(This is not a homework question.)

Let $(X_1 \sim N(\mu_1,\sigma_1), X_2 \sim N(\mu_2, \sigma_2))$ be a bivariate normal random variable with the correlation between $X_1$ and $X_2$ given by $\rho$. Let $Y_i = \exp(X_i)$ be a log-normal variable. What is the correlation between $X_1$ and $Y_2$?

The hard part of this seems to be calculating $E(X_1Y_2)$ in order to compute the covariance. Using the PDF for a bivariate normal distribution, one way to get at this is:

$$\begin{split}E(X_1Y_2) & = \int_{-\infty}^\infty \int_{-\infty}^\infty f(x,y)\cdot x \cdot e^y\:\mathrm{d}x\:\mathrm{d}y \\ &= \int_{-\infty}^\infty \int_{-\infty}^\infty \frac{x}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}\cdot\exp\left(\frac{-z}{2(1-\rho)^2} + y\right)\:\mathrm{d}x\:\mathrm{d}y \\ &=\text{?} \end{split}$$


$$z = \frac{(x-\mu_1)^2}{\sigma_1^2} - \frac{2\rho(x-\mu_1)(y-\mu_2)}{\sigma_1\sigma_2} + \frac{(y-\mu_2)^2}{\sigma_2^2}$$

I've found sources (e.g. this paper, or this question) tackling the correlation between two log-normal variables, and others (e.g. this question) addressing the correlation between $X_i$ and $Y_i$ (i.e. between a normal variable and its own exponent), but none tackling the more general case of the latter (i.e. the correlation between $X_i$ and $Y_{j\neq i}$).

I'm not currently mathematically knowledgeable enough to adapt the derivations given in those other cases to this one by myself, and I'm not sure how to finish solving the integral above without getting buried in ever-expanding chains of brute-force calculus, so any help finding a smarter approach would be much appreciated.

  • 1
    $\begingroup$ Without any loss of generality you may assume $\mu_1=\mu_2=0$ and $\sigma_1=1,$ because these effect only changes of location and scale, leaving the correlation unaltered. This simplifies the algebra of completing the square. My comment at your last reference, stats.stackexchange.com/questions/274853/…, readily applies to your case and makes short work of the calculations. $\endgroup$
    – whuber
    Jun 3, 2020 at 12:26

1 Answer 1


As is often the case, precisely formulating the question helped me work out the answer.

My approach makes use of the marginal expectation of the bivariate normal:

$$E_X(y) = E(X|Y=y) = \mu_x + \rho\frac{\sigma_x}{\sigma_y}(y-\mu_y)$$

Returning to my notation from the question above, this gives us:

$$\begin{split}E(X_1Y_2) & = \int_{-\infty}^\infty \int_{-\infty}^\infty f(x,y)\cdot x \cdot e^y\:\mathrm{d}x\:\mathrm{d}y = \int_{-\infty}^\infty e^y \left(\int_{-\infty}^\infty x \cdot f(x,y) \:\mathrm{d}x\right)\mathrm{d}y\\ &= \int_{-\infty}^\infty e^y \cdot h(y) \cdot E_X(y)\:\mathrm{d}y \:=\,\int_{-\infty}^\infty e^y \cdot h(y) \cdot \left[\mu_1 + \rho\frac{\sigma_1}{\sigma_2}(y-\mu_2)\right]\:\mathrm{d}y\\ &= \mu_1\int_{-\infty}^\infty e^y \cdot h(y)\:\mathrm{d}y + \rho\frac{\sigma_1}{\sigma_2}\int_{-\infty}^\infty y \cdot e^y \cdot h(y)\:\mathrm{d}y - \mu_2\rho\frac{\sigma_1}{\sigma_2}\int_{-\infty}^\infty e^y \cdot h(y)\:\mathrm{d}y\\ &= \mu_1E(Y_2) + \rho\frac{\sigma_1}{\sigma_2}E(X_2Y_2) - \mu_2\rho\frac{\sigma_1}{\sigma_2}E(Y_2) \end{split}$$

The answer to a previous question gives us $E(X_2Y_2) = (\mu_2 + \sigma_2^2) \cdot E(Y_2)$, which gives us:

$$\begin{split}E(X_1Y_2) & = \left[\mu_1 + \rho\frac{\sigma_1}{\sigma_2}(\mu_2 + \sigma_2^2) - \mu_2\rho\frac{\sigma_1}{\sigma_2}\right] \cdot E(Y_2) \\ &= \left[\mu_1 + \mu_2\rho\frac{\sigma_1}{\sigma_2} + \rho\sigma_1\sigma_2 - \mu_2\rho\frac{\sigma_1}{\sigma_2}\right] \cdot E(Y_2) \\ &= \left[\mu_1 + \rho\sigma_1\sigma_2\right] \cdot E(Y_2) \end{split}$$

where $h(x)$ is the marginal PDF of $X_2 \sim N(\mu_2,\sigma_2)$. This then gives us

$$\begin{split}\mathrm{Cov}(X_1,Y_2) & = E(X_1Y_2) - E(X_1)E(Y_2) \\ &= \left[\mu_1 + \rho\sigma_1\sigma_2\right] \cdot E(Y_2) - \mu_1 \cdot E(Y_2)\\ &= \rho\sigma_1\sigma_2 \cdot E(Y_2) \end{split}$$

And hence

$$\begin{split}\mathrm{Corr}(X_1,Y_2) & = \frac{\mathrm{Cov}(X_1,Y_2)}{\mathrm{sd}(X_1)\cdot\mathrm{sd}(Y_2)} = \frac{\rho\sigma_1\sigma_2 \cdot E(Y_2)}{\sigma_1\cdot\mathrm{sd}(Y_2)} = \boldsymbol{\rho\sigma_2\frac{E(Y_2)}{\mathrm{sd}(Y_2)}} \end{split}$$

The formulae for $E(Y_2)$ and $\mathrm{sd}(Y_2)$ can be found here.

  • $\begingroup$ That final expression can be simplified using $E[Y_2]/sd(Y_2) = \sqrt{\exp(\sigma_2^2)-1}$. The only parameters that matter are $\rho$ and $\sigma_2$. $\endgroup$ Dec 9, 2023 at 9:28

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