I have two random variables $X > 0$ and $Y > 0$.

Given that I can estimate $$\text{Cov}(X, Y),$$ how can I estimate $$\text{Cov}(\log(X), \log(Y))?$$


One could take the approach of Taylor expansion:



Take $U=\log(X)$, $V=\log(Y)$.

Use multivariate Taylor expansion to compute an approximation to $\rm{E}(UV)$ (in similar fashion to the example at the end of "First Moment" in the link which does the simpler case of $\rm{E}(X.1/Y))$, and use univariate expansions to compute approximations to $\rm{E}(U)$ and $\rm{E}(V)$ (as given in the first part of the same section) to similar accuracy. From those things, compute the (approximated) covariance.

Expanding out to similar degree of approximation as the example in the link, I think you end up with terms in the mean and variance of each (untransformed) variable, and their covariance.

Edit 2:

But here's a little trick that may save some effort:

Note that $\rm{E}(XY) = \rm{Cov}(X,Y) + \rm{E}(X)\rm{E}(Y)$ and $X=\exp(U)$ and $Y=\exp(V)$.

Given $$ \operatorname{E}\left[f(X)\right]\approx f(\mu_X) +\frac{f''(\mu_X)}{2}\sigma_X^2 $$ we have $$ \operatorname{E}(\exp(U)) \approx \exp(\mu_U) + \frac{\exp(\mu_U)}{2}\sigma_U^2 \approx \exp(\mu_U+\frac{1}{2}\sigma_U^2) $$

Edit: That last step follows from Taylor approximation $\exp(b) \approx 1 + b$, which is good for small $b$ (taking $b =\frac{1}{2}\sigma_U^2$).

(that approximation is exact for $U$, $V$ normal: $\operatorname{E}(\exp(U))=\exp(\mu_U+\frac{1}{2}\sigma_U^2)$)

Let $W = U + V$

$$ \operatorname{E}(XY) = \operatorname{E}(\exp(U).\exp(V)) = \operatorname{E}(\exp(W)) $$

$$ \approx \exp(\mu_{W}) + \frac{exp(\mu_{W})}{2}\sigma_W^2 \approx \exp(\mu_W+\frac{1}{2}\sigma_W^2) $$

and given $\operatorname{Var}(W) = \operatorname{Var}(U) + \operatorname{Var}(V) + 2 \operatorname{Cov}(U,V)$, then


$$ 1+\frac{\operatorname{Cov}(X,Y)}{\operatorname{E}(X)\operatorname{E}(Y)} = \frac{\operatorname{E}(XY)}{\operatorname{E}(X)\operatorname{E}(Y)} $$ $$ \approx \frac{\exp(\mu_W+\frac{1}{2}\sigma_W^2)}{\exp(\mu_U+\frac{1}{2}\sigma_U^2).\exp(\mu_V+\frac{1}{2}\sigma_V^2)} $$ $$ \approx \frac{\exp(\mu_U+\mu_V+\frac{1}{2}(\sigma_U^2+\sigma_V^2+2 \operatorname{Cov}(U,V)))}{\exp(\mu_U+\frac{1}{2}\sigma_U^2).\exp(\mu_V+\frac{1}{2}\sigma_V^2)} $$ $$ \approx \exp[\operatorname{Cov}(U,V)] $$

Hence $\operatorname{Cov}(U,V)\approx \log(1+\frac{\operatorname{Cov}(X,Y)}{\operatorname{E}(X)\operatorname{E}(Y)})$. This should be exact for $U,V$ bivariate gaussian.

If you used the first approximation rather than the second, you would get a different approximation here.

  • $\begingroup$ Could you give a bit more details please? Anyway, thx for the suggestion $\endgroup$ – user7064 Jan 8 '13 at 17:08
  • $\begingroup$ Edited for detail. $\endgroup$ – Glen_b -Reinstate Monica Jan 9 '13 at 0:02
  • $\begingroup$ Thanks @Glend_b. I will accept when details will be added. In the meanwhile, +1 :-) $\endgroup$ – user7064 Jan 18 '13 at 7:41
  • $\begingroup$ No worries; I was busy at the time, then totally forgot. Now fixed $\endgroup$ – Glen_b -Reinstate Monica Jan 18 '13 at 12:48
  • $\begingroup$ It generally works better for non-Gaussian variables if the variances of $U$ and $V$ are small (equivalently, if the coefficients of variation of $X$ and $Y$ are small). $\endgroup$ – Glen_b -Reinstate Monica Jan 22 '13 at 7:44

Without any additional assumptions on $X$ and $Y$, it is not possible to deduce the covariance of the log knowing the initial covariance. In the other hand, if you were able to compute $\mathrm{Cov}(X,Y)$ from $X$ and $Y$, what prevents you from calculating $\mathrm{Cov}(\log(X), \log(Y))$ from $\log(X)$ and $\log(Y)$ directly?


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