Question

A random variable \(X\) is said to follow the Weibull distribution with shape parameter \(\alpha\) and scale parameter \(\beta\), written \(W(\alpha, \beta)\) if its p.d.f. is given by

$$ f(x)=\frac{\alpha}{\beta^{\alpha}} x^{\alpha-1} e^{-\left(\frac{g}{3}\right)^{\alpha}} $$

for \(x>0\). The Weibull distribution is used to model lifetime of item subject to failure. If \(\alpha \in(0,1),\) it is used to model decreasing failure rate overtime, whereas if \(\alpha>1,\) one models increasing failure rate over time. It is easy to show that the c.d.f. of \(X\) is given by

$$ F(x)=1-e^{-\left(\frac{x}{b}\right)^{\alpha}} $$

(a) Show that \(Y=X^{\alpha} \sim \operatorname{EXP}\left(\beta^{\alpha}\right),\) that is \(E(Y)=\beta^{\alpha}\).

(b) Let \(X_{1}, \ldots, X_{n}\) be a random sample of size \(n\) from \(X\). What is the distribution of

$$ Q=\frac{2 n \sum_{i=1}^{n} Y_{i}}{\beta^{\alpha}} $$

Conclude that \(Q\) is a pivot.

(c) It is of interest to compare the failure rate of two different types of electronics whose lifetimes \(X\) and \(Y\) are assumed to follow \(W\left(\alpha_{1}, 2\right)\) and \(W\left(\alpha_{2}, 2\right)\) respectively. To compare the two failure times, one can compare the failure rates \(2^{\alpha_{1}}\) and \(2^{\alpha_{2}}\) via their ratio or difference. To that end, consider a random sample of size \(n_{1}\left(X_{1}, \ldots X_{n_{2}}\right)\) and \(n_{2}\left(Y_{1}, \ldots, Y_{m_{2}}\right)\) from Population 1 and 2 of electronics respectively. Show that

$$ \frac{2^{\infty_{2}} 2 n_{1} \sum_{i=1}^{n_{1}} X_{i}^{\alpha_{1}}}{2^{\alpha_{1}} 2 n_{2} \sum_{i=1}^{m_{2}} Y_{i}^{\alpha_{2}}} $$

is a pivotal quantity and give its distribution.

(d) Based on your answer in (c), derive a \((1-\alpha) \%\) confidence interval for \(\frac{2^{\infty_{2}}}{2^{\prime \prime}}\).

(e) From your answer in (d), deduce a confidence interval for \(\left(\alpha_{2}-\alpha_{1}\right)\) by passing to logarithm and interpret it.

(f) Application: A random sample of size \(n_{1}=20\) and \(n_{2}=17\) electronics from Type 1 and Type 2 electronics are taken respectively. Derive and interpret the \(95 \%\) confidence interval for \(\left(\alpha_{2}-\alpha_{1}\right)\) and interpret it. What can you really say about that interval in terms of the rate of failure of one type versus the other?

Answer 1

(a) The cdf of \(Y\) is

$$ F_{Y}(y)=P(Y \leq y)=P\left(X^{\alpha} \leq y\right)=P\left(X \leq y^{\frac{1}{\alpha}}\right)=1-\exp \left(-\frac{y}{\beta^{\alpha}}\right) ; y>0 $$

The pdf of \(Y\) is \(f_{Y}(y)=\frac{\mathrm{d} F_{Y}(y)}{\mathrm{d} y}=\frac{1}{\beta^{\alpha}} \exp \left(-\frac{y}{\beta^{\alpha}}\right) ; y>0\)

\(=0\) otherwise

Hence \(Y \sim\) Exponential distribution with \(E(Y)=\beta^{a}\)

(b) Now transform \(Y \rightarrow Z\) where \(z=\frac{2 y}{\beta^{\alpha}} ; \frac{\mathrm{d} z}{\mathrm{~d} y}=\frac{2}{\beta^{\alpha}} ; \mid\) Jacobian \(\mid=\frac{\beta^{\alpha}}{2}\)

