I'm interested to see a result where for large degree of freedom $m,$ the [chi distribution](https://en.wikipedia.org/wiki/Chi_distribution) $\chi\_m$ is increasingly well approximated by a family of normal distributions with parameters depending on $m.$ The motivation comes from the fact that for large $m, \chi^2\_m \sim \mathcal{N}(m,2m)$ approximately and the difference between the corresponding CDF's approach zero as the d.f. $m \to \infty$, as asked and proved in [this question](https://stats.stackexchange.com/questions/429782/why-is-the-limit-of-a-chi-squared-distribution-a-normal-distribution/483807#483807). The problem is: we can't apply some sort of [continuous mapping theorem](https://www.statlect.com/asymptotic-theory/continuous-mapping-theorem) here to pass onto square root. I said "some sort of" because we'd be looking here for a case where $X\_m - Y\_m \to\_d 0 \implies g(X\_m) - g(Y\_m)\to\_{d}0$ with $g$ being a continuous, real-valued function (in our case $X\_m:=\chi^2\_m, Y\_m:=\mathcal{N}(m,2m), g:= \sqrt{}$), and this is **not** the continuous mapping theorem.


Regardless, the [wiki page of chi-squared distribution](https://en.wikipedia.org/wiki/Chi-square_distribution#Asymptotic_properties) seems to refer to this theorem, proved by [Ronald Fisher](https://en.wikipedia.org/wiki/Ronald_Fisher):


**Approximately,** $\sqrt{2\chi^2\_m} \sim \mathcal{N}(\sqrt{2m-1},1) \implies \chi\_m \sim \mathcal{N}(\sqrt{m-\frac{1}{2}},\frac{1}{\sqrt{2}}).$


Despite the reference to a book on that wiki page, I'm unable to access it and am also unable to otherwise get a reference to this theorem. Do you happen to know the proof or could you refer me to it please? Thank you!



Asymptotics of $\chi_m$-distribution where the degree of freedom $m \to \infty?$
Such a result can be obtained by the so-called [delta method](https://en.wikipedia.org/wiki/Delta_method), which yields, in particular, the following: if $X\_m\sim\chi^2\_m$, then the distribution of $\sqrt X\_m$ is approximately $N(\sqrt m,1/\sqrt2)$ (for large $m$).


*Details:* By the central limit theorem, $\overline X\_m:=X\_m/m$ is approximately normal with mean $\mu:=1$ and standard deviation $\sqrt{2/m}$. Hence, by the delta method, for $g(x)\equiv\sqrt x$, $\sqrt{\overline X\_m}$ is approximately normal with mean $g(\mu)=\sqrt1=1$ and standard deviation $|g'(\mu)|\sqrt{2/m}=1/\sqrt{2m}$. Thus, $\sqrt X\_m=\sqrt m\,\sqrt{\overline X\_m}$ is approximately normal with mean $\sqrt m$ and standard deviation $1/\sqrt2$, as claimed.




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On the uniform and nonuniform bounds on the rate of convergence to normality in the (possibly multivariate) delta method, see [this and references there](https://projecteuclid.org/euclid.ejs/1460463653).


The convergence of the distribution of $\sqrt X\_m$ to normality is in a sense monotonic; cf. [formula (2.6)](https://projecteuclid.org/euclid.aos/1176325373).


