这是一份northeastern东北大学(美国) MATH 4681 作业代写的成功案例
$$
X_{A(\ell)}^{2}=\frac{T_{\ell}^{2}}{\widehat{V}\left(T_{\ell}\right)}, \quad \ell=r, s
$$
where
$$
T_{\ell}=\sum_{j} \widehat{w}{\ell j}\left(p{1 j}-p_{2 j}\right), \quad \ell=r_{1} s
$$
$$
\widehat{V}\left(T_{\ell}\right)=\widehat{V}\left(T_{\ell} \mid H_{0}\right)=\sum_{j} \widehat{w}{\ell j}^{2} \widehat{V}{0 j} \quad \ell=r, s
$$
and
$$
\widehat{w}{\ell j}=\frac{1}{g{\ell}^{\prime}\left(p_{j}\right) \widehat{V}{0 j}}, \quad \ell=r{1} s .
$$
MATH 4681COURSE NOTES ：
$$
V\left(\widehat{\mu}{\theta}\right)=\frac{\sum{j} \tau_{j}^{2}\left(\sigma_{\widehat{\theta}{j}}^{2}+\sigma{\theta}^{2}\right)}{\left(\sum_{j} \tau_{j}\right)^{2}}
$$
Thus
$$
V\left(\widehat{\mu}{\theta}\right) \sum{j} \tau_{j}=\frac{\sum_{j} \tau_{j}^{2}\left(\sigma_{\hat{\theta}{j}}^{2}+\sigma{\theta}^{2}\right)}{\sum_{j} \tau_{j}}
$$
so that
$$
E\left(X_{H, C}^{2}\right)=\sum_{j} \tau_{j}\left(\sigma_{\hat{\theta}{j}}^{2}+\sigma{\theta}^{2}\right)-\frac{\sum_{j} \tau_{j}^{2}\left(\sigma_{\widehat{\theta}{j}}^{2}+\sigma{\theta}^{2}\right)}{\sum_{j} \tau_{j}}
$$
Noting that $\tau_{j}=\sigma_{\hat{\theta}{j}}^{-2}$, it is readily shown that $$ E\left(X{H, C}^{2}\right)=(K-1)+\sigma_{\theta}^{2}\left[\sum_{j} \tau_{j}-\frac{\sum_{j} \tau_{j}^{2}}{\sum_{j} \tau_{j}}\right]
$$