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It is known that a variable $X$ has a log normal distribution if $\ln(X)$ is normally distributed. What is the distribution of $\ln(X)$ if is $X$ is normally distributed? Can someone provide a reference? I haven't been able to google my way to an answer.

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    $\begingroup$ You might first think about $\ln x $ when $ x < 0 $. That might provide either some illumination or the opportunity to modify your question. $\endgroup$
    – cardinal
    Dec 1 '14 at 17:34
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Every normal distribution has a finite proportion of its distribution that lies below 0 (i.e. $P(X\leq 0)>0$).

For the transformation to be meaningful, you'd have to avoid trying to take the log there, which means either modifying the "$\log$" part or modifying the "normal" part (e.g. by truncating everything at or below zero).

enter image description here

$\quad$ (This is a particular example, a $N(3.3,1.5^2)$ with everything at 0 or below truncated.)

Once you've done one or the other, it would then be necessary to work out the distribution of the transformed variable. In the case of the log of a truncated normal variable, you can write down an expression for the density by taking the density of a truncated normal and then applying the usual formula for the density of a transformed variable $g(y) = f(x(y)) \left|\frac{dx}{dy}\right|$. It doesn't seem to be especially enlightening in this case:

$f(x)=\frac{1}{\sqrt{2\pi}\,\sigma\,[1-\Phi(\frac{0-\mu}{\sigma})]} \exp(-\frac{1}{2\sigma^2}(x-\mu)^2)\,;\quad x>0,\sigma>0$

$\hspace{1cm}Y=\log(X)$, $X=\exp(Y)$, $dx=\exp(y)\,dy$

$g(y) = f(x(y)) \left|\frac{dx}{dy}\right|$

$\hspace{1cm}= \frac{1}{\sqrt{2\pi}\sigma\,[1-\Phi(\frac{0-\mu}{\sigma})]} \exp(y-\frac{1}{2\sigma^2}(\exp(y)-\mu)^2)\,;\quad-\infty< y<\infty\,.$

enter image description here

And here's some simulated values obtained by truncating a large sample from a $N(3.3,1.5^2)$ density and taking logs, for comparison:

enter image description here

(There's actually a longish very thin tail that extends down far enough that you get sample values further left than shown here, but I trimmed that so you could see the bulk of the density more clearly for comparison.)

The precise shape will depend on how much of the density was truncated, of course.

To my knowledge it doesn't have a particular name, at least not any in common use.

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