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

Chapter 9, Problem 53E

(choose chapter or problem)

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

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

Show that $Y_{(n)}=\min \left(Y_{1}, Y_{2}, \ldots, Y_{n}\right)$ is sufficient for $\theta$.

Equation Transcription:

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

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

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

$\theta{}$

Text Transcription:

Y_1, Y_2, …., Y_n

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

Y_(n) = min (Y_1, Y_2, …., Y_n)

theta

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