Noting that, for a lognormally distributed random variable $x$,

$$Var[x]=E[x]^2(e^{\sigma_x^2}-1) \tag{1}$$

(where $\sigma_x$ is the standard deviation of $\log(x)$)

And, in general,

$$Var[x]=E[x^2]-E[x]^2 \tag{2}$$

So that, setting (1) and (2) equal, we can say

$$E[x]^2(e^{\sigma_x^2}-1)=E[x^2]-E[x]^2 \tag{3}$$ $$E[x]^2 e^{\sigma_x^2}-E[x]^2=E[x^2]-E[x]^2 \tag{4}$$ $$\rightarrow E[x]^2 e^{\sigma_x^2}=E[x^2] \tag{5}$$

I am tempted to extrapolate to the covariance as follows:

$$Var(x) = Cov(x,x) = E[x]E[x]e^{\sigma_x \sigma_x}-E[x]E[x]$$

$$Cov[x,y]=E[x]E[y]e^{\sigma_x \sigma_y}-E[x]E[y]$$

($y$ being another lognormally distributed random variable correlated with $x$)

which, using the same logic as in the case of variance, would imply the following joint expected value:

$$\rightarrow E[x]E[y] e^{\sigma_x \sigma_y} = E[xy]$$

Is this a valid conclusion? I suspect it is not, but please point out why. What am I missing?


Or perhaps

$$Cov[x,y]=E[x]E[y]e^{\sigma_x \sigma_y \rho_{xy}}-E[x]E[y]$$ $$\rightarrow E[x]E[y] e^{\sigma_x \sigma_y \rho_{xy}} = E[xy]$$

Where $\rho_{xy}$ is the correlation of $\log(x)$ with $\log(y)$

  • $\begingroup$ You should not fall for that temptation. Observe that your covariance is not depending on a correlation coefficient, so that it makes no distinction between $x,y$ independent (hence covariance $=0$, or $x,y$ (highly) dependent (e.g. $x=y$). $\endgroup$ – drhab Nov 7 '17 at 16:10
  • $\begingroup$ @drhab Ok, what about $E(x)E(y)e^{\sigma_x \sigma_y \rho_{xy}} = E[xy]$? $\endgroup$ – ben Nov 7 '17 at 18:48
  • $\begingroup$ Sounds better, but I don't know and have not the opportunity to check right now. Maybe later. $\endgroup$ – drhab Nov 7 '17 at 18:53

What you mention in your edit is correct.

Here another path to find it, that gives me more comfort.

Let $R_1,R_2$ have joint normal distribution with $R_i\sim\text{Norm}(\mu_i,\sigma_i^2)$ and $\rho=\rho(R_1,R_2)$, i.e. the correlation coefficient of $R_1,R_2$.

Note that $R_1+R_2\sim\text{Norm}(\mu_1+\mu_2,\sigma_1^2+\sigma_2^2+2\rho\sigma_1\sigma_2)$

Let $X=e^{R_1}$ and $Y=e^{R_2}$.

Then $X\sim\text{LogNormal}(\mu_1,\sigma_1^2)$ and $Y\sim\text{LogNormal}(\mu_2,\sigma_2^2)$.

We have $XY=e^{R_1+R_2}$ and conclude that $XY\sim\text{LogNormal}(\mu_1+\mu_2,\sigma_1^2+\sigma_2^2+2\rho\sigma_1\sigma_2)$

It is well-known and not difficult to prove that $e^{\mu+\frac12\sigma^2}$ is the expectation of a random variable that has lognormal distribution with parameters $\mu$ and $\sigma^2$.

This leads easily to: $$\mathsf EXY=e^{\mu_1+\mu_2+\frac12\sigma_1^2+\frac12\sigma_2^2+\rho\sigma_1\sigma_2}=\mathsf EX\mathsf EYe^{\rho\sigma_1\sigma_2}$$and of course:$$\text{Cov}(X,Y)=\mathsf EX\mathsf EY(e^{\rho\sigma_1\sigma_2}-1)$$

Remark: your notations $\sigma_x,\sigma_y,\rho_{x,y}$ can confuse (that's why I used indices). Someone might think easily that $\sigma_x^2$ stands for the variance of $x$. This while in your question it stands for the variance of $\ln(x)$.


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