This is probably a silly question, but I do not have a ready answer for it, so I thought I'd get some opinions on it.

Let Y ~ Normal($ {\bf \unicode[Times]{x3Bc}} $, $ {\bf \unicode[Times]{x3C3}}^2 $). We take a large random sample of 1000 observations from Y. We want to make an inference about the value of $ {\bf \unicode[Times]{x3C3}}^2 $. The standard approach is to make use of the result that the standardized sample variance below has a chi-square distribution with n-1 degrees of freedom.

enter image description here

What if instead, we did the following. We use the fact that for large n, a binomial distribution is well approximated by the Normal distribution, and use the sample at hand to infer the value of the probability parameter p of the approximating Binomial distribution. In other words, we know that Y is Normally distributed, but (for large enough n) there must be a Binomial distribution that approximates this Normal distribution very well. Usually this fact is used in the other direction: Normal approximation to the Binomial instead of the Binomial approximation to the Normal, but nothing says we cannot use this fact in the other direction. Finally, knowing that the variance of the binomial is np(1-p), using basic algebra we have ourselves an interval estimate of the population variance, without relying on the chi-square distribution.

My questions are:

  1. Does this methodology not make sense because I made a logical error somewhere or misstated some fact?
  2. Assuming this methodology makes sense, is it not used because it is more tedious than using the chi-squared distribution or because it produces inferior estimates of the population variance? If it is more tedious, which step is?
  • 3
    $\begingroup$ Re "there must be a Binomial distribution that approximates:" Not so. You need to qualify that as a "possibly scaled and/or shifted Binomial distribution." That ought to make clear that this avenue of investigation is fruitless. $\endgroup$
    – whuber
    Commented Apr 10, 2020 at 16:50
  • 1
    $\begingroup$ See my brief 'answer' for a slight restatement of @whuber's Comment. $\endgroup$
    – BruceET
    Commented Apr 10, 2020 at 17:33
  • $\begingroup$ Thank you both. Interesting. I learned something. $\endgroup$ Commented Apr 11, 2020 at 14:45

1 Answer 1


Consider a normal distribution with $\mu = 100, \sigma^2 = 144, \sigma =12.$ Then the relationships for mean and variance of the 'matching' binomial would be $\mu = np, \sigma^2 = np(1-p)$ so that $\sigma^2/\mu = (1-p) = 144/100 > 1,$ which would lead to impossible negative $p.$

Also, even in cases where you could find binomial $n$ and $p,$ any inference about $\sigma^2$ would be only approximate. By contrast, inferences about $\sigma^2$ that are based on the relationship $$\frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(n-1)$$ are "optimal" is carefully defined and practically useful senses.

In statistics and probability there are many useful relationships between binomial distributions and their normal approximations, but this is not one of them.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.