3

What is $F_{k,\infty}$, i.e., $F$ distribution when the second degree of freedom approaches infinity? I'm wondering if there is a known distribution(such as $\chi_k^2$) that it converges to.

ExcitedSnail
  • 2,768
  • 3
    See this discussed in the following answers:
    1. https://stats.stackexchange.com/questions/216053/why-does-the-j-test-yield-a-chi2-distribution/216055#216055

    2. https://stats.stackexchange.com/questions/140876/distribution-of-ratio-of-2-chi-squared/140928#140928 (including mention of it following from Slutsky's theorem)

    3. https://stats.stackexchange.com/questions/278293/what-do-you-do-if-your-degrees-of-freedom-goes-past-the-end-of-your-tables/278294#278294

    – Glen_b Apr 15 '22 at 06:41

1 Answers1

6

The F-distribution is the distribution of the ratio of two independent scaled chi-squared random variables, so we have the general form $F_{k_1,k_2} \sim (k_2 \chi^2_{k_1})/ (k_1 \chi^2_{k_2})$. As $k_2 \rightarrow \infty$ we have the limit $\chi^2_{k_2}/k_2 \rightarrow 1$, so if we apply Slutsky's theorem we see that the F-distribution converges to a scaled chi-squared distribution:

$$F_{k_1,k_2} \rightarrow \frac{\chi^2_{k_1}}{k_1}.$$

Note that the scaled chi-squared distribution is an extremely useful distribution in statistical analysis (so much so that it should probably have taken the place of the chi-squared distribution; see discussion here).


Direct Proof: The above heuristic explanation uses the properties of the Snedecor F distribution based on its derivation from chi-squared random variables. However, it is simple to give a direct proof the above result by taking limits of the density kernel. The density kernel for the Snedecor F distribution can be written as:

$$\begin{align} \text{Snedecor-F}(x|k_1,k_2) &\propto \frac{1}{x} \sqrt{\frac{k_1^{k_1} k_2^{k_2} x^{k_1}}{(k_1x+k_2)^{k_1+k_2}}} \\[6pt] &\propto x^{k_1/2 - 1} \bigg( \frac{k_2}{k_1x+k_2} \bigg)^{(k_1 + k_2)/2} \\[6pt] &= x^{k_1/2 - 1} \bigg( 1 + \frac{k_1x}{k_2} \bigg)^{-(k_1 + k_2)/2} \\[6pt] &= x^{k_1/2 - 1} \bigg( 1 + \frac{k_1+k_2}{k_2} \cdot \frac{k_1x}{2} \cdot \frac{1}{(k_1+k_2)/2} \bigg)^{-(k_1 + k_2)/2}. \\[6pt] \end{align}$$

Taking $k_2 \rightarrow \infty$ we then obtain the limit:

$$\begin{align} \text{Snedecor-F}(x|k_1,k_2) &\propto x^{k_1/2 - 1} \bigg( 1 + \frac{k_1+k_2}{k_2} \cdot \frac{k_1x}{2} \cdot \frac{1}{(k_1+k_2)/2} \bigg)^{-(k_1 + k_2)/2} \\[6pt] &\rightarrow x^{k_1/2 - 1} \bigg( 1 + \frac{k_1x}{2} \cdot \frac{1}{(k_1+k_2)/2} \bigg)^{-(k_1 + k_2)/2} \\[6pt] &\rightarrow x^{k_1/2 - 1} \exp \bigg( - \frac{k_1 x}{2} \bigg) \\[12pt] &\propto \text{ScaledChiSq}(x|k_1). \\[6pt] \end{align}$$

As can be seen, in the limit the kernel of the Snedecor F density approaches the kernel of the scaled chi-squared density. Since convergence of the density kernels implies convergence in distribution, this is sufficient to prove that the Snedecor F distribution converges to the scaled chi-squared distribution. $\blacksquare$

Ben
  • 124,856
  • 2
    Thank you so much! Is there a reference or proof for this? – ExcitedSnail Apr 15 '22 at 04:38
  • 2
    @ExcitedSnail A reference for this fact (without proof) would be Wikipedia. – frank Apr 15 '22 at 06:33
  • 2
    Proof follows from https://en.wikipedia.org/wiki/Slutsky's_theorem – Glen_b Apr 15 '22 at 06:41
  • @Ben Thank you very much! This is very helpful! A last question: why $\frac{\chi^2_{k_2}}{k_2}\rightarrow 1$, it seems not quite intuitive. – ExcitedSnail Apr 17 '22 at 08:26
  • @Glen_b Thank you very much! Still have have some difficulty seeing the convergence to 1. – ExcitedSnail Apr 17 '22 at 08:27
  • @frank Thanks a lot. This is very helpful! – ExcitedSnail Apr 17 '22 at 08:28
  • Consider its mean and variance. More formally, what happens to the left and right limits of $F_n$ the cdf of that scaled chi square r.v. in the sequence as $n\to\infty$ – Glen_b Apr 18 '22 at 00:56
  • I have added a direct proof of this result using the density function of the F distribution; this is the simplest method of proof I'm aware of, and it does not require use of the intermediate properties of the distribution. – Ben Apr 18 '22 at 02:00