# 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?

## 1 Answer

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 $1/2$.

• To be clear, I restricted myself at the end to a simple question about which order statistic it will be. Are you ever seeing that the order statistic can (with higher probability) be the minimum? It seems to me that there might be a good argument for this. – Helmut Jan 31 '15 at 1:41
• Yes. In my simulations, for very small sample sizes (e.g., $n = 2$), it is more likely that $|X_{(1)}/k - 1| > |X_{(n)}/k - 1|$. This appears to be true for all degrees of freedom, although as $k \to \infty$, this probability tends to $1/2$. – heropup Jan 31 '15 at 2:42
• I observed the asymptotic result, but not the small-k result. Do you have a sense for why it happens for large k? – Helmut Jan 31 '15 at 2:49
• After more investigation, it appears that only when $n \in \{2, 3\}$ that $\Pr[X_\alpha = X_{(1)}] > 1/2$, regardless of $k$. – heropup Jan 31 '15 at 2:53
• What happened to the gamma and incomplete gamma functions in your derivation of the order statistics? – Helmut Jan 31 '15 at 18:35