统计代写|多元统计分析代写Multivariate Statistical Analysis代考|A Short Excursion into Matrix Algebra

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

统计代写|多元统计分析代写Multivariate Statistical Analysis代考|Derivatives

For later sections of this book, it will be useful to introduce matrix notation for derivatives of a scalar function of a vector $x$, i.e. $f(x)$, with respect to $x$. Consider $f: \mathbb{R}^{p} \rightarrow \mathbb{R}$ and a $(p \times 1)$ vector $x$, then $\frac{\partial f(x)}{\partial x}$ is the column vector of partial derivatives $\left{\frac{\partial f(x)}{\partial x_{j}}\right}, j=1, \ldots, p$ and $\frac{\partial f(x)}{\partial x T}$ is the row vector of the same derivative $\left(\frac{\partial f(x)}{\partial x}\right.$ is called the gradient of $\left.f\right)$.

We can also introduce second order derivatives: $\frac{\partial^{2} f(x)}{\partial x \partial x^{\top}}$ is the $(p \times p)$ matrix of elements $\frac{\partial^{2} f(x)}{\partial x_{i} \partial x_{j}}, i=1, \ldots, p$ and $j=1, \ldots, p\left(\frac{\partial^{2} f(x)}{\partial x \partial x}\right.$ is called the Hessian of $\left.f\right)$. Suppose that $a$ is a $(p \times 1)$ vector and that $\mathcal{A}=\mathcal{A}^{\top}$ is a $(p \times p)$ matrix. Then
$$\frac{\partial a^{\top} x}{\partial x}=\frac{\partial x^{\top} a}{\partial x}=a$$
The Hessian of the quadratic form $Q(x)=x^{\top} \mathcal{A} x$ is:
$$\frac{\partial^{2} x^{\top} \mathcal{A} x}{\partial x \partial x^{\top}}=2 . \mathcal{A}$$
Example $2.8$ Consider the matrix
$$\mathcal{A}=\left(\begin{array}{ll} 1 & 2 \ 2 & 3 \end{array}\right)$$

From formulas $(2.24)$ and $(2.25)$ it immediately follows that the gradient of $Q(x)=$ $x^{\top} \mathcal{A} x$ is
$$\frac{\partial x^{\top} \mathcal{A} x}{\partial x}=2, A x=2\left(\begin{array}{ll} 1 & 2 \ 2 & 3 \end{array}\right) x=\left(\begin{array}{ll} 2 x & 4 x \ 4 x & 6 x \end{array}\right)$$
and the Hessian is
$$\frac{\partial^{2} x^{\top} \mathcal{A x}}{\partial x \partial x^{\top}}=2, A=2\left(\begin{array}{ll} 1 & 2 \ 2 & 3 \end{array}\right)=\left(\begin{array}{ll} 2 & 4 \ 4 & 6 \end{array}\right)$$

统计代写|多元统计分析代写Multivariate Statistical Analysis代考|Partitioned Matrices

Very often we will have to consider certain groups of rows and columns of a matrix $\mathcal{A}(n \times p)$. In the case of two groups, we have
$$\mathcal{A}=\left(\begin{array}{ll} \mathcal{A}{11} & \mathcal{A}{12} \ \mathcal{A}{21} & \mathcal{A}{22} \end{array}\right)$$
where $\mathcal{A}{i j}\left(n{i} \times p_{j}\right), i, j=1,2, n_{1}+n_{2}=n$ and $p_{1}+p_{2}=p$.
If $\mathcal{B}(n \times p)$ is partitioned accordingly, we have:
\begin{aligned} \mathcal{A}+\mathcal{B} &=\left(\begin{array}{ll} \mathcal{A}{11}+\mathcal{B}{11} & \mathcal{A}{12}+\mathcal{B}{12} \ \mathcal{A}{21}+\mathcal{B}{21} & \mathcal{A}{22}+\mathcal{B}{22} \end{array}\right) \ \mathcal{B}^{\top} &=\left(\begin{array}{l} \mathcal{B}{11}^{\top} \mathcal{B}{21}^{\top} \ \mathcal{B}{12}^{\top} \mathcal{B}{22}^{\top} \end{array}\right) \ \mathcal{A B} &=\left(\begin{array}{l} \mathcal{A}{11} \mathcal{B}{11}^{\top}+\mathcal{A}{12} \mathcal{B}{12}^{\top} \mathcal{A}{11} \mathcal{B}{21}^{\top}+\mathcal{A}{12} \mathcal{B}{22}^{\top} \ \mathcal{A}{21} \mathcal{B}{11}^{\top}+\mathcal{A}{22} \mathcal{B}{12}^{\top} \mathcal{A}{21} \mathcal{B}{21}^{\top}+\mathcal{A}{22} \mathcal{B}{22}^{\top} \end{array}\right) \end{aligned}
An important particular case is the square matrix $\mathcal{A}(p \times p)$, partitioned in such a way that $\mathcal{A}{11}$ and $\mathcal{A}{22}$ are both square matrices (i.e. $n_{j}=p_{j}, j=1,2$ ). It can be verified that when $\mathcal{A}$ is non-singular $\left(\mathcal{A}, \mathcal{A}^{-1}=\mathcal{I}{p}\right)$ : $$\mathcal{A}^{-1}=\left(\begin{array}{ll} \mathcal{A}^{11} & \mathcal{A}^{12} \ \mathcal{A}^{21} & \mathcal{A}^{22} \end{array}\right)$$ where $$\left{\begin{array}{l} \mathcal{A}^{11}=\left(\mathcal{A}{11}-\mathcal{A}{12} \mathcal{A}{22}^{-1} \mathcal{A}{21}\right)^{-1} \stackrel{\text { def }}{=}\left(\mathcal{A}{11 \cdot 2}\right)^{-1} \ \mathcal{A}^{12}=-\left(\mathcal{A}{11 \cdot 2}\right)^{-1} \mathcal{A}{12} \mathcal{A}{22}^{-1} \ \mathcal{A}^{21}=-\mathcal{A}{22}^{-1} \mathcal{A}{21}\left(\mathcal{A}{11 \cdot 2}\right)^{-1} \ \mathcal{A}^{22}=\mathcal{A}{22}^{-1}+\mathcal{A}{22}^{-1} \mathcal{A}{21}\left(\mathcal{A}{11 \cdot 2}\right)^{-1} \mathcal{A}{12} \mathcal{A}{22}^{-1} \end{array}\right.$$

