I am trying to do some toy Monte Carlo sampling, to calculate the uncertainty of the product $A\times B$ of two random variables $A=a\pm\sigma_A$ and $B=b\pm\sigma_B$.

I also assume that these variables are fully correlated.

I use a multivariate normal distribution, specifying the covariance $\text{cov}=\left(\begin{array}{cc} \sigma_A^2 & \sigma_A \sigma_B\\ \sigma_A \sigma_B & \sigma_B^2\\ \end{array}\right)$ to correspond to unit correlation coefficients, using e.g.

numpy.random.multivariate_normal(mean=[a, b], cov=[[sA**2,sA*sB],[sA*sB,sB**2]], size=1)

to sample values $a^\prime$ and $b^\prime$ to get the distribution of $a^\prime \times b^\prime $, for which I then calculate mean and standard deviation.

It seems that, for arbitrary $a, \sigma_A, b, \sigma_B$, the mean of the $a^\prime \times b^\prime$ distribution is biased, and that this bias leads to higher values the bigger the uncertanties $\sigma_A$ and $\sigma_B$ are with respect to $a$ and $b$, respectively.

What is the reason behind this?

P.S. In the uncorrelated case, i.e. using $\text{cov}=\left(\begin{array}{cc} \sigma_A^2 & 0\\ 0 & \sigma_B^2\\ \end{array}\right)$, the mean of the $a^\prime \times b^\prime$ distribution does not seem to be similarly biased.

  • $\begingroup$ This classic paper looks like it would be of help to you. $\endgroup$
    – Ed V
    Jan 28 at 19:55
  • $\begingroup$ Thank you very much for this comment. In this paper, there's a difference in the definition of the expectation value of x*y depending on whether x and y are independent or not. I am not an expert in studying statistics, however, so I am still not sure if that's indeed the case here! $\endgroup$ Jan 29 at 9:46
  • $\begingroup$ Not much I can say, but I remember doing some simple Monte Carlo sims of products of independent Gaussian variates. Then I found the Goodman paper and it looks (to me, not a stats expert by any means) like it covers the cases of independent and correlated Gaussians. The real experts here can certainly answer, but maybe consider asking another question focussed exactly on whether Goodman’s results address the bias (or non-bias) of the product. $\endgroup$
    – Ed V
    Jan 29 at 13:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.