# Deriving confidence interval for scale parameter of Weibull distribution

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

• 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. $\mathbb P(a\leq \theta X_{(1)}\leq b)=0.95$.
– NCh
Mar 4, 2020 at 17:14
• 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 $\theta X_{(1)}$. Mar 4, 2020 at 21:18
• Looking at the CDF of X. It looks like $EXP(X^2 \, \theta^3)$.
Mar 4, 2020 at 23:49
• Does the answer below help? Mar 8, 2020 at 15:26
• @StubbornAtom it does help. I have been waiting for to someone vote up so I can confirm it is correct. Thanks!
Mar 9, 2020 at 20:57

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

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