If $Y \sim N(\mu,\sigma^2)$ is normally distributed, then $X=\mathrm{e}^Y$ is lognormally distributed. To get the log-$\mu$ and log-$\sigma$ of this lognormal distribution you calculate $$\sigma^2 = \ln\left( \frac{\mathit{Var}}{E^2} + 1 \right)$$ and $$\mu = \ln(E)-\frac{\sigma^2}{2}$$.

But how is this in the case of a bivariate Normaldistribution, $Y \sim N((\mu_1, \mu_2),\Sigma)$, $\Sigma=((\sigma_1,\rho),(\rho,\sigma_2))$ How do I get $\mu, \sigma, \rho$? And more general in the case of a multivariate distribution?

  • $\begingroup$ What is $E$ in your notation in the expressions for $\sigma^2$ and $\mu$? $\endgroup$
    – Confounded
    Commented Apr 16, 2020 at 9:29
  • $\begingroup$ I presume that $E$ is meant to be the population mean of $X$, i.e. $E=E[X]$ and similarly $Var$ is meant to be $\text{Var}[X]$ $\endgroup$
    – Glen_b
    Commented Nov 8, 2021 at 2:05

1 Answer 1




Now the distribution of $Y_1+Y_2$ is normal (and straightforward), so $E(e^{Y_1+Y_2})$ is just the expectation of a univariate lognormal.

The $E(X_1)E(X_2)$ term you can already do.

As a result, it's straightforward to write $\text{Cov}(X_1,X_2)$ in terms of $\mu,\sigma$ and $\rho$ and thereby to solve for $\rho$.

$Y_1+Y_2\sim N(\mu_1+\mu_2,\sigma_1^2+\sigma_2^2+2\rho\sigma_1\sigma_2)$, so

$e^{Y_1+Y_2}\sim logN(\mu_1+\mu_2,\sigma_1^2+\sigma_2^2+2\rho\sigma_1\sigma_2)$,

which has expectation $\exp[\mu_1+\mu_2+\frac{1}{2}(\sigma_1^2+\sigma_2^2+2\rho\sigma_1\sigma_2)]$.


So $\text{Cov}(X_1,X_2)=E(X_1)E(X_2)[\exp(\rho\sigma_1\sigma_2)-1]$

And hence:



You can extend this approach to calculating $\rho_{ij}$ from $\text{Cov}(X_i,X_j)$ and the other quantities.

However, if you're trying to do this to estimate parameters from a sample, using sample moments of a lognormal to do parameter estimation (i.e. method-of-moments) doesn't always perform all that well. (You might consider MLE if you can.)

  • $\begingroup$ It seems from your formulas for the covariance of two log-normal RVs that when $\rho=1$, the covariance does not reduce to the product of two standard deviations, i.e. $\sqrt{Var(X_1)Var(X_2)}$, and so the correlation between the log-normals is not 1. Is this correct? What would $\rho$ have to be for the log-normals to be 100% correlated then? $\endgroup$
    – Confounded
    Commented Aug 25, 2020 at 9:25
  • $\begingroup$ In general when $\rho=1$ the relationship between the lognormals is curved and so the Pearson correlation won't generally reach 1. You'd need $\rho=1$ and $\sigma_1=\sigma_2$ for it to be straight. $\endgroup$
    – Glen_b
    Commented Aug 25, 2020 at 10:15
  • $\begingroup$ In short, the Pearson correlation of the $X$'s is different from the Pearson correlation of the $Y$'s (the latter is denoted $\rho$ above) $\endgroup$
    – Glen_b
    Commented Sep 22, 2022 at 1:01

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