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The Wikipedia article of the log-normal distribution says

If $X$ is a random variable with a normal distribution, then $Y = \exp(X)$ has a log-normal distribution; likewise, if $Y$ is log-normally distributed, then $X = \log(Y)$ is normally distributed. (This is true regardless of the base of the logarithmic function: if $\log_a(Y)$ is normally distributed, then so is $\log_b(Y)$, for any two positive numbers $a, b ≠ 1$.)

(The general form of this has described the log-normal distribution on Wikipedia since at least 2006.)

I have two questions.

1) is it the case that the logarithm can be taken to any base (i.e., it need not be the natural log), while the exponential must have Euler's number $e$ as the number which is raised to a power? Or, is it the case that I could use other bases to describe the log normally distributed variable, $X = a^{\mu+Z\sigma}$, where $Z$ is a standard normal and $a$ is any positive, real number?

And, conditional on the answer to #1, 2) why the asymmetry?

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    $\begingroup$ Did you know that $a^x$ is defined as $\exp(x \log(a))$? This relationship provides straightforward answers to the first question. As to the second question, you would need to ask the Wikipedia authors about that. $\endgroup$ – whuber May 28 '12 at 19:59
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    $\begingroup$ I hope you don't mind that I suggested an edit to use the phrase "Euler's number" instead of Euler's constant, which is not $e$ but $\gamma$. $\endgroup$ – Neil G May 28 '12 at 20:14
  • $\begingroup$ @whuber no I didn't! Thanks. And, Neil yes the edit is fine, thanks. $\endgroup$ – Andy McKenzie May 28 '12 at 20:45
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    $\begingroup$ I updated the wikipedia page to reflect these suggestions so that people won't be confused in the future like I was. Thanks all. en.wikipedia.org/wiki/Log-normal_distribution $\endgroup$ – Andy McKenzie May 28 '12 at 21:36
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    $\begingroup$ @AndyMcKenzie: Good idea. Although, you should probably move the note from the lede to the body of the article since the lede should have the highest level information rather than details. $\endgroup$ – Neil G May 31 '12 at 1:51
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The reason that the base doesn't matter is because if $X = \log Y \sim \mathcal N(\mu, \sigma^2)$ then

\begin{align} X_b &= \log_b Y \\ &= \frac{\log Y}{\log b} \\ &= \frac{X}{\log b} \sim \mathcal N\left(\frac{\mu}{\log b}, \left(\frac{\sigma}{\log b}\right)^2\right) \end{align} is normally distributed.

Going the other way, if $Y = e^X \sim \log \mathcal N(\mu, \sigma^2)$, \begin{align} Y_a &= a^X \\ &= e^{(\log a)X} \sim \mathcal \log N\left({\mu}{\log a}, \left({\sigma}{\log a}\right)^2\right) \end{align}

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  • $\begingroup$ thanks, this helps give mathematical intuition as to the symmetry. The only part I don't quite understand is why you wrote $Y_b$ in the second part, but I suppose that doesn't matter much. $\endgroup$ – Andy McKenzie May 28 '12 at 21:04
  • $\begingroup$ @AndyMcKenzie: Good catch. $\endgroup$ – Neil G May 28 '12 at 21:06
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It is just common to use e as the exponent in the negative exponential. Certainly the distribution could be defined using any othe postive number. But the normal distribution is certainly defined in terms of e in the density. The normal density is $\dfrac{1}{\sqrt{2\pi}\sigma}\exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right]$ where $\mu$ is the mean and $\sigma$ is the standard deviation. The density would have to be written differently if you use a different base. Referring to the lognormal specifically as log base e is also a traditional way to do it though it is not the only way to do it.

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    $\begingroup$ Thnaks for going to the trouble to improve the look of my answer Procrastinator. $\endgroup$ – Michael R. Chernick May 28 '12 at 20:11

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