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I read about a so-called "Kahn Pseudo Normal" distribution in a forum post.

It says when $U$ is uniformly distributed over $[0,1]$, then $\log(1/U-1)$ is approximately normal distributed, that is

Averages of the ... distribution converge to a normal distribution (central limit theorem).

I searched for "Kahn Normal distribution" over the web, but found nothing relevant.

Can someone give me some more explanation or background of this method?

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This is merely the standard logistic distribution.

We can derive this in a few lines. Let $U\sim \text{Unif}(0,1)$

Let $X = 1/U - 1$.

$F_X(x)=P(X\leq x) = P(1/U-1 \leq x) = P(U\geq 1/(x+1)) = 1-1/(x+1)\,,x>0$

Hence $f_X(x) = 1/(x+1)^2\,,\: x>0\,.$ (This is a shifted Pareto(1))

Now let $Y=\log(X)$. Then

$P(Y\leq y) = P(\log(X)\leq y) = P(X \leq \exp(y)) = 1-\frac{1}{1+\exp(y)} = \frac{\exp(y)}{1+\exp(y)}$

which is the cdf of a standard logistic distribution.

(Such a method of generating values from the logistic distribution corresponds to the inverse-cdf method.)

It is symmetric but has exponential tails. It is fairly similar to a normal in the between the quartiles - or perhaps out a bit further, say the middle 75% or so - but not at all close to the normal distribution in the tails:

quantile plot of logistic vs normal

(the line has slope 1.6, approximately the ratio of the interquartile ranges)

The statement that averages converge* to a normal is hardly news; you don't need to be at all close to a normal to have that; the CLT applies quite widely.

* however the statement as given is strictly incorrect. Standardized averages converge to a standard normal by the CLT. Raw averages actually converge to a spike at 0.

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  • $\begingroup$ In some sources I see $\log(x/(1-x))$ as the inverse CDF of the logistic distribution. This differs from $\log(1/x-1)=-\log(x/(1-x))$ in a sign. Which is the correct one? $\endgroup$
    – plasmacel
    May 10 '17 at 13:58
  • $\begingroup$ I nowhere say that $\log(1/x-1)$ is the inverse cdf (it isn't), However, I do say the method corresponds to the inverse cdf method. Note that $\log(1/v-1) = \log((1-v)/v)= -\log(v/(1-v))$, Keep in mind that the standard logistic is symmetric about $0$. Alternatively (but equivalently), note that if $u=1-v$ $\log(1/v-1)=\log(u/(1-u))$, and if $V$ is uniform on $(0,1)$ then so is $U=1-V$. $\endgroup$
    – Glen_b
    May 10 '17 at 14:06

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