# Asymptotics of maxima of i.i.d. chi-square random variables

How to find the following:

Let $X_1$, $X_2$, $X_3$,..., $X_n$, be i.i.d with chi-square distribution with one-degree of freedom. Find $a_n$ and $b_n$ such that $a_n(\max_i X_i - b_n)$ converges in distribution to a nondegenerate random variable.

I thought about Central limit theorem but i dont think here is the case??

Thanks a lot!

-
You need the tails of a chi-squared 1. If they were exponential (chi-square 2) the argument goes $\mathbb P (max < a) = (1-e^{-a})^n$ and to get a nontrivial limit you want $e^{-a} = x/n, a = log(n) - log(x)$. In this case $b_n = log n , a_n = 1$ and $P(max X_i - log n < x) \rightarrow e^{-e^x}$ or something similiar. Details are harder for chi-squared 1 but the idea is the same. – mike Jul 23 '13 at 16:32

Now, chi-squared distribution with one degree of freedom $\chi^2_1$ is Gamma distribution $\Gamma(1/2,2)$. The distribution of the maximum of $n$ Gamma-distributed random variables converges to Gumbel distribution as $n\rightarrow\infty$, and Table 3.4.4 (see page 156) of the aforementioned reference states that $a_n(\max Y_i-b_n)\rightarrow \Lambda$, where $Y_i\sim\Gamma(k,\theta)$, $\theta$ expresses the scale (rather than the rate) of Gamma distribution, $a_n=1/\theta$, and $b_n=\theta(\ln n+(k-1)\ln \ln n-\ln\Gamma(k))$, and $P(\Lambda\leq x)=e^{-e^{-x}}$.
Thus, in your case $a_n=2$ and $b_n=\ln n-\frac{1}{2}\ln\ln n-\ln\Gamma(\frac{1}{2})$.