这是一份liverpool利物浦大学MATH363的成功案例

$$
\mathbf{e}{2}^{}=\left[I{n_{2}}-h\right]^{-1} \mathbf{e}{2}, $$ so that becomes $$ \hat{\boldsymbol{\beta}}^{}-\hat{\boldsymbol{\beta}}=-\mathbf{M}^{-1} \mathbf{X}{2}^{\prime}\left[\mathbf{I}{n{2}}-\mathbf{h}\right]^{-1} \mathbf{e}{2} $$ Finally then, $$ \hat{\mathbf{Y}}^{}-\hat{\mathbf{Y}}=\mathbf{X}\left(\hat{\boldsymbol{\beta}}^{}-\hat{\boldsymbol{\beta}}\right)=-\mathbf{X} \mathbf{M}^{-\mathbf{1}} \mathbf{X}{2}^{\prime}\left[\mathbf{I}{n{2}}-\mathbf{h}\right]^{-1} \mathbf{e}{2} $$ The $n \times n{2}$ matrix $\mathbf{X} \mathbf{M}^{-1} \mathbf{X}{2}^{\prime}$ on the right consists of the last $n{2}$ columns of $\mathbf{H}$. Matrix computation then gives
$$
|\hat{\mathbf{Y}} *-\hat{\mathbf{Y}}|^{2}=\mathbf{e}{2}^{\prime}\left(\mathbf{I}{n_{2}}-\mathbf{h}\right)^{-1} \mathbf{h}\left(\mathbf{I}{n{2}}-\mathbf{h}\right)^{-1} \mathbf{e}_{2}
$$

MATH363 COURSE NOTES ：

$$
\mathbf{M}=\left[\begin{array}{cc}
M & 0 \
0 & S_{x x}
\end{array}\right], \quad S_{x x}=\left|\mathbf{x}{1}\right|^{2}, \quad \overline{\mathbf{x}}{i}=\left(1, x_{i}-\bar{x}\right)
$$$$
\hat{\boldsymbol{\beta}}{-i}-\hat{\boldsymbol{\beta}}=-\left[\begin{array}{c} 1 / n \ \left(x{i}-\vec{x}\right) / S_{x x}
\end{array}\right]\left[e_{-i} /\left(1-h_{i}\right)\right]
$$
for
$$
h_{i}=\frac{1}{n}+\left(x_{i}-\bar{x}\right)^{2} / S_{x x}
$$
The $j$ th element of $\hat{\mathbf{Y}}{-i}-\hat{\mathbf{Y}}$ is $$ -\left[\frac{1}{n}+\frac{\left(x{i}-\bar{x}\right)\left(x_{j}-\bar{x}\right)}{S_{x x}}\right]\left[e_{-i} /\left(1-h_{i}\right)\right]
$$