统计代写|多元统计分析代写Multivariate Statistical Analysis代考|Column Space and Null Space of a Matrix

Define for $\mathcal{X}(n \times p)$
$$\operatorname{Im}(\mathcal{X}) \stackrel{\text { def }}{=} C(\mathcal{X})=\left{x \in \mathbb{R}^{n} \mid \exists a \in \mathbb{R}^{p} \text { so that } \mathcal{X} a=x\right},$$
the space generated by the columns of $\mathcal{X}$ or the column space of $\mathcal{X}$. Note that $C(\mathcal{X}) \subseteq \mathbb{R}^{n}$ and $\operatorname{dim}{C(\mathcal{X})}=\operatorname{rank}(\mathcal{X})=r \leq \min (n, p)$
$$\operatorname{Ker}(\mathcal{X}) \stackrel{\text { def }}{=} N(\mathcal{X})=\left{y \in \mathbb{R}^{p} \mid \mathcal{X} y=0\right}$$
is the null space of $\mathcal{X}$. Note that $N(\mathcal{X}) \subseteq \mathbb{R}^{p}$ and that $\operatorname{dim}{N(\mathcal{X})}=p-r$.
Remark $2.2 N\left(\mathcal{X}^{\top}\right)$ is the orthogonal complement of $C(\mathcal{X})$ in $\mathbb{R}^{n}$, i.e. given a vector $b \in \mathbb{R}^{n}$ it will hold that $x^{\top} b=0$ for all $x \in C(\mathcal{X})$, if and only if $b \in N\left(\mathcal{X}^{\top}\right)$.
Example $2.12$ Let $\mathcal{X}=\left(\begin{array}{lll}2 & 3 & 5 \ 4 & 6 & 7 \ 6 & 8 & 6 \ 8 & 2 & 4\end{array}\right)$. It is easy to show (e.g. by calculating the determinant of $\mathcal{X})$ that $\operatorname{rank}(\mathcal{X})=3$. Hence, the column space of $\mathcal{X}$ is $C(\mathcal{X})=\mathbb{R}^{3}$. The null space of $\mathcal{X}$ contains only the zero vector $(0,0,0)^{\top}$ and its dimension is equal to $\operatorname{rank}(\mathcal{X})-3=0$.

统计代写|多元统计分析代写Multivariate Statistical Analysis代考|Derivatives

∂一个⊤X∂X=∂X⊤一个∂X=一个

∂2X⊤一个X∂X∂X⊤=2.一个

∂X⊤一个X∂X=2,一个X=2(12 23)X=(2X4X 4X6X)

∂2X⊤一个X∂X∂X⊤=2,一个=2(12 23)=(24 46)

统计代写|多元统计分析代写Multivariate Statistical Analysis代考|Column Space and Null Space of a Matrix

\operatorname{Im}(\mathcal{X}) \stackrel{\text { def }}{=} C(\mathcal{X})=\left{x \in \mathbb{R}^{n} \mid \存在一个 \in \mathbb{R}^{p} \text { 使得 } \mathcal{X} a=x\right}，\operatorname{Im}(\mathcal{X}) \stackrel{\text { def }}{=} C(\mathcal{X})=\left{x \in \mathbb{R}^{n} \mid \存在一个 \in \mathbb{R}^{p} \text { 使得 } \mathcal{X} a=x\right}，

\operatorname{Ker}(\mathcal{X}) \stackrel{\text { def }}{=} N(\mathcal{X})=\left{y \in \mathbb{R}^{p} \mid \mathcal{X} y=0\right}\operatorname{Ker}(\mathcal{X}) \stackrel{\text { def }}{=} N(\mathcal{X})=\left{y \in \mathbb{R}^{p} \mid \mathcal{X} y=0\right}

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MATLAB代写

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