# Kahn Pseudo Normal distribution

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?

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:

(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.

• 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? May 10 '17 at 13:58
• 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 $U=1-V$. May 10 '17 at 14:06