Theoretical Statistics and Mathematics Unit, ISI Delhi

Weibull Distribution: Some Stochastic Comparisons Results

by Baha-Eldin Khaledi and Subhash Kochar

Let $X_1, \ldots, X_n$ be independent random variables such that $X_i$ has Weibull distribution with shape parameter $\alpha $ and scale parameter $\lambda_i$, $i = 1,\ldots, n$. Let $X_1^*, \ldots, X^*_n$ be another set of independent Weibull random variables with the same common shape parameter $\alpha $, but with scale parameters as $\mbox {\boldmath$\ \lambda ^*$}=(\lambda_1^*, \ldots, \lambda_n^*)$. Suppose that $\mbox {\boldmath$\ \lambda $} \stackrel {m} \succeq \mbox {\boldmath$\ \lambda ^*$}$. We prove that when $0 < \alpha < 1,$ $(X_{(1)}, \ldots, X_{(n)}) \stackrel {st} \succeq (X^*_{(1)}, \ldots, X^*_{(n)})$. For $\alpha \ge 1$ we prove that $ X_{(1)} \le_{hr} X^*_{(1)}$ whereas the inequality is reversed when $\alpha \le 1$. Let $Y_1, \ldots, Y_n$ be a random sample of size $n$ from a Weibull distribution with shape parameter $\alpha $ and scale parameter $\tilde{\lambda} =(\prod_{i=1}^{n}\lambda_i)^{1/n}$, the geometric mean of the 's. It is shown that $X_{(n)}\ge_{hr} Y_{(n)}$ for all values of $\alpha $ and in case $\alpha \le 1$, we also have that $X_{(n)}$ is greater than $Y_{(n)}$ according to dispersive ordering. In the process we also prove some new results on stochastic comparisons of order statistics for the proportional hazards family.

isid/ms/2004/22 [fulltext]

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