Then the pdf of \(Z\) is \(f_{Z}(z)=\frac{1}{2} e^{-\frac{z}{2}} ; z>0\)

\(=0\) otherwise Hence \(Z=\frac{2 Y}{\beta^{\alpha}} \sim \chi_{2}^{2}\)

Since \(Y_{i}^{\prime}\) s are independent so from additive property of \(\chi^{2}\), \(Q=\frac{2 \sum_{i=1}^{n} Y_{i}}{\beta^{\alpha}}=\frac{2 n \bar{Y}_{n}}{\beta^{\alpha}} \sim \chi_{2 n}^{2}\) which is independent of \(\alpha\) and \(\beta\)

Hence \(Q\) is pivot. (Note: I think the expression of \(\left.Q=\frac{2 \sum_{i=1}^{n} Y_{i}}{\beta^{\alpha}}=\frac{2 n \bar{Y}_{n}}{\beta^{\alpha}}\right)\)

(c) \(\frac{2 \sum_{i=1}^{n_{1}} X_{i}}{2^{\alpha_{1}}}=\frac{2 n_{1} \bar{X}_{n_{1}}}{2^{\alpha_{1}}} \sim \chi_{2 n_{1}}^{2} ; \frac{2 \sum_{i=1}^{n_{2}} Y_{i}}{2^{\alpha_{2}}}=\frac{2 n_{2} \bar{Y}_{n_{2}}}{2^{\alpha_{2}}} \sim \chi_{2 n_{2}}^{2}\)

and \(X\) and \(Y^{\prime}\) s are independent \(\mathrm{so}\) \(\frac{\frac{2 n_{1} \bar{X}_{n_{1}}}{2^{\alpha_{1}}} /\left(2 n_{1}\right)}{\frac{2 n_{1} \bar{Y}_{n_{2}}}{2^{\alpha_{2}}} /\left(2 n_{2}\right)}=\frac{2^{\alpha_{2}} 2 n_{1} \sum_{i=1}^{n_{1}} X_{i}^{\alpha_{1}}}{2^{\alpha_{2}} 2 n_{2} \sum_{i=1}^{n_{2}} Y_{i}^{\alpha_{2}}} \sim F_{2 n_{1}, 2 n_{2}}\) and it is al so free from \(\left(\alpha_{1}, \alpha_{2}\right),\) so it is pivot.

$$ \begin{aligned} &\text { (d) } P\left(F_{1-\alpha / 2,2 n_{1}, 2 n_{2}} \leq \frac{2^{\alpha_{2}} 2 n_{1} \sum_{i=1}^{n_{1}} X_{i}^{\alpha_{1}}}{2^{\alpha_{2}} 2 n_{2} \sum_{i=1}^{n_{2}} Y_{i}^{\alpha_{2}}} \leq F_{\alpha / 2,2 n_{1}, 2 n_{2}}\right)=1-\alpha\\ &\text { So } 100(1-\alpha) \% \text { C.I. for } \frac{2^{\alpha_{2}}}{2 \alpha_{1}} \text { is }\\ &\left(F_{1-\alpha / 2,2 n_{1}, 2 n_{2}} \times \frac{\bar{Y}_{n}}{\bar{X}_{n}}, F_{\alpha / 2,2 n_{1}, 2 n_{2}} \times \frac{\bar{Y}_{n}}{\bar{X}_{n}}\right) \end{aligned} $$

(e) \(100(1-\alpha) \%\) C.I. for \(\alpha_{2}-\alpha\) is \(\left(\frac{1}{\log 2} \times \log \left(F_{1-\alpha / 2,2 n_{1}, 2 n_{2}} \times \frac{\bar{Y}_{n}}{\bar{X}_{n}}\right), \frac{1}{\log 2} \times \log \left(F_{\alpha / 2,2 n_{1}, 2 n_{2}} \times \frac{\bar{Y}_{n}}{\bar{X}_{n}}\right)\right)\)