这是一份oxford牛津大学作业代写的成功案例
$$
\phi_{0 j}^{2}=\frac{\pi_{j}\left(1-\pi_{j}\right)}{\zeta_{j} \xi_{1 j} \xi_{2 j}}
$$
Therefore, the test statistic can be expressed as
$$
X_{A(g)}^{2}=\frac{T^{2}}{V\left(T \mid H_{0}\right)}=\frac{\left.\sum_{j} \mid \tilde{w}{j}\left(p{1 j}-p_{2 j}\right)\right]^{2}}{\sum_{j} \tilde{w}{j}^{2} \sigma{0 j}^{2}}=\frac{\tilde{T}^{2}}{V\left(\tilde{T} \mid H_{0}\right)}
$$
$$
F(x) \leq\left|y^{}\right||x| $$ and $$ -F(x)=F(-x) \leq p(-x)=\left|y^{}\right||x|
$$
that is,
$$
|F(x)| \leq\left|y^{}\right||x| . $$ This shows that $F:=x^{} \in X^{}$ and $\left|x^{}\right| \leq\left|y^{}\right|$. Since the reversed inequality is trivial for any linear extension of $y^{}$, the theorem is proved in the case of real scalars.
Oxford COURSE NOTES ：
using the weights
$$
\tilde{w}{j}=\frac{1}{\phi{0 j}^{2} g^{\prime}\left(\pi_{j}\right)} .
$$Under $H_{0}: \pi_{1 j}=\pi_{2 j}=\pi_{j}$, then $\tilde{T} \approx \mathcal{N}\left(0, \sigma_{\bar{T}{0}}^{2}\right)$, where $E(\tilde{T})=0$ and $$ \begin{aligned} V\left(\tilde{T} \mid H{0}\right) &=\sigma_{T_{0}}^{2}=\sum_{j}^{2} \tilde{w}^{2} \sigma_{0 j}^{2}=\sum_{j} \tilde{w}{j}^{2} \frac{\phi{0 j}^{2}}{N} \
&=\frac{1}{N} \sum_{j j}\left(\frac{1}{\phi_{0 j}^{2} g^{\prime}\left(\pi_{j}\right)^{2}}\right)=\frac{\phi_{\widetilde{T}_{0}}^{2}}{N}
\end{aligned}
$$