Question about asymmetry of chi-square distribution Let $X_1,\dots,X_n$ be a set of i.i.d. chi-square random variables with $k$ degrees of freedom. Consider the statistic $\arg\max_i\{|X_i/k - 1|\}=X_{\alpha}$. I wonder about the probability that $X_{\alpha} = X_{\text{max}}$, rather than $X_{\text{min}}$? Does it change if $k$ gets large?
I suspect that it may be very, very difficult to get a precise answer to the above questions. 
So I would like to ask the simpler question: can we at least say that it is more likely that $X_{\alpha} = X_{\text{max}}$? Simulations seem to indicate yes. Is there a heuristic reason why?
 A: From my simulations, it is clear that the answer is going to depend strongly on $k$ and $n$.  For all $k$, as $n$ increases, $\Pr[X_\alpha = X_{(n)}]$ increases:  the probability that the given statistic is the maximum order statistic becomes increasingly likely.  The effect of increasing $k$ is to make the rate at which this probability increases as a function of $n$ slower.
Given some more thought, one might be able to prove these claims mathematically.

We can get an exact probability for the case $k = 2$ (which is exponential), and for general $n$:  the joint distribution of $(X_{(1)}, X_{(n)})$ is $$f(x,y) = \frac{1}{2}\binom{n}{2}e^{-(x+y)/2}(e^{-x/2} - e^{-y/2})^{n-2}, \quad 0 \le x < y.$$  The probability that $|X_{(1)} - 2| > |X_{(n)} - 2|$ is equivalent to $$p(n) = \Pr[X_{(1)} + X_{(n)} < 4] = \int_{x=0}^2 \int_{y=x}^{4-x} f(x,y) \, dy \, dx.$$  For specific values of $n$, we have $$\begin{align*} p(2) &= 1-3e^{-2} \\ p(3) &=  1 - 6e^{-2} + 8e^{-3} - 3e^{-4} \\ p(4) &= 1 - 6e^{-2} + 15e^{-4} + 2e^{-6} \\ p(5) &= 1 - \tfrac{20}{3}e^{-2} + 30 e^{-4} - \tfrac{128}{3}e^{-5} + 20e^{-6} - \tfrac{5}{3}e^{-8}. \end{align*}$$  Only the first two are greater than $1/2$.
