Fisher convergence as sample goes to infinity.

I was wondering how does the Fisher law behave when $n => \inf.$. $F_{q, n-p} = \frac{SSE_0-SSE/q}{SSE/(n-p)}$ I expect the test statistics goes lower as n goes up but does the pdf of the law converges to something particular like a gaussian or something ?

An $F$ distribution with degrees of freedom $m$ and $n$ is the ratio of two independent random variables. The numerator is $Q_m$ with $mQ_m \sim \mathsf{Chisq}(m).$ That is, $Q_m$ is a chi-squared random variable divided by its degrees of freedom. Similarly, the denominator is $Q_n$ with $nQ_n \sim \mathsf{Chisq}(n).$
As $n \rightarrow \infty,$ the denominator random variable $Q_n$ converges in probability to $1.$ Thus, as $n \rightarrow \infty$ and $m$ is held fixed, $F = Q_m/Q_n$ converges in distribution to $Q_m.$
The figure below shows the density function of $\mathsf{F}(5, 1000)$ in black and the density function of $\mathsf{Chisq}(5)$ divided by $5$ as a broken red curve. Of course $n = 1000$ is not quite $\infty,$ but (I hope) large enough to make the point.
The figure below shows a sequence of densities of chi-squared distributions, divided by degrees of freedom, to illustrate increasing concentration about $1$ as DF increases. Degrees of freedom are 10, 20, 50, 100, and 500, plotted in blue, green, orange, red, and black, respectively.
• So that means that as n goes to infinity the denominator basically becomes 1 but the numerator gets to be $X1/n1$ $with X1 = Chisq(n1)$ ? It looks like it's converging to a gaussian seeing the black curve. (altough not centered nor scaled) Jan 4, 2018 at 16:29