I have three random variables: A, B and C. I know their pairwise correlations. In other words, I know what is correlation between A and B, B and C, and finally, A and C. I also know means of all three variables.

Is there a chance to estimate covariance between D and C, where D is a product of A and B:

$ \text{cov}(D, C) = \text{cov}(A \cdot B, C) $

In addition to that, I would like to be able to estimate the following covariation:

$ \text{cov}(A \cdot B, B \cdot C) $

I guess that it might be impossible, than I would like to know, if it is possible to give the upper and low bounds or estimate those covariances approximately.

  • $\begingroup$ I assume that there are no independences between $A, B$ and $C$?? $\endgroup$
    – Fiodor1234
    Apr 21, 2021 at 9:49
  • $\begingroup$ @Fiodor1234, A, B and C are mutually correlated. Doesn't it means that they are not independent? $\endgroup$
    – Roman
    Apr 21, 2021 at 11:13
  • $\begingroup$ What do you know about the (marginal) distributions of your random variables in addition to their means? $\endgroup$
    – g g
    Apr 21, 2021 at 16:35
  • 2
    $\begingroup$ Because covariances are proportional to scale, and you supply no information about that, the situation is as bad as possible: the covariance between $D$ and $C$ could be literally any value unless $A$ and $B$ are independent of $C.$ $\endgroup$
    – whuber
    Apr 21, 2021 at 16:52
  • $\begingroup$ Is there a specific reason that you deal with this expression $\text{cov}(A \cdot B, B \cdot C)$ in general? If you have a particular distribution then this information might help to solve your specific case. In what sort of use case do you know pairwise correlations but not this special covariance? How did you get to know the correlations? Are they some model/distribution parameters (in which case you can share the entire model) or are they estimated from data (in which case you could just as well estimate the covariance of the product on the same way you estimated correlations). $\endgroup$ Apr 23, 2021 at 7:25

1 Answer 1


Let's start with the first covariance: $$\text{Cov}(A\cdot B,C)=\mathbb{E}[A\cdot B\cdot C]-\mathbb{E}[A\cdot B]\mathbb{E}[C]$$ You know the means of $A,B,C$, namely $\mu_A,\mu_B,\mu_C$, and the pairwise correlation between $A,B$, which I will denote as $\rho_{A,B}$. In case you also have access to the standard deviations of $A,B$, written $\sigma_A,\sigma_B$, you can rewrite

$$\mathbb{E}[A\cdot B]=(\rho_{A,B}\sigma_A\sigma_B+\mu_A\mu_B)$$

Now your desired covariance can be written as: $$\text{Cov}(A\cdot B,C)=\mathbb{E}[A\cdot B\cdot C]-(\rho_{A,B}\sigma_A\sigma_B+\mu_A\mu_B)\mu_C$$

Similarly, you can write the second covariance as: $$\text{Cov}(A\cdot B,B\cdot C)=\mathbb{E}[A\cdot B^2\cdot C]-(\rho_{A,B}\sigma_A\sigma_B+\mu_A\mu_B)(\rho_{B,C}\sigma_B\sigma_C+\mu_B\mu_C)$$

In both cases you have a term that cannot be computed with the information at disposal. In particular, note that we cannot say anything about $\mathbb{E}[A\cdot B\cdot C]$ and $\mathbb{E}[A\cdot B^2\cdot C]$. You can take a look at this example in Mathematics Stack Exchange, where the particular case of your problem when $A,B,C$ are pairwise independent is discussed. In general, note that factorizing the joint distribution requires information about the joint behavior of two of the random variables conditioned on the other (e.g. $A,B|C$), which cannot be inferred with your information.

Since you were talking about estimating the autocorrelations, in case you have access to samples of $A,B,C$ (namely $a_i,b_i,c_i$) one option is to use the sample average. That is

$$ \mathbb{E}[A\cdot B\cdot C]\approx \frac{1}{N}\sum_{i=1}^{N}a_i\cdot b_i\cdot c_i$$

You can bound the quality of the estimator above as a function of $N$ using Chebyshev's inequality. Given that you also don't have access to the variance of $A\cdot B\cdot C$ (but you can compute the sample variance) you can take a look at this post.

  • 1
    $\begingroup$ Your second formula given mean of product. From this formula follows that if A and B are uncorrelated, then mean of product is equal to minus product of means. But shouldn't it be just product of means? $\endgroup$
    – Roman
    Apr 23, 2021 at 9:51
  • $\begingroup$ @Roman thanks a lot for spotting this mistake! It should be corrected now. Note that the second equation is obtained by just rearranging the terms in the definition of the correlation coefficient. $\endgroup$
    – Oriol B
    Apr 28, 2021 at 15:58

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