# Chi distribution approaches the normal distribution?

Wikipedia states here that the chi distribution $$\chi _{k}$$ with $$k$$ degrees of freedom converges to the standard normal distribution

$$\lim _{k\to \infty }{\tfrac {\chi _{k}-\mu _{k}}{\sigma _{k}}}{\xrightarrow {d}}\ N(0,1)\,}$$

I cannot find any references or equivalent statements on the internet. Is this statement true and, if yes, does someone know a reference or a proof?

• See here, which (since $\chi_1^2$ has a mean of 1 and variance of 2) implies $\frac{\chi_k-\sqrt{k-1/2}}{1/2}$ converges in distribution to a standard normal distribution $\mathcal{N}(0,1)$ as $k$ increases. Dec 18, 2020 at 11:36

The $$x\ge0$$ PDF $$\tfrac{1}{2^{k/2-1}\Gamma(k/2)}x^{k-1}e^{-x^2/2}$$ is asymptotic to $$e^{-S}$$ with$$S:=x^2/2-(k-1)\ln x+(k/2-1)\ln2+\ln\Gamma(k/2).$$As per Laplace's method, for large $$k$$ we can approximate the log-PDF as a quadratic, based on its first two derivatives. Since $$S_x=x-\tfrac{k-1}{x}$$ is $$0$$ at $$x=\sqrt{k-1}$$, and $$S_{xx}=1+\tfrac{k-1}{x^2}>0$$, this value of $$x$$ minimizes $$S$$ and maximizes the PDF. At this value of $$x$$,$$S=(k-1)/2-\tfrac12(k-1)\ln(k-1)+(k/2-1)\ln2+\ln\Gamma(k/2)$$and $$S_{xx}=2$$, so in general$$S\approx (k-1)/2-\tfrac12(k-1)\ln(k-1)+(k/2-1)\ln2+\ln\Gamma(k/2)+\left(x-\sqrt{k-1}\right)^2,$$and we've approximated the $$\chi_k$$ distribution as $$N(\sqrt{k-1},\,\tfrac12)$$. This Gaussian approximation can be restated as $$\frac{\chi_k-\mu_k}{\sigma_k}$$ being approximately Gaussian, with mean $$0$$ and variance $$1$$, and hence $$\approx N(0,\,1)$$.