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 $\left(\frac{1}{\log 2} \times \log \left(F_{1-\alpha / 2,2 n_{1}, 2 n_{2}} \times \frac{\bar{Y}_{n}}{\bar{X}_{n}}\right), \frac{1}{\log 2} \times \log \left(F_{\alpha / 2,2 n_{1}, 2 n_{2}} \times \frac{\bar{Y}_{n}}{\bar{X}_{n}}\right)\right)$