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# 数学代写|Gaussian random variables 数论代考

## 数论代考

By definition, a random vector $X$ with values in $\mathrm{R}^{m}$ is called a (centered) gaussian vector if there exists a non-negative quadratic form $\mathrm{Q}$ on $\mathrm{R}^{m}$ such that the characteristic function $\varphi_{X}$ of $X$ is of the form
$$\varphi_{\mathrm{X}}(t)=e^{-\mathrm{Q}(t) / 2}$$
for $t \in \mathrm{R}^{m}$. The quadratic form can be recovered from $\mathrm{X}$ by the relation
$$\mathrm{Q}\left(t_{1}, \ldots, t_{m}\right)=\sum_{1 \leqslant i, j \leqslant m} a_{i, j} t_{i} t_{j}$$
with $a_{i, j}=\mathbf{E}\left(\mathrm{X}{i} \mathrm{X}{j}\right)$, and the (symmetric) matrix $\left(a_{i, j}\right){1 \leqslant i, j \leqslant m}$ is called the correlation matrix of $X$. The components $X{i}$ of $X$ are independent if and only if $a_{i, j}=0$ if $i \neq j$, i.e., if and only if the components of $X$ are orthogonal.
If $X$ is a gaussian random vector, then $X$ is mild, and in fact
$$\sum_{\boldsymbol{k}} \mathrm{M}{m}(\mathrm{X}) \frac{t{1}^{k_{1}} \cdots t_{m}^{k_{m}}}{k_{1} ! \cdots k_{m} !}=\mathbf{E}\left(e^{t \cdot \mathrm{X}}\right)=e^{\mathrm{Q}(t) / 2}$$
for $t \in \mathbf{R}^{m}$, so that the power series converges on all of $\mathrm{C}^{m}$. The Laplace transform $\psi_{\mathrm{X}}(z)=\mathbf{E}\left(e^{z \cdot \mathrm{X}}\right)$ is also defined for all $z \in \mathbf{C}^{m}$, and in fact
(B.9)
$$\mathbf{E}\left(e^{z \cdot \mathrm{X}}\right)=e^{\mathrm{Q}(z) / 2} .$$
For $m=1$, this means that a random variable is a centered gaussian if and only if there exists $\sigma \geqslant 0$ such that
$$\varphi_{\mathrm{X}}(t)=e^{-\sigma^{2} t / 2}$$
and in fact we have
$$\mathbf{E}\left(\mathrm{X}^{2}\right)=\mathbf{V}(\mathrm{X})=\sigma^{2}$$
If $\sigma=1$, then we say that $\mathrm{X}$ is a standard gaussian random variable (also sometimes called a standard normal random variable). We then have
$$\mathbf{P}(a<\mathrm{X}<b)=\frac{1}{\sqrt{2 \pi}} \int_{a}^{b} e^{-x^{2} / 2} d x$$
for all real numbers $a<b$.
EXERCISE B.7.1. We recall a standard proof of the fact that the measure on $\mathbf{R}$ given by
$$\mu=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2} d x$$
is indeed a gaussian probability measure with variance $1 .$
(1) Define
$$\varphi(t)=\varphi_{\mu}(t)=\frac{1}{\sqrt{2 \pi}} \int_{\mathbf{R}} e^{i t x-x^{2} / 2} d x$$
for $t \in \mathbf{R}$. Prove that $\varphi$ is of class $C^{1}$ on $\mathbf{R}$ and satisfies $\varphi^{\prime}(t)=-t \varphi(t)$ for all $t \in \mathbf{R}$ and $\varphi(0)=1$.
(2) Deduce that $\varphi(t)=e^{-t^{2} / 2}$ for all $t \in \mathbf{R}$. [Hint: This is an elementary argument with ordinary differential equations, but because the order is 1 , one can define $g(t)=$ $e^{t^{2} / 2} \varphi(t)$ and check by differentiation that $g^{\prime}(t)=0$ for all $t \in \mathbf{R}$.]
148
We will use the following simple version of the Central Limit Theorem:
THEOREM B.7.2. Let $\mathrm{B} \geqslant 0$ be a fixed real number. Let $\left(\mathrm{X}{n}\right)$ be a sequence of independent real-valued random variables with $\left|\mathrm{X}{n}\right| \leqslant \mathrm{B}$ for all $n$. Let
$$\alpha_{n}=\mathrm{E}\left(\mathrm{X}{n}\right), \quad \beta{n}=\mathrm{V}\left(\mathrm{X}{n}^{2}\right)$$ Let $\sigma{\mathrm{N}} \geqslant 0$ be defined by
$$\sigma_{\mathrm{N}}^{2}=\beta_{1}+\cdots+\beta_{\mathrm{N}}$$
for $\mathrm{N} \geqslant 1$. If $\sigma_{\mathrm{N}} \rightarrow+\infty$ as $n \rightarrow+\infty$, then the random variables
$$\mathrm{Y}{\mathrm{N}}=\frac{\left(\mathrm{X}{1}-\alpha_{1}\right)+\cdots+\left(\mathrm{X}{\mathrm{N}}-\alpha{\mathrm{N}}\right)}{\sigma_{\mathrm{N}}}$$
converge in law to a standard gaussian random variable.
Proof. Although this is a very simple case of the general Central Limit Theorem for sums of independent random variables (indeed, even of Lyapunov’s well-known version), we give a proof using Lévy’s criterion for convenience. First of all, we may assume that $\alpha_{n}=0$ for all $n$ by replacing $\mathrm{X}{n}$ by $\mathrm{X}{n}-\alpha_{n}$ (up to replacing B by $2 \mathrm{~B}$, since $\left|\alpha_{n}\right| \leqslant \mathrm{B}$ ). by

