That is, if X is a (asymptotic) normal random variable, is $ln\left(\frac{X}{1-X} \right)$ also (asymptotically) normally distributed?

From this question, i suppose it isn't the case but this paper seems to state otherwise (although i'm not sure i've understood it as it is)

at page 137 is the proof that the logit transformation of the relative frequency of a binomial random variable (which, asymptotically follows normal distribution) is asymptotically normally distributed.

what i've understood it says is as follows: let $X\sim B(n,p)$. Then as n goes to infinity, the relative frequency $f=\frac{X}{n} \sim N\left(p,\frac{p(1-p)}{n}\right)$ asymptotically. Since the logit estimater $\hat L = ln(\frac{f}{1-f})$ is a function of $f$ which has a continuous second-order derivative in an interval that contains $f=p$ as an interior point, $\hat L$ is asymptotically normally distributed with a mean equal to the true logit $L=ln(\frac{p}{1-p})$ and a variance equal to the asymptotic variance of $f$ multiplied by the square of the derivative of $\hat L$ with respect to $f$ evaluated at $f=p$.

So basically i don't understand how the fact that $\hat L$ is a function of an asymptotically normal $f$ also makes $\hat L$ asymptotically normal, and also where the asymptotic variance being as such came from. What am I missing here?

  • $\begingroup$ The variances of the estimators decrease to zero, implying eventually you can approximate $x\to\log(1/(1-x))$ as a linear function. This approximation requires only that this function be differentiable, with nonzero derivative, in a neighborhood of the limiting value of the estimator. $\endgroup$
    – whuber
    Feb 9, 2020 at 18:08
  • $\begingroup$ @whuber Oh.....so is it like ln(x/(1-x)) is asymptotically normal but only as the distribution of ln(x/(1-x)) degenerates to a constant? $\endgroup$ Feb 10, 2020 at 7:49

1 Answer 1


This can be shown by applying twice in a row the Delta method, first starting from $f_n$ and considering the non-linear function $h_n = h(f_n) = f_n/(1-f_n)$ and then examining the non-linear function $g(h_n) = \ln(h_n)$.

Just remember that "asymptotic distribution" usually refers to a properly centered and scaled function of the core quantity, not the core quantity itself. As $n$ goes to infinity $f_n$ will collapse to a constant. It is the function $\sqrt{n}(f_n - p)$ that will maintain a non-degenerate distribution at the limit.


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