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I searched the internet alot . The only relevant clue is in Wikipedia:
F-distribution
Beside that I didn't find any proof for this theory.

If $Y$ has $B\left(\frac{d_1}{2}, \frac{d_2}{2}\right)$ distribution, then show that $X$ with given formula has $F\left(d_1,d_2\right)$ distribution.
$$X = \frac{d_2Y}{d_1\left(1-Y\right)}$$

I tried this but I can't do anything because of $\Gamma{}$ integral:
$$Y = \frac{\Gamma\left({\frac{d_1 + d_2}{2}}\right)}{\Gamma\left(\frac{d_1}{2}\right)\Gamma\left(\frac{d_2}{2}\right)}x^{\frac{d_1}{2} - 1}\left(1-x\right)^{\frac{d_2}{2}-1}$$

so:
$$X = \frac{d_2Y}{d_1\left(1-Y\right)}$$ $$X = \frac{d_2\frac{\Gamma\left({\frac{d_1 + d_2}{2}}\right)}{\Gamma\left(\frac{d_1}{2}\right)\Gamma\left(\frac{d_2}{2}\right)}x^{\frac{d_1}{2} - 1}\left(1-x\right)^{\frac{d_2}{2}-1}}{d_1\left(1-\frac{\Gamma\left({\frac{d_1 + d_2}{2}}\right)}{\Gamma\left(\frac{d_1}{2}\right)\Gamma\left(\frac{d_2}{2}\right)}x^{\frac{d_1}{2} - 1}\left(1-x\right)^{\frac{d_2}{2}-1}\right)}$$

$$X = x^{\frac{d_1}{2} - 1}\frac{\Gamma\left({\frac{d_1 + d_2}{2}}\right)}{\Gamma\left(\frac{d_1}{2}\right)\Gamma\left(\frac{d_2}{2}\right)}\left(\frac{d_2\Gamma\left(\frac{d_1}{2}\right)\Gamma\left(\frac{d_2}{2}\right)\left(1-x\right)^{\frac{d_2}{2}-1}}{d_1\Gamma\left(\frac{d_1}{2}\right)\Gamma\left(\frac{d_2}{2}\right) - \Gamma\left({\frac{d_1 + d_2}{2}}\right)x^{\frac{d_1}{2} - 1}\left(1-x\right)^{\frac{d_2}{2}-1}}\right)$$

compare it to $F\left(d_1,d_2\right)$:

$$F\left(d_1,d_2\right)=x^{\frac{d_1}{2} - 1}\frac{\Gamma\left({\frac{d_1 + d_2}{2}}\right)}{\Gamma\left(\frac{d_1}{2}\right)\Gamma\left(\frac{d_2}{2}\right)}\left(\frac{d_1}{d_2}\right)^{\frac{d_1}{2}-1}\frac{1}{\left(1+\frac{d_1}{d_2}x\right)^{\frac{d_1+d_2}{2}}}$$

so:

$$\left(\frac{d_1}{d_2}\right)^{\frac{d_1}{2}-1}\frac{1}{\left(1+\frac{d_1}{d_2}x\right)^{\frac{d_1+d_2}{2}}} = \left(\frac{d_2\Gamma\left(\frac{d_1}{2}\right)\Gamma\left(\frac{d_2}{2}\right)\left(1-x\right)^{\frac{d_2}{2}-1}}{d_1\Gamma\left(\frac{d_1}{2}\right)\Gamma\left(\frac{d_2}{2}\right) - \Gamma\left({\frac{d_1 + d_2}{2}}\right)x^{\frac{d_1}{2} - 1}\left(1-x\right)^{\frac{d_2}{2}-1}}\right)$$

I can't go on anymore because of $\Gamma$ integral!

If possible, give me a hint to prove this.
Any help will be appreciated.

jameselmore
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Amin
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  • The assertion is wrong, one should center the Xs to get convergence in distribution. – Did Dec 14 '18 at 07:11
  • My English is not good. can you explain more? do you mean question is wrong? or my answer? if my answer is wrong, can you hint me for correct way? I use this question in my prove (in the question text): https://math.stackexchange.com/questions/843660/square-root-of-chi-square-distribution-tends-to-n0-1 – Amin Dec 14 '18 at 07:20
  • The question is wrong. It seems you miscopied the other page. – Did Dec 14 '18 at 07:33
  • It's my teacher question. Are you sure that is wrong? do you mean that my prove is correct but my teacher question is wrong? – Amin Dec 14 '18 at 07:35
  • Please see previous comment. Whether I am sure or not (and yes, I am sure), do you understand the reason in my first comment why the question must be wrong? – Did Dec 14 '18 at 07:36
  • I didn't understand your reason in first comment, because of that, I ask to explain more. My English is not good. I'm sorry. – Amin Dec 14 '18 at 07:39
  • Sorry but my impression is that this is not a language problem as much as you would like us to believe... What does the CLT assert, already? – Did Dec 14 '18 at 07:49
  • Let us continue this discussion in chat. – Amin Dec 14 '18 at 07:55

1 Answers1

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I ask my teacher and it was wrong. The question is:

Problem

Show that: If $X_1,X_2,...,X_n$ be independent random variable with $\chi^2_1$ (df=1) and $W_n=X_1+X_2+...+X_n$ then $$\lim \limits_{n \to \infty}{\frac{(W_n - n)\sqrt{n}}{n\sqrt{2}}} \sim N_{(0,1)}$$

Answer

based on my last proof in question:
I start with this fact that $\chi^2_1$ has $\mu=1$ and $\sigma^2=2$ and based on CLT (central limit theorem), when $n \to \infty$ we have: $$W_n=X_1+X_2+...+X_n \sim N_{(n\mu, n\sigma^2)} \sim N_{(n,2n)}$$ (based on my teacher lesson, we write $\sigma^2$ in second paramter of $N$).

So we know that: $$Z=\frac{W_n-n}{\sqrt{2n}} \sim N_{(0,1)}$$ and we can write it: $$W_n=\sqrt{2n}N_{(0,1)}+n$$

Now, we go to solve the limit: $$\lim \limits_{n \to \infty}{\frac{(W_n - n)\sqrt{n}}{n\sqrt{2}}} = \lim \limits_{n \to \infty}{\frac{(\sqrt{2n}N_{(0,1)})\sqrt{n}}{n\sqrt{2}}} = N_{(0,1)}\lim \limits_{n \to \infty}{\frac{(\sqrt{2n})\sqrt{n}}{n\sqrt{2}}} = N_{(0,1)}$$

proof finished!
I'm sorry but it is a mistake from my teacher. However thanks for help.

Amin
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