# Conditional expectation between lognormal random variables

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?

• Are you assuming that $\log (X)$ and $\log(Y)$ are jointly normal? Commented Sep 7, 2022 at 15:42
• I haven't thought about that, but I believe yes. If i have two normals, then they should be jointly normal because any linear combination gives me a normal by reproducing property of the normals, right? Commented Sep 7, 2022 at 15:46
• Actually no, and that's the tricky part. The classic counterexample is: take $U$ a normal random variable, and define $V := U \cdot (2B-1)$, with $B$ a Bernoulli$\big(\frac{1}{2}\big)$. You can show that $V$ is also normal, but $U+V$ is most certainly not as it has a $\frac{1}{2}$ probability of being $0$. Look for "copulas" for other examples in this vein where each marginal distribution is normal. The "jointly normal" case is a rare thing among the possibilities Commented Sep 7, 2022 at 16:39
• These three assertions are true: 1: Any linear combinations of a pair of jointly normal variables is normal. 2: Two independent normal variables are jointly normal. 3: Two normal variables are not necessarily jointly normal. I assume your textbook assumes either joint normality or independence (which implies joint normality) Commented Sep 8, 2022 at 12:27
• I see, yes, that makes perfect sense. I missed independence or joint normality assumption. Thanks Commented Sep 8, 2022 at 12:47

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}
• @DavidMark this is just a standard modelling scenario (esp. in finance) in which we can find the closed form of $E[Y|X]$ easily because of how we constructed the joint distribution of $\ln(X),\ln(Y)$. I am not aware of general results of that kind for the multivariate lognormal distribution. Commented Sep 9, 2022 at 9:46