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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2$\begingroup$ That's not the reason that the ratio of independent normals has no (central) moments. Indeed $\mathbb P(Y = 0) = 0$ in both cases. $\endgroup$– cardinalCommented Jan 26, 2012 at 1:21
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$\begingroup$ What are the reasons that the ratio of independent normals has no (central) moments? $\endgroup$– KuJCommented Jan 26, 2012 at 3:25
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2$\begingroup$ There's too much probability near the origin of the denominator. $\endgroup$– whuber ♦Commented Jan 26, 2012 at 4:14
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1 Answer
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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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2$\begingroup$ 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. :) $\endgroup$– cardinalCommented Jan 26, 2012 at 3:03
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$\begingroup$ 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? $\endgroup$– KuJCommented Jan 26, 2012 at 5:31
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5$\begingroup$ The coefficient of the covariance term should be $-2$ not $+2$. Your formula gives the variance of the sum, not the difference. $\endgroup$ Commented Jan 26, 2012 at 6:49
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2$\begingroup$ I have made what I believe to be a couple of minor typo corrections. Please make sure I haven't, instead, introduced extra ones. :) $\endgroup$– cardinalCommented Jan 26, 2012 at 13:15
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$\begingroup$ @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... $\endgroup$– jbowmanCommented Jan 26, 2012 at 16:04