Question

Suppose X1, X2, .., Xn is an iid sample from where >0.

Answer 1

$(a)\\The~likelihood~function~of~\theta=L(\theta)=\theta^ne^{-\theta\sum_{i=1}^nx_i}\\ l(\theta)=\log L(\theta)=n\log \theta-\theta\sum_{i=1}^nx_i\\ \frac{\mathrm{d}l(\theta) }{\mathrm{d} \theta}=\frac{n}{\theta}-\sum_{i=1}^nx_i=0\Rightarrow \hat{\theta}=\frac{1}{\bar{x}},~where,~\bar{x}=\frac{1}{n}\sum_{i=1}^nx_i\\ Likelihood~ratio=\Lambda=\frac{L(\theta_0)}{L(\hat{\theta})}=(\bar{x}\theta_0)^n\exp(-n\theta_0\bar{x}&plus;n)<k\\ or,~\bar{x}^n\exp(-n\theta_0\bar{x})<k^*~or~\sum_{i=1}^nx_i>c\\ where,~c~is~obtained~from~size~condition~i.e.\\ P\left(\sum_{i=1}^nX_i>c|H_0 \right )=P\left(2\theta_0\sum_{i=1}^nX_i>2\theta_0 c|H_0 \right )=\alpha\Rightarrow c=\frac{1}{2\theta_0}\chi^2_{\alpha,2n}\\ Since~we~know,~2\theta X_i\sim \chi^2_2~and~since~X_i's~are~independent~then~from~additive~\\ property~of~\chi^2,~2\theta\sum_{i=1}^vX_i\sim \chi^2_{2n}\\ Power~function=P(\sum_{i=1}^nX_i>c|\theta)=P\left(2\theta\sum_{i=1}^nX_i>\frac{\theta}{\theta_0}\chi^2_{\alpha,2n}\right)=\beta(\theta).\\$

$(b)~\chi^2_{0.10,20}=28.4120\\ Power=P(4\sum_{i=1}^nX_i>2*28.4120)=0.00002$