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Let $$f_\theta(x_1,\dots,x_n)=\frac{1}{24^n}\prod_{i=1}^nx_i^4\theta^{-5n}e^{\sum\limits_{i=1}^n\frac{-x_i}{\theta}}\hspace{0.5cm}\underset{\forall i=1\dotsm n}{x_i\in\mathbb{R}^+}\;\theta\in\mathbb{R^+}$$

The distribution belongs to a regular exponential family because

$f_\theta(X_1,\dots,X_n)=\frac{1} {24^n}\prod_{i=1}^nx_i^4\theta^{-5n}e^{\sum\limits_{i=1}^n\frac{-x_i}{\theta}}=h(X_1,\dots,X_n)c(\theta)e^{Q(\theta)T(X_1,\dots,X_n)}$

And

$-\frac{c'(\theta)}{c(\theta)Q'(\theta)}=-\frac{-5n\theta^{-5n-1}}{-\theta^{-2}\theta^{-5n}}=-5n\theta$

So $E[T]=-5n\theta$ where $T=\sum_{i=1}^n-Xi$

Then $G=-\frac{T}{5n}$ is the UMVUE for $\theta$

How can I find the UMVUE for $\frac{1}{\theta}$

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    $\begingroup$ If the distribution belongs to a regular exponential family, regularity conditions will automatically hold. You must mention the sample space (support) and parameter space. $\endgroup$ – StubbornAtom May 20 at 13:40
  • $\begingroup$ Thank you @StubbornAtom and for this problem, how can I find the UMVUE for $h(\theta)=\theta^{-1}$ I found the UMVUE for $\theta$ $\endgroup$ – Jhon Knows May 20 at 14:07
  • $\begingroup$ Yes I forgot the minus sign $\endgroup$ – Jhon Knows May 20 at 14:51
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    $\begingroup$ Find $E(1/\sum X_i)$ where $\sum X_i$ has a Gamma distribution. Note that $X_i/\theta$ is a Gamma variate. This should help: math.stackexchange.com/q/28779/321264. $\endgroup$ – StubbornAtom May 20 at 14:59
  • $\begingroup$ Thank you @StubbornAtom, can you give me some hint to find a cofidence interval with $(1-\alpha)100\%$ please? $\endgroup$ – Jhon Knows May 23 at 12:13
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$T(X_1,\dots,X_n)=\sum\limits_{i=1}^nX_i\sim\text{Gamma}(5n,\theta)$

Then

\begin{align} E\left[\frac{1}{T}\right]&=\int_0^\infty\frac{1}{x}\cdot\frac{x^{5n-1}e^{-\frac{x}{\theta}}}{\Gamma(5n)\theta^{5n}}\;dx \\&=\frac{1}{\Gamma(5n)\theta^{5n}}\int_0^\infty x^{5n-2}e^{-\frac{x}{\theta}}\;dx \\&=\frac{\theta^{5n-1}}{\Gamma(5n)\theta^{5n}}\cdot\int_0^\infty \frac{x^{5n-2}}{\theta^{5n-1}}e^{-\frac{x}{\theta}}\;dx \\&=\frac{1}{\Gamma(5n)\theta}\cdot\Gamma(5n-1) \\&=\frac{1}{\theta(5n-1)} \end{align}

So $$H(X_1,\dots,X_n)=\frac{5n-1}{T}$$

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  • $\begingroup$ Thank you and sorry for the cross-post $\endgroup$ – Jhon Knows May 23 at 18:19

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