Mode of the $ \chi^2 $ distribution 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?
 A: 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.
A: 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.
