### 计算机代写|机器学习代写machine learning代考|COMP4702

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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

## 计算机代写|机器学习代写machine learning代考|Distance and Inner-Product Random Kernel Matrices

The most widely used kernel model in machine learning applications is the heat kernel $\mathbf{K}=\left{\exp \left(-\left|\mathbf{x}i-\mathbf{x}_j\right|^2 / 2 \sigma^2\right)\right}{i, j=1}^n$, for some $\sigma>0$. It is thus natural to start the large-dimensional analysis of kernel random matrices by focusing on this model.
As mentioned in the previous sections, for the Gaussian mixture model above, as the dimension $p$ increases, $\sigma^2$ needs to scale as $O(p)$, so say $\sigma^2=\tilde{\sigma}^2 p$ for some $\tilde{\sigma}^2=O(1)$, to avoid evaluating the exponential at increasingly large values for $p$ large. As such, the prototypical kernel of present interest is
$$\mathbf{K}=\left{f\left(\frac{1}{p}\left|\mathbf{x}i-\mathbf{x}_j\right|^2\right)\right}{i, j-1}^n,$$
for $f$ a sufficiently smooth function (specifically, $f(t)=\exp \left(-t / 2 \tilde{\sigma}^2\right)$ for the heat kernel). As we will see though, it is much desirable not to restrict ourselves to $f(t)=\exp \left(-t / 2 \tilde{\sigma}^2\right)$ so to better appreciate the impact of the nonlinear kernel function $f$ on the (asymptotic) structural behavior of the kernel matrix $\mathbf{K}$.

## 计算机代写|机器学习代写machine learning代考|Euclidean Random Matrices with Equal Covariances

In order to get a first picture of the large-dimensional behavior of $\mathbf{K}$, let us first develop the distance $\left|\mathbf{x}_i-\mathbf{x}_j\right|^2 / p$ for $\mathbf{x}_i \in \mathcal{C}_a$ and $\mathbf{x}_j \in \mathcal{C}_b$, with $i \neq j$.

For simplicity, let us assume for the moment $\mathbf{C}_1=\cdots=\mathbf{C}_k=\mathbf{I}_p$ and recall the notation $\mathbf{x}_i=\boldsymbol{\mu}_a+\mathbf{z}_i$. We have, for $i \neq j$ that “entry-wise,”
\begin{aligned} \frac{1}{p}\left|\mathbf{x}_i-\mathbf{x}_j\right|^2= & \frac{1}{p}\left|\boldsymbol{\mu}_a-\boldsymbol{\mu}_b\right|^2+\frac{2}{p}\left(\boldsymbol{\mu}_a-\boldsymbol{\mu}_b\right)^{\top}\left(\mathbf{z}_i-\mathbf{z}_j\right) \ & +\frac{1}{p}\left|\mathbf{z}_i\right|^2+\frac{1}{p}\left|\mathbf{z}_j\right|^2-\frac{2}{p} \mathbf{z}_i^{\top} \mathbf{z}_j . \end{aligned}
For $\left|\mathbf{x}_i\right|$ of order $O(\sqrt{p})$, if $\left|\mu_a\right|=O(\sqrt{p})$ for all $a \in{1, \ldots, k}$ (which would be natural), then $\left|\mu_a-\mu_b\right|^2 / p$ is a priori of order $O(1)$ while, by the central limit theorem, $\left|\mathbf{z}_i\right|^2 / p=1+O\left(p^{-1 / 2}\right)$. Also, again by the central limit theorem, $\mathbf{z}_i^{\top} \mathbf{z}_j / p=$ $O\left(p^{-1 / 2}\right)$ and $\left(\mu_a-\mu_b\right)^{\top}\left(\mathbf{z}_i-\mathbf{z}_j\right) / p=O\left(p^{-1 / 2}\right)$

As a consequence, for $p$ large, the distance $\left|\mathbf{x}i-\mathbf{x}_j\right|^2 / p$ is dominated by $| \boldsymbol{\mu}_a-$ $\boldsymbol{\mu}_b |^2 / p+2$ and easily discriminates classes from the pairwise observations of $\mathbf{x}_i, \mathbf{x}_j$, making the classification asymptotically trivial (without having to resort to any kernel method). It is thus of interest consider the situations where the class distances are less significant to understand how the choices of kernel come into play in such more practical scenario. To this end, we now demand that $$\left|\mu_a-\mu_b\right|=O(1),$$ which is also the minimal distance rate that can be discriminated from a mere Bayesian inference analysis, as thoroughly discussed in Section 1.1.3. Since the kernel function $f(\cdot)$ operates only on the distances $\left|\mathbf{x}_i-\mathbf{x}_j\right|$, we may even request (up to centering all data by, say, the constant vector $\frac{1}{n} \sum{a=1}^k n_a \mu_a$ ) for simplicity that $\left|\mu_a\right|=O(1)$ for each $a$.

# 机器学习代考

## 计算机代写|机器学习代写machine learning代考|Euclidean Random Matrices with Equal Covariances

$$\frac{1}{p}\left|\mathbf{x}_i-\mathbf{x}_j\right|^2=\frac{1}{p}\left|\boldsymbol{\mu}_a-\boldsymbol{\mu}_b\right|^2+\frac{2}{p}\left(\boldsymbol{\mu}_a-\boldsymbol{\mu}_b\right)^{\top}\left(\mathbf{z}_i-\mathbf{z}_j\right) \quad+\frac{1}{p}\left|\mathbf{z}_i\right|^2+\frac{1}{p}\left|\mathbf{z}_j\right|^2-\frac{2}{p} \mathbf{z}_i^{\top} \mathbf{z}_j$$

$$\left|\mu_a-\mu_b\right|=O(1)$$

## 有限元方法代写

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## MATLAB代写

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