I've been studying a statistics textbook that makes a claim that the $ \chi^2 $ distribution with $ k $ degrees of freedom has a mode at $ k - 2 $ without proof. (Wikipedia seems to agree) Why is this? Is there a geometric, or even algebraic way, to understand this statement?

  • 1
    $\begingroup$ Look at the wikipedia page for mode. The mode occurs at the highest peak of a continuous distribution which can be found using calculus. $\endgroup$ May 3, 2016 at 0:44
  • $\begingroup$ True. That doesn't seem very intuitive, though. (I'll be honest, taking a derivative of the gamma function is pretty nasty; I can't say I can do it. Thank goodness Wolfram Alpha can.) $\endgroup$ May 3, 2016 at 1:01
  • $\begingroup$ You don't need to take the derivate of the Gamma function, since it is not a function of $x$. $\endgroup$ May 3, 2016 at 1:03
  • $\begingroup$ Good point; $ k $ is a parameter. $\endgroup$ May 3, 2016 at 1:06

2 Answers 2


The pdf of a $\chi^2_k$ distribution is, $$f(x) = 2^{-k/2} \Gamma{(k/2)}^{-1} x^{k/2 - 1}e^{-x/2}. $$

We need to find $x^*$ such that $x^* = \arg \max_\limits{x > 0} f(x)$. Then $x^*$ is the mode. Note that $\arg \max_\limits{x > 0} f(x) = \arg \max_\limits{x > 0} \log f(x)$, so we will find the mode by maximizing the log of the pdf instead of maximizing the pdf (this turns out to be easier).

\begin{align*} \log f(x) &= -\dfrac{k}{2} \log 2 - \log \Gamma(k/2) + \left(\dfrac{k}{2} - 1 \right) \log x - \dfrac{x}{2}\\ \dfrac{\log f(x)}{dx} &= \left(\dfrac{k}{2} - 1 \right) \dfrac{1}{x} - \dfrac{1}{2} \overset{set}{=} 0\\ \Rightarrow x^* &= k-2 \end{align*}

Thus we get that the mode is $x^* = k-2$. If $k \leq 2$, then the mode is $0$, since the $\chi^2$ pdf in that case is decreasing on the positives.

EDIT: To verify that the second derivate is negative, look at @MatthewGunn's comment below.

  • 6
    $\begingroup$ Elaborating a bit, the 2nd derivative of $\log f$ is $-\left(\frac{k}{2} - 1 \right) x^{-2}$ hence, the pdf is log-concave everywhere iff $k \geq 2$. Thus the first order condition $x^* = k - 2$ achieves a maximum when $k \geq 2$. $\endgroup$ May 3, 2016 at 1:28
  • $\begingroup$ +1 Thank you for this. I couldn't find it on multiple online searches. $\endgroup$ May 3, 2016 at 5:26
  • 1
    $\begingroup$ +1. Plus, we can of course disregard the $2^{-k/2}\Gamma(k/2)^{-1}$ multiplicative constant from the very start (although it also drops out in the differentiation). $\endgroup$ May 3, 2016 at 6:30
  • $\begingroup$ I can't disagree that you're right. I guess I should have phrased it as a question about intuition, rather than a geometry/algebra argument. I'll accept this answer though since it answers the question. :) $\endgroup$ May 4, 2016 at 1:06

The solution by Greenparker is correct and the double derivative can be proved to be less than 0. Substitute x=k-2 -> k=x+2 in the 2nd derivative solution obtained (i.e. -(((x+2)/2)-1)*x^-2)

Thus we get, (-x/2)*(x^-2) which is negative. hence double derivative is negative.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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