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Let $X_1, \ldots, X_n$ be random variables such that $X_i$ has Weibull distribution with shape parameter $3$ and scale parameter $\theta$ and define $X_{(1)}=\min\{X_1, \ldots ,X_n\}$. Derive $95 \%$ confidence interval for $\theta$ based on $X_{(1)}$.

I know that, but not sure where to go with it :

The PDF of Weibull distribution is $$ f(x \mid k,\theta) = \begin{cases} k \, \theta \left(x \, \theta \right)^{k-1}e^{-(x \, \theta)^{k}} & x\geq0 ,\\ 0 & x<0, \end{cases} $$ for $\theta >0$, $k > 0$ is said to have a Weibull distribution, denoted $WEI( k , \theta)$; $k$ and $\theta$ are called its shape and scale parameters, respectively.

The CDF of Weibull distribution is

$$F_{X}(x) = 1- e^{-\left(x \theta \right)^{k}}$$

Now the distribution of $X_{(1)}$ :

$$F_{X_{(1)}}(x) = \min\{\,X_1,\ldots,X_n\,\} = 1- [ 1 - F_{X}(x) ]^{n}$$

However, I am not really sure how to approach this question. I would really appreciate it if someone could give me some direction.

Edit Attempts according to @NCh

$$ f(x \mid\theta) = \begin{cases} 3 \, \theta \left(x \, \theta \right)^{2}e^{-(x \, \theta)^{3}} & x\geq0 ,\\ 0 & x<0, \end{cases} $$

The CDF of Weibull distribution is

$$F_{X}(x) = 1- e^{-\left(x \theta \right)^{3}}$$

Now, let $W= \theta \, X_{(1)}$ $$ \begin{align} F_{W}(w) &= P \left( W \le w \right) \\ &= P(X_{(1)} \le \frac{w}{\theta}) \\ &= 1- \left[ 1 - F_{X}(\frac{w}{\theta}) \right]^{n} \\ &= 1- \left[e^{-w^3} \right]^{n} \end{align} $$

What I am supposed to do with the $n$.

Now, $$ \begin{align} P\left(a \le W \le b \right) = P(W \le b) - P(W \le a)=0.95 \end{align} $$

Where to go from here. Thank you

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    $\begingroup$ Substitute $k=3$ and find cdf of $\theta X_{(1)}$. It does not depend on $\theta$ and you can then find $a$ and $b$ s.t. $\mathbb P(a\leq \theta X_{(1)}\leq b)=0.95$. $\endgroup$
    – NCh
    Mar 4, 2020 at 17:14
  • $\begingroup$ Check the CDF of $X$ from en.m.wikipedia.org/wiki/Weibull_distribution. Weibull is related to exponential, which helps in recognising the distribution of the pivot $\theta X_{(1)}$. $\endgroup$ Mar 4, 2020 at 21:18
  • $\begingroup$ Looking at the CDF of X. It looks like $EXP(X^2 \, \theta^3)$. $\endgroup$
    – ADAM
    Mar 4, 2020 at 23:49
  • $\begingroup$ Does the answer below help? $\endgroup$ Mar 8, 2020 at 15:26
  • $\begingroup$ @StubbornAtom it does help. I have been waiting for to someone vote up so I can confirm it is correct. Thanks! $\endgroup$
    – ADAM
    Mar 9, 2020 at 20:57

1 Answer 1

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Assuming $X_1,X_2,\ldots,X_n$ are independent with common distribution $\mathsf{Weibull}(3,\theta)$.

You have correctly shown that CDF of $W=\theta X_{(1)}$ is $P(W\le w)=1-e^{-n w^3}$ for $w\ge 0$.

So we can take $W$ as a pivot for deriving a confidence interval for $\theta$.

Suppose that for some $k\,(>0)$,

$$P_{\theta}(W\ge k)=P_{\theta}\left(\theta\ge \frac k{X_{(1)}}\right)=0.95\quad,\,\forall\,\theta>0$$

That is, $k$ is so chosen that $$e^{-nk^3}=0.95 $$

This gives $k^3=-\frac1n\ln(0.95)$, or $k\approx\frac{0.371553}{n^{1/3}}$ (taking the real solution).

So a (one-sided) $95\%$ confidence interval for $\theta$ is simply $\left[\frac k{X_{(1)}},\infty\right)\approx\left[\frac{0.371553}{n^{1/3}X_{(1)}},\infty\right)$.


If you want a two-sided confidence interval, you can use the pivot $2nW^3\sim \chi^2_2$. (Note that $W^3$ has an exponential distribution with mean $1/n$.)

Suppose $\chi^2_{\alpha,2}$ is the upper $100\alpha\%$ point of $\chi^2_2$ distribution, i.e. $P(\chi^2_2>\chi^2_{\alpha,2})=\alpha$ for $\alpha\in(0,1)$.

Then

$$P_{\theta}\left[\chi^2_{1-\alpha/2,2}\le 2nW^3\le\chi^2_{\alpha/2,2}\right]=1-\alpha\quad,\forall\,\theta>0$$

In your case, this reduces to

$$P_{\theta}\left[\left(\frac{\chi^2_{0.975,2}}{2n X_{(1)}^3}\right)^{1/3}\le \theta\le \left(\frac{\chi^2_{0.025,2}}{2n X_{(1)}^3}\right)^{1/3}\right]=0.95\quad,\forall\,\theta$$

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