# What are the mean and variance of the ratio of two lognormal variables?

Ratios of two normally distributed variables (e.g X/Y) have no moments (e.g. means and variances) because Y can have zero values. However, lognormal variables have no zero values. How can I calculate the mean and variance of the ratio of two lognormal variables?

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That's not the reason that the ratio of independent normals has no (central) moments. Indeed $\mathbb P(Y = 0) = 0$ in both cases. – cardinal Jan 26 '12 at 1:21
You have asked nine questions so far, several of them with multiple answers. You should consider going back through them and accepting the best answer for each question provided that at least one suitable one is present. This can be done by clicking on the check mark to the left of the answer. – cardinal Jan 26 '12 at 1:23
What are the reasons that the ratio of independent normals has no (central) moments? – Jinn-Yuh Guh Jan 26 '12 at 3:25
There's too much probability near the origin of the denominator. – whuber Jan 26 '12 at 4:14

Note that $\log(X/Y) = \log(X) - \log(Y)$. Since $X$ and $Y$ are lognormally distributed, $\log(X)$ and $\log(Y)$ are Normally distributed.

I'll assume that $\log(X)$ and $\log(Y)$ have means $\mu_X$ and $\mu_Y$, variances $\sigma^2_X$ and $\sigma^2_Y$, and covariance $\sigma_{XY}$ (equal to zero if $X$ and $Y$ are independent) and are jointly normally distributed. The difference $Z$ is then normally distributed with mean $\mu_Z = \mu_X - \mu_Y$ and variance $\sigma^2_Z = \sigma^2_X + \sigma^2_Y - 2\sigma_{XY}$.

To get back to $X/Y$, note that $X/Y = \exp Z$, showing that $X/Y$ is itself lognormally distributed with parameters $\mu_Z$ and $\sigma^2_Z$. The relationship between the mean and variance of a lognormal variate and the mean and variance of the corresponding normal variate is:

$\mathbb E(X/Y) = \mathbb E e^Z = \exp \{\mu_Z + \frac{1}{2}\sigma^2_Z \}$

$\mathrm{Var}(X/Y) = \mathrm{Var}(e^Z) = \exp \{2\mu_Z + 2\sigma^2_Z\} - \exp \{2\mu_Z + \sigma^2_Z\} \>.$

This can be rather easily derived by considering the moment-generating function of the normal distribution with mean $\mu_Z$ and variance $\sigma^2_Z$.

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A somewhat important point here is that $(\log X, \log Y)$ be jointly normally distributed. As you essentially note, there is a contrast here with the case of constructing a Cauchy by ratios of normals in that the result in your answer holds in much greater generality than the latter. :) – cardinal Jan 26 '12 at 3:03
1. Should your formula be used (instead of the naïve estimator such as the arithmetic mean and naïve variance of X/Y) if I want to use the ratio variable (X/Y) as an independent variable in linear regression analyses? If yes, how to do it by the statistical softwares? 2. Should X/Y ever be used as a dependent variable in linear regression analyses? – Jinn-Yuh Guh Jan 26 '12 at 5:31
The coefficient of the covariance term should be $-2$ not $+2$. Your formula gives the variance of the sum, not the difference. – probabilityislogic Jan 26 '12 at 6:49
I have made what I believe to be a couple of minor typo corrections. Please make sure I haven't, instead, introduced extra ones. :) – cardinal Jan 26 '12 at 13:15
@probabilityislogic - Thanks, and cardinal too (+1 both), I was a little hurried when I did it. Should have not hit the post button. A lesson to be learned... – jbowman Jan 26 '12 at 16:04