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 $1$?

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    $\begingroup$ 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. $\endgroup$ – Glen_b Jul 8 '17 at 0:52
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    $\begingroup$ $\sqrt{n/4}(X-1)$ converges to a standard Normal distribution. $\endgroup$ – whuber Jul 8 '17 at 1:54
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    $\begingroup$ @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. $\endgroup$ – Flowsnake Jul 8 '17 at 2:43
  • $\begingroup$ @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. $\endgroup$ – Alecos Papadopoulos Jul 30 '17 at 21:38
  • $\begingroup$ Thank you, I will do that, although it feels strange accepting my own answer. $\endgroup$ – Flowsnake Jul 31 '17 at 3:07

Sorry for the bump, but I will do what was suggested in the comments and post this as answer instead of editing the question:

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.


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