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Supose that $S_n$ has a $\chi^2$ distribution with $n$ degrees of freedom. Show that $$ \mathbb{P}(S_n \le x) = f\left(\sqrt{2x}-\sqrt{2n}\right) $$ where $f(u)$ is the normal distribution.

I tried this: $\mathbb{P}(S_n \le x)= \mathbb{P}(\sqrt{2 S_n}-\sqrt{2 n} <= \sqrt{2x}-\sqrt{2n})$ now I am trying to show that $\sqrt{2 S_n} - \sqrt{2n}$ converges in law to $\mathcal{N}(0,1)$.

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Please check your spelling. –  Rasmus Dec 13 '11 at 21:07

1 Answer 1

A result due to Fisher states that the distribution of $\sqrt{2S_n}$ is approximately normal with mean $\sqrt{2n-1}$ and unit variance. Hence $\sqrt{2S_n}=\sqrt{2n-1}+Z_n$ where the distribution of $Z_n$ is approximately standard normal. Let $G$ denote the standard normal CDF. This means that $$ [S_n\leqslant x]=[\sqrt{2S_n}\leqslant \sqrt{2x}]=[Z_n\leqslant \sqrt{2x}-\sqrt{2n-1}], $$ hence $$ \mathrm P(S_n\leqslant x)\approx G(\sqrt{2x}-\sqrt{2n-1}). $$

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