By definition, a random vector $X$ with values in $\mathrm{R}^{m}$ is called a (centered) gaussian vector if there exists a non-negative quadratic form $\mathrm{Q}$ on $\mathrm{R}^{m}$ such that the characteristic function $\varphi_{X}$ of $X$ is of the form
$$\varphi_{\mathrm{X}}(t)=e^{-\mathrm{Q}(t) / 2}$$
for $t \in \mathrm{R}^{m}$. The quadratic form can be recovered from $\mathrm{X}$ by the relation
$$\mathrm{Q}\left(t_{1}, \ldots, t_{m}\right)=\sum_{1 \leqslant i, j \leqslant m} a_{i, j} t_{i} t_{j}$$
with $a_{i, j}=\mathbf{E}\left(\mathrm{X}{i} \mathrm{X}{j}\right)$, and the (symmetric) matrix $\left(a_{i, j}\right){1 \leqslant i, j \leqslant m}$ is called the correlation matrix of $X$. The components $X{i}$ of $X$ are independent if and only if $a_{i, j}=0$ if $i \neq j$, i.e., if and only if the components of $X$ are orthogonal.
If $X$ is a gaussian random vector, then $X$ is mild, and in fact
$$\sum_{\boldsymbol{k}} \mathrm{M}{m}(\mathrm{X}) \frac{t{1}^{k_{1}} \cdots t_{m}^{k_{m}}}{k_{1} ! \cdots k_{m} !}=\mathbf{E}\left(e^{t \cdot \mathrm{X}}\right)=e^{\mathrm{Q}(t) / 2}$$
for $t \in \mathbf{R}^{m}$, so that the power series converges on all of $\mathrm{C}^{m}$. The Laplace transform $\psi_{\mathrm{X}}(z)=\mathbf{E}\left(e^{z \cdot \mathrm{X}}\right)$ is also defined for all $z \in \mathbf{C}^{m}$, and in fact
(B.9)
$$\mathbf{E}\left(e^{z \cdot \mathrm{X}}\right)=e^{\mathrm{Q}(z) / 2} .$$
For $m=1$, this means that a random variable is a centered gaussian if and only if there exists $\sigma \geqslant 0$ such that
$$\varphi_{\mathrm{X}}(t)=e^{-\sigma^{2} t / 2}$$
and in fact we have
$$\mathbf{E}\left(\mathrm{X}^{2}\right)=\mathbf{V}(\mathrm{X})=\sigma^{2}$$
If $\sigma=1$, then we say that $\mathrm{X}$ is a standard gaussian random variable (also sometimes called a standard normal random variable). We then have
$$\mathbf{P}(a<\mathrm{X}<b)=\frac{1}{\sqrt{2 \pi}} \int_{a}^{b} e^{-x^{2} / 2} d x$$
for all real numbers $a<b$.
EXERCISE B.7.1. We recall a standard proof of the fact that the measure on $\mathbf{R}$ given by
$$\mu=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2} d x$$
is indeed a gaussian probability measure with variance $1 .$
(1) Define
$$\varphi(t)=\varphi_{\mu}(t)=\frac{1}{\sqrt{2 \pi}} \int_{\mathbf{R}} e^{i t x-x^{2} / 2} d x$$
