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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