##### Question

In: Computer Science

# A random variable X is said to follow the Weibull distribution with shape parameter

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?

## Solutions

##### Expert Solution

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

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