# Question Concerning the Invariance of a Log-Transformed Normal Random Variable under Reciprocal Transformations

So I just started looking through through E.J. Gumbel's "Statistics of Extremes" (1958) and I came across a rather strange problem that I had never seen before. The problem is phrased as follows:

(1) Show that the distribution obtained from a logarithmic transformation of a normal variate is invariant under a reciprocal transformation. (2) How is the mode of the transformed distribution related to the transformation of the initial mode?

I could understand going in the direction of

$$Y = e^{X}$$,

where $$X \sim N(\mu,\sigma)$$, to obtain the pdf of the log-normal distribution (since there is at least the special case of $$\mu = 0$$ where it is indeed invariant under reciprocal transformations); however, it seems like Gumbel wants the reader to work with

$$Y = \log{X}$$,

where $$X \sim N(\mu, \sigma)$$, to obtain the pdf of the resulting distribution and then obtain the pdf of

$$W = Y^{-1}$$

to show that we still obtain the same pdf as before. My initial thought was "wouldn't I need to do some kind of truncation to restrict X for log(X) to make sense?", but I figured that maybe Gumbel was just assuming that I would know this, so I went ahead and just assumed that we are only dealing with a truncated normal distribution that bounds X below by 0 but allows it to be arbitrarily large. The answers I'm getting don't seem to reward me with "invariance under reciprocal transformations" (I can't tell if he means that we are staying within the same family of distributions or whether I am literally still getting the exact same distribution).

Anyways, Gumbel has a tendency to use different terms for familiar concepts, so it may be that I am misunderstanding what he asks; otherwise, how is this question not fundamentally non-sensical? If I am missing something really obvious, then I would happy with any advice that I can get.

P.S.: I am aware of other books on extreme values. I just wanted to try this specific textbook out.

• You don't have to truncate anything, because $Y$ is almost surely positive. The symmetry of $Y$ is an immediate consequence of the (obvious) fact that $X$ and $2\mu-X$ have the same distribution.
– whuber
Jun 28, 2022 at 13:04
• @whuber My biggest issue here is that $X$ can be negative and the domain of a logarithmic function is positive. Even if I could get past this, I am still not convinced that $Y$ is almost surely positive. The interval $(0,1]$ will still yield non-positive values and that doesn't seem like a probability that is zero to me, though correct me if I am wrong. It's not clear to me how I avoid this issue without some form of truncation. Jun 28, 2022 at 16:20
• Although the language in the quotation is poor, it is clear that by "logarithmic transformation" of a Normal variable $X$ it intends $Y = \exp(X)$ and $X = \log(Y).$ That is the only interpretation that leads to the stated result.
– whuber
Jun 28, 2022 at 17:14
• Yeah, you are probably right. I felt the same at first, but I suppose I felt like trying to interpret the question in that way felt arrogant and out of place for someone at my skill level. I wanted to at least pose the question to people who are more experienced than myself before going in that direction to make sure that I wasn't missing something really obvious. I'd still like to keep this question up to see if anyone else comes along, but your input is greatly appreciated and I suppose I will proceed as though Gumbel was trying to nudge the reader in the direction of the log-normal dist. Jun 29, 2022 at 0:46

After reading further into the book, I am able to confirm that @whuber is right when they say that Gumbel meant $$Y = exp(X)$$ and $$X = log(Y)$$. Here is the evidence from chapter 1.1.9. The Logarithmic Normal Distribution which is on page 16 of the 2004 Dover version of the text.
Let f(y) and F(y) be the normal functions. Then the distribution g(z) and the probability G(z) obtained from the transformation $$y = lg\ z$$ are $$g(z)=\frac{1}{z \sqrt{2 \pi}} exp(-\frac{lg^{2}z}{2});\ G(z) = F(lg\ z).$$