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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!

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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

1 Answer 1

up vote 0 down vote accepted

This is in the domain of the extremal value theory and I found that a good reference for it is this book. Indeed, its authors point out the nice symmetry between the laws of large numbers and central limit theorems and extreme value theory.

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 $b_n=\ln n-\frac{1}{2}\ln\ln n-\ln\Gamma(\frac{1}{2})$.

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