2
$\begingroup$

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 $\mathcal{N}(0,1)$.

  • 0
    Please check your spelling.2011-12-13

1 Answers 1