Suppose $X_1 , ..., X_n$ are random samples of exponential distribution with mean $\theta$. Determine $a$ and $b$ such that $a\sum_{i=1}^n X_i +b$ be admissible under the loss function $L(\theta,\delta)=\frac{(\delta - \theta)^2}{\theta ^2}$.
All my efforts fail, I can't get my head around this, any help would be very much appreciated. Thanks.
Let me just give a little suggestion:
If $X_1,..,X_n \sim Exp(\frac{1}{\theta})$ ($\frac{1}{\theta}>0$)Then
$Y$ = $\sum_{i=1}^n X_i \sim \Gamma(n,\frac{1}{\theta})$, ($n>0)$
Then, $Z=a*Y-b$. Now proceed finding the distribution of $Z$.. (https://onlinecourses.science.psu.edu/stat414/node/157)
After bounty:
Let me complete how I would proceed:
$Y=g(Z)= \frac{Z+b}{a}$
$g'(Z)=\frac{1}{a}$
$f_Z(z)=f_Y(g(Z))*|g'(Z)|$
= $\frac{(\frac{1}{\theta})^n}{\Gamma(n)} * (\frac{z+b}{a})^{n-1}*exp(-\frac{z+b}{\theta*a}) * \frac{1}{a}$ ..
At this point I would integrate this distribution with the integral equal to 1, namely:
$\int_{-\infty}^{+\infty} f_Z(z)d(z) = 1$
I think a change of variables to get rid of the $b$ inside the integral could be an idea.
I think this problem is rather similar to Example 8.2.8 in my book, The Bayesian Choice.
Example 8.2.8 (Example 8.2.4 continued) For the chi-squared distribution,the natural parameterization is $$ \theta={1\over \sigma^2}, \quad T(y)=-{1\over 2}y, \quad \psi(\theta)=-{p\over 2}\log(\theta), $$ and $$ \int_0^c e^{-\gamma_0\lambda\theta} \theta^{-\lambda p/2} \,d\theta $$ is infinite if $\lambda p \ge 2$. Similarly, $$ \int^{+\infty}_c e^{-\gamma_0\lambda\theta} \theta^{-\lambda p/2} \,d\theta = +\infty $$ if $\gamma_0\lambda<0$ or $\gamma_0\lambda=0$ and $\lambda p \le 2$. Therefore, the Bayes estimator $$ \delta^\pi(y) = {\gamma_0\lambda\over 1+\lambda} - {1\over 1+\lambda} {y \over 2} $$ is admissible if $\gamma_0=0$ and $\lambda=2/p$ or $\gamma_0<0$ and $\lambda\ge 2/p$; these conditions lead to the estimators $$ \varphi_1(y) = {p\over p+2} \left( {-y\over 2} \right) \quad\hbox{ and } \quad \varphi_2(y) = {\gamma_0 \lambda \over 1+\lambda} +{1\over 1+\lambda} \left( {-y\over 2} \right), $$ for the estimation of $\mathbb{E}_\sigma(-y/2)=-{p\over 2}\sigma^2$, that is, to the following admissible Bayes estimators of $\sigma^2$: $$ \delta_1(y)={y\over p+2} \quad \hbox{ and } \quad \delta_2(y)=ay+b,\quad b>0,\ 0\le a\le {1\over p+2}. $$
