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I have a problem in numerically evaluating the PDF of $Y=\min(X_1,X_2,\cdots,X_N)$ where $N=\binom{M}{K}$, the binomial coefficient and $X_i$s are iid Chi-square random variables.

The CDF of $Y$ is $F_Y(y)=1-[1-F_X (y)]^N$, where $F_X (y)$ is the CDF of an $X_i$.

Then the PDF of $Y$ is $P_Y (y)=\frac{d}{dy}F_Y (y)=N [1-F_X (y)]^{N-1}P_X (y)$.

I have tired to plot $P_Y (y)$ for $M=100$ and $K=10$, I do not have any problem. When $K$ increases keeping $M$ as constant, the PDF of $P_Y (y)$ exhibit a sharp transition from 0 to 1 in the lower tail and the area under the PDF does not equal to 1 numerically. However, analytically the area can be shown to be equal to 1, that is, $\int_{0}^{\infty} P_Y (y) dy=1.$

My questions are

  1. Why does the area does not equal to one when N increases? Is there a discontinuity happening as N increases?

We note that $0 \leq F_X (y) \leq 1$. Then, the term $[1-F_X (y)]^{N-1}$ in $P_Y (y)$ for large $N$ becomes a binary threshold function. Is there is a problem with the binary threshold function such as discontinuity?

Please help me, I do not know where I am making mistake.

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  • $\begingroup$ Your problem with the computation are clearly numerical. As a first try, defgine the tail function $G_Y(y)= 1-F_Y(y)=(1-F_X(y))^N$. Then you can first try to calculate the logarithm of the tail function! $\endgroup$ Mar 25, 2014 at 9:47
  • $\begingroup$ Thank you. I will try an let you know. $\endgroup$
    – Oliver
    Mar 25, 2014 at 10:36
  • $\begingroup$ As per your suggestion, $ln G_Y (y) = N*ln(1-F_X (y))$. Since $(1-F_X (y))$ lies between o and 1, $ln(1-F_X (y))$ is an negative number. But I still have to multiply by $N$, a very large number. any other idea. I still have the difficulty with the numerical computation. $\endgroup$
    – Oliver
    Mar 27, 2014 at 6:15
  • $\begingroup$ Lets say $\ln(1-F_X(y))$ is $-10^{3}$ and $N=10^n$ is very large. The product becomes $-10^{3+n}$. That exponent $3+n$ must become around $400$ (or a little smaller) before that is out of the dynamical range of dpuble precision numbers! That is a very large $N$! How large is your $N$? If it is really that large, you need to use higher-precioson floats. Doubt that. $\endgroup$ Mar 27, 2014 at 8:26
  • $\begingroup$ Thank you for your comments. My $N=\binom{M}{K}$, where $M=1000$ and $K=20$. The problem happens even for $M=100$ and $K=15$. Today I saw something about extreme order statistics. Do you know anything about that? does it help in my case? $\endgroup$
    – Oliver
    Mar 27, 2014 at 10:30

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