E}\{v^2\}-\mathbb{E}^2\{y\}\mathbb{E}^2\{v\}$ (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&=(\sigma_y^2 + \mathbb E^2\{y\})(\sigma_v^2 + \mathbb E^2\{v\}) - \mathbb E^2\{y\}\mathbb E^2\{v\}\\ &= \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 &=\mathbb{E}\left[(yv-\mathbb{E}[yv])^2\right] &{\scriptstyle{\text{(by definition)}}}\\ &=\mathbb{E}\left[(yv-\mathbb{E}[y]\mathbb{E}[v])^2\right] & {\scriptstyle{E[yv]=E[y]E[v] ~\text{since}~ y ~\text{and}~v~\text{are uncorrelated}}}\\ &=\mathbb{E}\left[y^2v^2\right] &{\scriptstyle{\mathbb E[y] = \mathbb E[v] = 0~\text{by assumption}}}\\ &=\mathbb{E}[y^2]\mathbb{E}[v^2] &{\scriptstyle{E[y^2v^2]=E[y^2]E[v^2] ~\text{since}~ y^2 ~\text{and}~v^2~\text{are uncorrelated}}}\\ &= \sigma_y^2\sigma_v^2 &{\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).

  • $\begingroup$ Hi @Sofiane, thanks for answer, sorry for forgetting to mention that y[n] and v[n] are zero mean and statistically independent $\endgroup$
    – ViniLL
    Feb 17, 2019 at 20:49
  • 1
    $\begingroup$ Actually, you also need to state that the independence of $y$ and $v$ also guarantees that $y^2$ and $v^2$ are independent and thus we can factor $E\big[r^2[n]\big]$ as $$E\big[r^2[n]\big] = E\big[y^2[n]v^2[n]\big] = E\big[y^2[n]\big]E\big[v^2[n]\big].$$ With regard to the "if and only if", note that it suffices if $y$ and $v$ are uncorrelated and so are $y^2$ and $v^2$; independence is not necessary. $\endgroup$ Feb 17, 2019 at 21:52
  • $\begingroup$ @DilipSarwate thanks , I see, but considering what you said where does $$m_r^2 $$ term in my last equation dissapears? $\endgroup$
    – ViniLL
    Feb 17, 2019 at 23:29
  • $\begingroup$ @DilipSarwate $x^2$ is measurable, so if $y$ and $v$ are independent, the squared processes are also independent. $\endgroup$
    – Sofiane
    Feb 18, 2019 at 8:57
  • 1
    $\begingroup$ @Sofiane I didn't claim that squaring preserved uncorrelation: I said "$y$ and $v$ being uncorrelated and $y^2$ and $v^2$ also being uncorrelated suffices," (I emphasized also in the quote above: I should have emphasized it in the original comment). I have added the proof for the other case to your answer. $\endgroup$ Feb 19, 2019 at 22:12


Goodman, Leo A. "On the exact variance of products." Journal of the American statistical association 55.292 (1960): 708-713.


Bohrnstedt, George W., and Arthur S. Goldberger. "On the exact covariance of products of random variables." Journal of the American Statistical Association 64.328 (1969): 1439-1442.

the variance of the product of $z=xy$ , $x$ and $y$ independent.

$$ \sigma_z^2=m_x^2 \sigma_y^2 + m_y^2 \sigma_x^2 + \sigma_x^2 \sigma_y^2 $$ where $E\{x\}=m_x$ and the same for $y$

Bohrstein and Goldberger's derivation is easier to follow but Goodman isn't behind a paywall.

The papers also cover the dependent case which is a bit more complicated.


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