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I'm interested to see a result where for large degree of freedom $m,$ the 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. The problem is: we can't apply some sort of 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 seems to refer to this theorem, proved by 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!

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    $\begingroup$ Related: stats.stackexchange.com/questions/241504/… which (since $\chi^2_1$ has a mean of $1$ and variance of $2$) implies $\dfrac{\chi_m -\sqrt{m-1/2}}{1/2}$ converges in distribution to a standard normal distribution $\mathcal N(0,1)$ as $m$ increases, the same result as you attribute to Fisher $\endgroup$ – Henry Aug 20 '20 at 14:40
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    $\begingroup$ @Henry Thanks for pointing this out - this is indeed of help here - appreciate it! $\endgroup$ – Learning math Aug 21 '20 at 10:25
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    $\begingroup$ @Henry BTW: a little typo I think: your denominator would be $1/\sqrt{2}$, not $1/2.$ $\endgroup$ – Learning math Sep 10 '20 at 14:50
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    $\begingroup$ Learning math: you are correct but I cannot edit comments. I should have written $\dfrac{\chi_m -\sqrt{m-1/2}}{\sqrt{1/2}}$ converges in distribution to a standard normal distribution $\mathcal N(0,1) $ as $m$ increases $\endgroup$ – Henry Sep 10 '20 at 15:17
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Such a result can be obtained by the so-called 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.


On the uniform and nonuniform bounds on the rate of convergence to normality in the (possibly multivariate) delta method, see this and references there.

The convergence of the distribution of $\sqrt X_m$ to normality is in a sense monotonic; cf. formula (2.6).

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  • $\begingroup$ @Ioself: thank you, I'll check out the delta method! Quick question before I do though: will this method imply that the distribution of the two CDF's converge to zero? $\endgroup$ – Learning math Aug 19 '20 at 21:23
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    $\begingroup$ @Learningmath : The uniform (Kolmogorov) distance between the two cdf's does converge to $0$, actually with rate $O(1/\sqrt m)$; stronger than this nonuniform bounds also hold. See the first linked paper of the two added ones. $\endgroup$ – Iosif Pinelis Aug 19 '20 at 21:51
  • $\begingroup$ thanks again! I do wonder if we can also say that the $\sqrt{X_m} = \chi_m$ converges in density to the density of $N_m \sim N(\sqrt{m-1/2}, 1/\sqrt{2}), i.e. lim_{m \to \infty} ||f_{\chi_m} - f_{N_m}||_{L^{\infty}(\mathbb{R})} \to 0, m \to \infty?$ Here $f_Z$ denotes the PDF of the random variable $Z.$ $\endgroup$ – Learning math Sep 4 '20 at 14:56
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    $\begingroup$ @Learningmath : Yes, the pdf convergence also holds, as can be checked directly. $\endgroup$ – Iosif Pinelis Sep 4 '20 at 21:21

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