4
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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 $g'(\theta)^2 = 1/4\theta$, 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.

$\endgroup$
0
$\begingroup$

Do you have a reference? It is probably an empirical approximation, which seems to be quite good. We make a plot to look at the quality of approximation, for one specific quantile (0.95):

enter image description here

For the record, here is the code used for plotting:

qapprox  <-  function(n, p=0.95) 0.5*(qnorm(p)+sqrt(2*n-1))^2
plot(qchisq(0.95, 1:100), qapprox(1:100), xlab="exact", ylab="approximate")
 abline(0, 1, col="red")
 text(80, 20, "axis labels are degrees of freedom")
 title("Comparison exact and approximate quantiles\nfor chisquare distribution")
$\endgroup$
0
$\begingroup$

Fisher gives this approximation in his monograph "Statistical Methods for Research Workers" p62 - I am not sure where it is derived though.

Zar gives other approximations - see Jerrold H. Zar "Approximations for the Percentage Points of the Chi-Squared Distribution" Journal of the Royal Statistical Society. Series C (Applied Statistics), Vol. 27, No. 3 (1978), pp. 280-290

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.