# Convergence of F(n,n) distribution to normal

Suppose that $X \sim F(n,n)$, an F distribution on $n$ and $n$ degrees of freedom. I'm trying to figure out why some literature state that $X$ converges in distribution to a normal distribution.

My confusion lies in the fact that $X$ can be written as a ratio of two chi-squares,

$$X = \frac{\chi_n^2/n}{\chi_n^2/n}$$

However, it is known that $\chi_n^2/n$ converges in probability to $1$. So by Slutsky's theorem, shouldn't the above converge to a degenerate distribution at $1$?

• An argument that it converges to a normal would require us to standardize in some suitable way; it'll be a standardized variate that converges to a normal. You'll still want to use Slutsky in there as well, I'd say. Jul 8, 2017 at 0:52
• $\sqrt{n/4}(X-1)$ converges to a standard Normal distribution.
– whuber
Jul 8, 2017 at 1:54
• @whuber Thank you. I tried reasoning out the asymptotic distribution in my edited post. Please let me know if there is anything missing in my argument. Jul 8, 2017 at 2:43
• @Flowsnake I suggest you take the "edit" you inserted in your question and post it as an answer and accept it too (it is totally within the rules of the site). In this way, the thread will be taken out of the "unanswered" queue. Jul 30, 2017 at 21:38
• Thank you, I will do that, although it feels strange accepting my own answer. Jul 31, 2017 at 3:07

Let $(X_1,..,X_n)$ and $(Y_1,..,Y_n)$ be independent chi square samples with 1 degree of freedom. Then by the CLT,
$$\sqrt{n}(\bar{X} - 1) \rightarrow N(0,2)$$ $$\sqrt{n}(\bar{Y} - 1) \rightarrow N(0,2)$$
It follows from the Delta Method with $g(x,y) = \frac{x}{y}$ and $\nabla g(x,y) = (\frac{1}{y} , -\frac{x}{y^2})$, that,
$$\sqrt{n}(g(\bar{X},\bar{Y}) - g(1,1)) \rightarrow N(0,4)$$
But $g(\bar{X},\bar{Y}) = \dfrac{\frac{1}{n}\sum\limits_{i=1}^nX_i}{\frac{1}{n}\sum\limits_{i=1}^nY_i} \sim F_{n,n}$. So, $\sqrt{n}(F - 1) \rightarrow N(0,4)$ as @whuber showed in the comments.