My question is whether the conditional expectation between lognormal random variables $Y$ and $X$, i.e $\mathbb{E}(Y|X)$ has a closed form linear (or non-linear) expression similar to Gaussian random variables. Recall that if $(Z,W)$ are jointly normal, then $\mathbb{E}(Z|W)=\beta_0 +\beta_1W$, where $\beta_0=\mathbb{E}(Z)-\beta_1\mathbb{E}(W)$ and $\beta_1=\frac{Cov(Z,W)}{Var(W)}$.
Can we use this result by transforming lognormals to normals and then after using the result for normals, transform it back to logs?
Best Answer
Suppose that we have $$X=e^{\mu_X+\sigma_XZ_1},\,Y=e^{\mu_Y+\sigma_Y\rho Z_1+\sigma_Y\sqrt{1-\rho^2}Z_2}$$ where $Z_1,Z_2\sim \mathcal{N}(0,1)$ are independent, $\mu_X,\mu_Y\in \mathbb{R}$, $\sigma_X,\sigma_Y>0$ and $\rho \in (-1,1)$. Then $\ln(X)\sim \mathcal{N}(\mu_X,\sigma_X^2)$, $\ln(Y)\sim \mathcal{N}(\mu_Y,\sigma_Y^2)$ and $\textrm{Cov}[\ln(X),\ln(Y)]=\sigma_X\sigma_Y\rho$. Also note that $$Z_1=\frac{\ln(X)-\mu_X}{\sigma_X}$$ So we can write, thanks to the independence of $Z_1$ and $Z_2$ and the fact that $X=f(Z_1)$ is independent of $Z_2$, $$\begin{aligned}E[Y|X]&=E[e^{\sigma_Y\sqrt{1-\rho^2}Z_2}]e^{\mu_Y+\frac{\sigma_y}{\sigma_X}\rho (\ln(X)-\mu_X)}=\\ &=e^{\frac{1}{2}\sigma_Y^2(1-\rho^2)+\mu_Y+\frac{\sigma_y}{\sigma_X}\rho (\ln(X)-\mu_X)}\end{aligned}$$