这是一份GLA格拉斯哥大学STATS4074_1/STATS5011_1 作业代写的成功案例

$$
\begin{aligned}
&d_{i, i}=0 \
&d_{i, j}=d_{j, i}>0 \text { for } i \neq j ; \text { and } \
&d_{i, j} \leq d_{i, k}+d_{j, k} \text { (this being the triangle inequality). }
\end{aligned}
$$
The first two are straightforward, each thing is identical to itself $\left(d_{i, i}=0\right)$; and the difference between $\mathrm{A}$ and $\mathrm{B}$ must be the same as between $\mathrm{B}$ and $\mathrm{A}$ (thus $d_{i, j}=d_{j, i}$ ); and the value must be positive when they differ. The third part of the definition, the triangle inequality is also a simple concept; the direct distance between London and Sydney cannot

STATS4074_1/STATS5011_1 COURSE NOTES ：

From regression theory, the vector of partial regression coefficients for predicting the value of $y$ given a vector of observations $\mathbf{z}$ is $\mathbf{P}^{-1} \sigma(\mathbf{z}, y)$, where $\mathbf{P}$ is the covariance matrix of $\mathbf{z}$, and $\sigma(\mathbf{z}, y)$ is the vector of covariances between the elements of $\mathbf{z}$ and the variable $y$. Since $\boldsymbol{S}=\sigma(\mathbf{z}, \omega)$, it immediately follows that
$$
\mathbf{P}^{-1} \sigma(\mathbf{z}, \omega)=\mathbf{P}^{-1} \mathbf{S}=\boldsymbol{\beta}
$$
is the vector of partial regression for the best linear regression of relative fitness $\omega$ on phenotypic value $\mathbf{z}$, viz.,
$$
\omega(\mathbf{z})=1+\sum_{j=1}^{n} \beta_{j}\left(z_{j}-\mu_{j}\right)=1+\beta^{\mathrm{T}}(\mathbf{z}-\boldsymbol{\mu}) .
$$