# 科学家和工程师的物理学|PHYS1001 Physics for Scientists and Engineers代写 UWA代写

For computational purposes, let
$$\mathbf{Y}^{}=p\left(\mathbf{Y} \mid \mathscr{L}\left((\mathbf{A B}){11}, \ldots,(\mathbf{A B}){a b}\right)\right)=\sum_{i j} \bar{Y}{i j} \cdot(\mathbf{A B}){i j}$$
Then $\hat{\mathbf{Y}}{A B}=\mathbf{Y}^{}-\left(\hat{\mathbf{Y}}{0}+\hat{\mathbf{Y}}{A}+\hat{\mathbf{Y}}{B}\right)$ and by the Pythagorean Theorem,
$$\left|\hat{\mathbf{Y}}{A B}\right|^{2}=\left|\hat{\mathbf{Y}}^{}-\hat{\mathbf{Y}}{0}\right|^{2}-\left[\left|\hat{\mathbf{Y}}{A}\right|^{2}+\left|\hat{\mathbf{Y}}{B}\right|^{2}\right]$$
But $\mathbf{Y}^{}-\hat{\mathbf{Y}}{0}=\sum{i j}\left(\bar{Y}{i j} \ldots-\bar{Y}{\ldots} \ldots\right)(\mathbf{A B}){i j}$ so $$\left|\hat{\mathbf{Y}}^{}-\hat{\mathbf{Y}}{0}\right|^{2}=\sum_{i j}\left(\bar{Y}{i j} \ldots-\bar{Y}{\ldots} \ldots\right)^{2}(c m)=\left|\hat{\mathbf{Y}}^{}\right|^{2}-\left|\hat{\mathbf{Y}}{0}\right|^{2}=\sum{i j} \vec{Y}{i j}^{2} \ldots(c m)-\bar{Y}{\ldots}^{2} \ldots n$$

## PHYS1001COURSE NOTES ：

Then
and
$$\mathrm{SSA}=\left|\hat{\mathbf{Y}}{A}\right|^{2}=\sum{1}^{l}\left[\left(a_{i}-\bar{a}\right)+\left(\bar{\varepsilon}{i} \cdot-\bar{\varepsilon} .\right)\right]^{2} J,$$ $$\mathrm{SSE}=|\mathbf{e}|^{2}=\sum{i j}\left(\varepsilon_{i j}-\bar{\varepsilon}{i}\right)^{2}$$ Let $W{i}=a_{i}+\vec{\varepsilon}{i, .}$ Then $W{i} \sim N\left(0, \sigma_{a}^{2}+\frac{\sigma^{2}}{J}\right)$ and the $W_{i}$ are independent. It follows that
$$\frac{\sum_{i}\left(W_{i}-\bar{W}\right)^{2}}{\sigma_{a}^{2}+\frac{\sigma^{2}}{J}}=\frac{\text { SSA }}{\sigma^{2}+J \sigma_{a}^{2}} \sim \chi_{I-1}^{2}$$
In addition, the $W_{i}$ are independent of the vector e, and therefore of SSE, which is the same as it is for the fixed effects model. Thus, SSE $/ \sigma^{2} \sim \chi_{(\Lambda-1)}^{2}$.