这是一份GLA格拉斯哥大MATHS5070_1 /MATHS4104_1作业代写的成功案例

$$
l_{\mathrm{p}}(\psi)=c-\frac{n}{2} \log (\mathrm{SSB}+\lambda \mathrm{SSW})+\frac{n-m}{2} \log \lambda,
$$
where $\lambda=1+k \psi$. The maximum of $l_{\mathrm{p}}$ is given by
$$
\hat{\lambda}=\left(1-\frac{1}{m}\right) \frac{\mathrm{MSB}}{\mathrm{MSW}},
$$
where MSB $=\operatorname{SSB} /(m-1)$ and MSW $=\operatorname{SSW} /(n-m)$, or
$$

MATHS5070_1 /MATHS4104_1 COURSE NOTES ：

$$
\hat{F}{r}(x)=\frac{1}{m{r}} \sum_{u=1}^{m_{r}} 1_{\left(\hat{\alpha}{r}, u \leq x\right)} \stackrel{P}{\longrightarrow} F{r}(x), \quad x \in C\left(F_{r}\right),
$$
where $\hat{\alpha}{r, u}$ is the $u$ th component of $\hat{\alpha}{r}, 1 \leq r \leq s$, and
$$
\hat{F}{0}(x)=\frac{1}{n} \sum{u=1}^{n} 1_{\left(\tilde{\epsilon}{u} \leq x\right)} \stackrel{P}{\longrightarrow} F{0}(x), \quad x \in C\left(F_{0}\right),
$$