Let $X_1$ and $X_2$ be identically distributed correlated lognormal random variables:

$$X_1,X_2 \sim \ln \mathcal{N}(\mu_X,\sigma_X^2)$$

(such that their logs are bivariate normal).

The correlation between $X_1$ and $X_2$ is given by:

$$\text{corr}(X_1,X_2) = \rho_X$$

Let $Y_1$ and $Y_2$ be the reciprical of these random variables:

$$Y_1 = \frac{1}{X_1}, Y_2 = \frac{1}{X_2}$$

Then $Y_1$ and $Y_2$ will also be lognormally distributed according to:

$$Y_1,Y_2 \sim \ln \mathcal{N}(-\mu_X,\sigma_X^2)$$

The correlation between them is given by $\rho_Y$:

$$\text{corr}(Y_1,Y_2) = \rho_Y$$

Can an expression for $\rho_Y$ be written in terms of $\rho_X$?

  • $\begingroup$ it's easy in the case where $(\log X_1,\log X_2)$ are bivariate normal. Did you intend that situation or did you mean for it to be a general question? $\endgroup$ – Glen_b Aug 2 '16 at 13:23
  • $\begingroup$ I think that I intend this to be the case. I am thinking about the reciprocal of a lognormal random process (which is obtained through the exponentiation of a multivariate Gaussian). I know what the marginal distribution of the reciprocal lognormal will be but I am not sure about the correlation structure. I chose lognormal since the reciprocal transformation is nice. The reciprocal of a lognormal variable is lognormal. I believe other distributions are not easy to work with when the reciprocal is taken. $\endgroup$ – 7Jack Aug 2 '16 at 14:18

If you have two variables, $Z_1$ and $Z_2$ that are bivariate normal, then the correlation of two corresponding lognormals is a simple function of the correlation of the corresponding normals and their standard deviations. For example, mpiktas' answer to this question gives the result.

Now $-Z_1$ and $-Z_2$ are bivariate normal with the same variances and correlation as $Z_1$ and $Z_2$, so $Y_1=e^{-Z_1}$ and $Y_2=e^{-Z_2}$ must have the same correlation as $X_1=e^{Z_1}$ and $X_2=e^{Z_2}$ do.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.