Solution: Let Y1, Y2, . . . , Yn be a random sample from a

Chapter 9, Problem 93E

(choose chapter or problem)

Let $Y_{1}, Y_{2}, \ldots, Y_{n}$ be a random sample from a population with density function

$f(y \mid \theta)=\left\{\begin{array}{ll}\frac{2 \theta^{2}}{y^{3}}, & \theta<y<\infty \\0, & \text { elsewhere }\end{array}\right.$

In Exercise 9.53, you showed that $Y_{(1)}=\min \left(Y_{1}, Y_{2}, \ldots, Y_{n}\right)$ is sufficient for $\theta$.

a Find the MLE for $\theta$. [Hint: See Example 9.16.]

b Find a function of the MLE in part (a) that is a pivotal quantity.

c Use the pivotal quantity from part (b) to find a 100( 1 - $\alpha$)% confidence interval for $\theta$.

Equation Transcription:

${Y}_{1},\ {Y}_{2},\ ....,\ {Y}_{n}$

$f\ (y\ |\ \theta{})=$ $\{{\ }_{0,\ \ \ \ \ \ \ \ \ \ \ \ elsewhere}^{\frac{2{\theta{}}^{2}}{{y}^{3}},\ \ \ \ \ \ \ \ \theta{}\ <\ y\ <\ \infty{},}$

${Y}_{(1)}$ = min $\left({{Y}_{1},\ {Y}_{2},\ ....,\ {Y}_{n}}\right)$

$\theta{}$

$\alpha{}$

$\theta{}$

Text Transcription:

Y_1, Y_2, …., Y_n

f(y | theta) = {frac{2 theta^{2}}{y^{3}}, & theta < y < infty 0, & elsewhere.

Y_(1) = min (Y_1, Y_2, …., Y_n)

theta

alpha

theta

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