for $t \in \mathbf{R}$. Prove that $\varphi$ is of class $C^{1}$ on $\mathbf{R}$ and satisfies $\varphi^{\prime}(t)=-t \varphi(t)$ for all $t \in \mathbf{R}$ and $\varphi(0)=1$.
(2) Deduce that $\varphi(t)=e^{-t^{2} / 2}$ for all $t \in \mathbf{R}$. [Hint: This is an elementary argument with ordinary differential equations, but because the order is 1 , one can define $g(t)=$ $e^{t^{2} / 2} \varphi(t)$ and check by differentiation that $g^{\prime}(t)=0$ for all $t \in \mathbf{R}$.]
148
We will use the following simple version of the Central Limit Theorem:
THEOREM B.7.2. Let $\mathrm{B} \geqslant 0$ be a fixed real number. Let $\left(\mathrm{X}{n}\right)$ be a sequence of independent real-valued random variables with $\left|\mathrm{X}{n}\right| \leqslant \mathrm{B}$ for all $n$. Let
$$\alpha_{n}=\mathrm{E}\left(\mathrm{X}{n}\right), \quad \beta{n}=\mathrm{V}\left(\mathrm{X}{n}^{2}\right)$$ Let $\sigma{\mathrm{N}} \geqslant 0$ be defined by
$$\sigma_{\mathrm{N}}^{2}=\beta_{1}+\cdots+\beta_{\mathrm{N}}$$
for $\mathrm{N} \geqslant 1$. If $\sigma_{\mathrm{N}} \rightarrow+\infty$ as $n \rightarrow+\infty$, then the random variables
$$\mathrm{Y}{\mathrm{N}}=\frac{\left(\mathrm{X}{1}-\alpha_{1}\right)+\cdots+\left(\mathrm{X}{\mathrm{N}}-\alpha{\mathrm{N}}\right)}{\sigma_{\mathrm{N}}}$$
converge in law to a standard gaussian random variable.
Proof. Although this is a very simple case of the general Central Limit Theorem for sums of independent random variables (indeed, even of Lyapunov’s well-known version), we give a proof using Lévy’s criterion for convenience. First of all, we may assume that $\alpha_{n}=0$ for all $n$ by replacing $\mathrm{X}{n}$ by $\mathrm{X}{n}-\alpha_{n}$ (up to replacing B by $2 \mathrm{~B}$, since $\left|\alpha_{n}\right| \leqslant \mathrm{B}$ ). by

## 数论代写

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## 编码理论代写

1. 数据压缩（或信源编码
2. 前向错误更正（或信道编码
3. 加密编码
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## 复分析代考

(1) 提到复变函数 ，首先需要了解复数的基本性左和四则运算规则。怎么样计算复数的平方根， 极坐标与 $x y$ 坐标的转换，复数的模之类的。这些在高中的时候囸本上都会学过。
(2) 复变函数自然是在复平面上来研究问题，此时数学分析里面的求导数之尖的运算就会很自然的 引入到复平面里面，从而引出解析函数的定义。那/研究解析函数的性贡就是关楗所在。最关键的 地方就是所谓的Cauchy一Riemann公式，这个是判断一个函数是否是解析函数的关键所在。
(3) 明白解析函数的定义以及性质之后，就会把数学分析里面的曲线积分 $a$ 的概念引入复分析中， 定义几乎是一致的。在引入了闭曲线和曲线积分之后，就会有出现复分析中的重要的定理: Cauchy 积分公式。 这个是易分析的第一个重要定理。