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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 don't think it is the useful here.

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The Central Limit Theorem won't be useful here since the random variable of interest $M_n := \max (X_1, \ldots, X_n)$ is not the sum/average of a series of random variables but rather their maximum.

The study of the maximum of a sequence of random variables is part of extreme value theory. The cornerstone result is the Extremal Types Theorem (also known as Fisher-Tippett-Gnedenko theorem) which provides the three possible non-degenerate limit distribution (note the slightly different role of $a_n$ compared to your notation). These three distribution are grouped in the Generalized extreme value (GEV) distribution. So, the Extremal Types Theorem is essentially the analogous of the Central Limit Theorem for maxima, where the non-degenerate limit, if it exists, is the GEV distribution.

One way to find the normalising sequences $a_n$ and $b_n$ is to use the sufficient (but not necessary) von Mises conditions.

Let $F$ denote the distribution function of $X_i, i=1, \ldots, n$. For a sufficiently smooth distribution $F$ with upper terminal $x_F = \sup\{ x : F(x) < 1 \}$ , define the reciprocal hazard function as $$ r(x) = \frac{1-F(x)}{f(x)} , $$ where $f$ denotes the density function. Then, with $b_n = F^{-1}(1 - 1/n)$, $a_n = r(b_n)$, and $\xi = \lim_{x\to x_F} r'(x)$ the distribution of $(M_n - b_n)/a_n$ is GEV with shape parameter $\xi$.

The von Mises condition could be used to find the sequences $a_n$ and $b_n$ but computations for the $\chi_1^2$ distribution might be painful.

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  • $\begingroup$ Thank you for the information. To clarify what does $F^{-1}$ stand for? Inverse or $1/F(x)$ ? $\endgroup$ – Salih Ucan Jul 23 '13 at 10:19
  • $\begingroup$ The notation $F^{-1}(x)$ denotes the inverse of $F(x)$. $\endgroup$ – QuantIbex Jul 23 '13 at 10:54

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