The Weibull pdf is given as the following.

$$f(x) = \begin{cases} \frac{1}{\alpha}mx^{m-1}e^{-x^m/\alpha} \quad \text{if } x > 0 \\ 0 \quad \text{else} \end{cases}$$

If a random sample of size $n$ is taken from a Weibull distribution, what is the pdf and cdf of the minimum of the sample, $Y_i$?

I realize $g_1(y_1) = n[1-F(y_1)]^{n-1}f(y_1)$; however how do I find the cdf $F(y_1)$? I tried integrating the pdf from 0 to $y$ but am having difficulty.


Let $X_1,X_2,\dots,X_n$ be iid Weibull, and let $W$ be the minimum of the $X_i$.

Then $W\gt w$ if and only if all the $X_i$ are $\gt w$. This has probability $(1-F_X(w))^n$, where $F_X$ is the cdf of one of the Weibulls. Thus the cdf of $W$ is given by $$F_W(w)=1-\Pr(W\gt w)=1-(1-F_X(w))^n$$ (for positive $w$).

To find $1-F_X(w)$, we need to calculate $$\int_w^\infty \frac{1}{\alpha}mx^{m-1}\exp(-x^m/\alpha)\,dx.$$ Make the substitution $\frac{x^m}{\alpha}=t$. Then $dt=\frac{1}{\alpha}mx^{m-1}\,dx$, and the integral becomes the easy $$\int_{t=w^m/\alpha}^\infty e^{-t}\,dt.$$

  • $\begingroup$ How would I find the pdf of the minimum of the sample? $\endgroup$ – idealistikz Jun 13 '13 at 6:29
  • $\begingroup$ I have shown you how to find the cdf (cumulative distribution function). To find the probability density function, differentiate. We get $nf_X(w)(1-F_X(w))^{n-1}$. But your post seemed to indicate you knew that. $\endgroup$ – André Nicolas Jun 13 '13 at 6:42

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