My Math Forum computing the mean of the ith order statistics for $n$ Rayleigh random variables
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 June 22nd, 2016, 02:55 PM #1 Newbie   Joined: Jun 2016 From: Jordan Posts: 1 Thanks: 0 computing the mean of the ith order statistics for $n$ Rayleigh random variables I am trying to compute the mean of the ith order statistics for $n$ Rayleigh random variables as follows: $\int_0^\infty A \, x \, F^{i-1} (1-F)^{n-i} \, f \, dx$ where $A = i \binom{n}{i}$ and $F=1-e^{\frac{-x^2}{2\sigma^2}}$ is the CDF of the Rayleigh RV and $f=\frac{x}{\sigma^2}e^{\frac{-x^2}{2\sigma^2}}$ is its pdf. I start by substituting $p=1-F$, changing the integral limits to between $0$ and $1$ and then substituting the values of $x=(-2\sigma^2 \ln{ P})^{\frac{1}{2}}$=$\sigma\sqrt{2}(\ln{\frac{1}{p} })^{1/2}$. The integral becomes: $$A \sqrt{2} \sigma\int_0^1 (\ln{\frac{1}{p}})^{1/2} (1-p)^{i-1} P^{n-i} dp$$ Then, by using integration by parts: $$-A \sqrt{2} \, \sigma \, \Gamma(3/2) \bigg[ (n-i) \beta (n-i,i) - (i-1) \beta(n-i+1, i=1)\bigg]$$ where $\Gamma$ is the Gamma function and $\beta$ is the Beta function. In the integration by parts, I use $u= (1-p)^{i-1} P^{n-i}$ and $dv=(\ln{\frac{1}{p}})^{1/2} \, dp$ The problem is that I have run a simulation for these values and the results are different from the theoretical results I derived. Could you please help me in finding any mistakes I may have made? Last edited by Saud; June 22nd, 2016 at 02:58 PM.

 Tags $n$, computing, integrals, ith, order, order_statistics, random, rayleigh, statistics, variables

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