I've been searching for weeks now but I can't find a proof for the following relationship between the quantiles of the chi-squared distribution and the quantiles of the standard normal distribution:

$$chi^2_{n;q} approx frac{1}{2}(z_q + sqrt{2n-1})^2$$

$z_q$ is the q-quantile of the standard normal distribution, whereas n are the degrees of freedom.

How do we come to this conclusion?

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#### Best Answer

A slight variation on this approximation can be derived using the **delta-method** via the fact that the chi-squared distribution converges to the normal distribution. Although both are asymptotically valid, an alternative is mentioned by Fisher, without subtracting one from the part in the square-root (i.e., as $chi_{n,q}^2 approx tfrac{1}{2} (z_q + sqrt{2n})^2$). Your approximation approaches the true critical point values from above, and the Fisher approximation approaches the true critical point values from below. The one you mention is more accurate in terms of the relative error, and it is probably derived as a variation from the one used by Fisher.

**Derivation:** It is well-known that the chi-squared distribution converges to the normal distribution as $n rightarrow infty$. More specifically, if $chi_{n}^2$ has a chi-squared distribution with $n$ degrees-of-freedom then we have the following convergence-in-distribution:

$$frac{chi_{n}^2-n}{sqrt{2n}} overset{text{Dist}}{longrightarrow} text{N}(0,1) quad quad quad text{as } n rightarrow infty.$$

(More generally, the gamma distribution converges to the normal as the shape approaches infinity.) Hence, with a simple change in the multiplying constant we have the corresponding result:

$$sqrt{2n} Big( frac{chi_{n}^2}{n}-1 Big) overset{text{Dist}}{longrightarrow} text{N}(0,4).$$

We now apply the delta-method using a continuously differentiable transformation $g$, to give us an alternative asymptotic result:

$$sqrt{2n} Big( g Big( frac{chi_{n}^2}{n} Big) – g(1) Big) overset{text{Dist}}{longrightarrow} text{N} Big( 0, 4 g'(theta)^2 Big).$$

Using the transformation $g(theta) = sqrt{theta}$ gives us $g'(theta)^2 = 1/4theta$, which yields the alternative result:

$$sqrt{2n} Big( sqrt{frac{chi_{n}^2}{n}} – 1 Big) overset{text{Dist}}{longrightarrow} text{N}(0, 1).$$

Hence, for large $n$ we have:

$$sqrt{2 chi_{n}^2} -sqrt{2 n} overset{text{Approx}}{sim} text{N}(0, 1).$$

(This approximation is mentioned in Fisher (1934) (p. 62), though he does not show the derivation.) Now, from the definition of the quantile you have:

$$begin{equation} begin{aligned} q &approx mathbb{P}( sqrt{2 chi_{n}^2} -sqrt{2 n} geqslant z_q) \[8pt] &= mathbb{P}( sqrt{2 chi_{n}^2} geqslant z_q + sqrt{2 n}) \[8pt] &= mathbb{P}( 2 chi_{n}^2 geqslant (z_q + sqrt{2 n})^2) \[8pt] &= mathbb{P} Big( chi_{n}^2 geqslant frac{1}{2} (z_q + sqrt{2 n})^2 Big). \[8pt] end{aligned} end{equation}$$

Hence, we have the approximate quantile:

$$chi_{n,q}^2 approx frac{1}{2} (z_q + sqrt{2 n})^2.$$

Your variation can be derived from the asymptotic result $sqrt{2 chi_{n}^2} -sqrt{2 n-1} overset{text{Approx}}{sim} text{N}(0, 1)$, which is also asymptotically valid, since the effect of the minus-one is vanishing as $n rightarrow infty$. By an analogous derivation, this alternative asymptotic form gives:

$$chi_{n,q}^2 approx frac{1}{2} (z_q + sqrt{2 n-1})^2.$$

**Remark:** The approximation used by Fisher differs slightly from the approximation in your question, since it does not subtract one in the part in the square root. Both are valid asymptotically, since the effect of subtracting one is vanishing as $n rightarrow infty$. It is worth noting that the approximation in your question is more accurate than the one used by Fisher, in terms of relative error.

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