Given $X_1$ and $X_2$ normal random variables with correlation coefficient $\rho$, how do I find the correlation between following lognormal random variables $Y_1$ and $Y_2$?

$Y_1 = a_1 \exp(\mu_1 T + \sqrt{T}X_1)$

$Y_2 = a_2 \exp(\mu_2 T + \sqrt{T}X_2)$

Now, if $X_1 = \sigma_1 Z_1$ and $X_2 = \sigma_1 Z_2$, where $Z_1$ and $Z_2$ are standard normals, from the linear transformation property, we get:

$Y_1 = a_1 \exp(\mu_1 T + \sqrt{T}\sigma_1 Z_1)$

$Y_2 = a_2 \exp(\mu_2 T + \sqrt{T}\sigma_2 (\rho Z_1 + \sqrt{1-\rho^2}Z_2)$

Now, how to go from here to compute correlation between $Y_1$ and $Y_2$?


1 Answer 1


I assume that $X_1\sim N(0,\sigma_1^2)$ and $X_2\sim N(0,\sigma_2^2)$. Denote $Z_i=\exp(\sqrt{T}X_i)$. Then

\begin{align} \log(Z_i)\sim N(0,T\sigma_i^2) \end{align} so $Z_i$ are log-normal. Thus

\begin{align} EZ_i&=\exp\left(\frac{T\sigma_i^2}{2}\right)\\ var(Z_i)&=(\exp(T\sigma_i^2)-1)\exp(T\sigma_i^2) \end{align} and \begin{align} EY_i&=a_i\exp(\mu_iT)EZ_i\\ var(Y_i)&=a_i^2\exp(2\mu_iT)var(Z_i) \end{align}

Then using the formula for m.g.f of multivariate normal we have

\begin{align} EY_1Y_2&=a_1a_2\exp((\mu_1+\mu_2)T)E\exp(\sqrt{T}X_1+\sqrt{T}X_2)\\ &=a_1a_2\exp((\mu_1+\mu_2)T)\exp\left(\frac{1}{2}T(\sigma_1^2+2\rho\sigma_1\sigma_2+\sigma_2^2)\right) \end{align} So \begin{align} cov(Y_1,Y_2)&=EY_1Y_2-EY_1EY_2\\ &=a_1a_2\exp((\mu_1+\mu_2)T)\exp\left(\frac{T}{2}(\sigma_1^2+\sigma_2^2)\right)(\exp(\rho\sigma_1\sigma_2T)-1) \end{align}

Now the correlation of $Y_1$ and $Y_2$ is covariance divided by square roots of variances:

\begin{align} \rho_{Y_1Y_2}=\frac{\exp(\rho\sigma_1\sigma_2T)-1}{\sqrt{\left(\exp(\sigma_1^2T)-1\right)\left(\exp(\sigma_2^2T)-1\right)}} \end{align}

  • $\begingroup$ Note that as long as the approximation $e^x \cong 1+ x$ is valid on the final formula found above one has $\rho_{Y_1Y_2} \cong \rho$. $\endgroup$
    – danbarros
    Aug 24, 2017 at 9:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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