Let $x_1 ,\ldots ,x_n $ be a random sample taken from the Weibull distribution $$f(x;\nu ,\theta)=\frac{\nu}{\theta}x^{\nu -1}\exp(-\frac{x^\nu}{\theta}),$$ where $x,\nu,\theta >0$. I got the Bayes estimator of $\theta$ under the linear exponential loss function (LLF) $$L(\Delta)=e^{a\Delta}-a\Delta-1, \quad a\neq 0, \quad \Delta=\frac{\hat{\theta}-\theta}{\theta},$$ with respect to the conjugate prior distribution of the form $g(\theta )=\frac{\beta^{\alpha}}{\Gamma (\alpha)}\theta^{-(\alpha+1)}\exp(-\frac{\beta}{\theta})$ (where $\alpha ,\beta ,\theta >0$) is as follows $$\hat{\theta}=\varphi (\beta +T),$$ where $\varphi =\frac{1}{a}(1-\exp(-\frac{a}{n+\alpha+1})),$ $T=\sum_{i=1}^{n}x_i^{\nu}$ and $\frac{2nT}{\theta}\sim \chi^2_{2n}$.
Why the risk function of $\hat{\theta}$ under the LLF is $$R(\hat{\theta})=exp\Big( a(\frac{\varphi \beta}{\theta}-1)\Big)(1-a\varphi)^{-n}-1-a(n+\frac{\beta}{\theta}-1)?$$
