# 多元统计分析|MA3066/STAT 505/STAT 530/EXST 7037/STAT 450/550/STA3200/MATH5855Multivariate Statistical Analysis代写

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Providing the breeders’ equation holds over several generations, the cumulative response to $m$ generations of selection is just
$$R^{(m)}=\sum_{t=1}^{m} R(t)=\sum_{t=1}^{m} \mathbf{G}(t) \beta(t) .$$
In particular, if the genetic covariance matrix remains constant, then
$$R^{(m)}=\mathbf{G}\left(\sum_{t=1}^{m} \beta(t)\right)$$Hence, if we observe a total response to selection of $\mathbf{R}{\text {total }}$, then we can estimate the cumulative selection differential as $$\beta{\text {wotal }}=\sum \beta(t)=\mathbf{G}^{-1} \mathbf{R}_{\text {wotal }} \text {. }$$

## MA3066/STAT 505/STAT 530/EXST 7037/STAT 450/550/STA3200/MATH5855 COURSE NOTES ：

If one has access to a sample of generation means (as opposed to a starting and end points), then notes that a more powerful test can be obtained (from the theory of random walks) by considering $\Delta \mu^{}$, the largest absolute deviation from the starting mean anywhere along the series of $t$ generation means. Here, the rate of evolution is too fast for drift if $$\sigma_{m}^{2}<\frac{\left(\Delta \mu^{}\right)^{2}}{2 t(2.50)^{2}}=0.080 \frac{\left(\Delta \mu^{}\right)^{2}}{t}$$ while it is too slow for drift when $$\sigma_{m}^{2}>\frac{\left(\Delta \mu^{}\right)^{2}}{2 t(0.56)^{2}}=1.59 \frac{\left(\Delta \mu^{*}\right)^{2}}{t} .$$
Note by comparison with that the tests for too fast a divergence are very similar, while Bookstein’s test for too slow a divergence (a potential signature of stabilizing selection) is much less stringent than the Turelli-Gillespie-Lande test.

# 多元统计分析| Multivariate Statistical Analysis 代写 STAT 505代考

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\begin{aligned}
&\text { Wilks’ lambda }=\prod_{i=1}^{f} \frac{1}{1+\tau_{i}}=\frac{|\mathbf{E}|}{|\mathbf{E}+\mathbf{H}|} \
&\text { Piliai’stracc }=\sum_{i=1}^{s} \frac{\eta_{1}}{1+\tau_{i}}=\operatorname{tr}\left[\mathbf{H}(\mathbf{H}+\mathbf{E})^{-1}\right]
\end{aligned}

\begin{gathered}
\text { Hotelling-Lawley trace }=\sum_{i=1}^{\pi} \eta_{i}=\mathrm{tr}{\left[\mathrm{HE}^{-1}\right]} \ \text { Roy’s greatest root }=\frac{\eta{1}}{1+\eta_{1}}
\end{gathered}

## STAT 505COURSE NOTES ：

\begin{array}{ll}
\lambda_{1}=5.83, & \mathbf{e}{1}^{\prime}=[383,-924,0] \ \lambda{2}=2.00, & \mathbf{e}{2}^{\prime}=[0,0,1] \ \lambda{3}=0.17, & \mathbf{e}_{3}^{\prime}=[.924, .383,0]
\end{array}