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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

## 物理代写|统计力学代写Statistical mechanics代考|Gaussian distribution

For which value of $x$ is $f(x)$ in Eq. (3.39) maximized? To answer that question, we develop an approximation assuming that $N$ and $x$ are sufficiently large that Stirling’s approximation is valid, but where no restriction is placed on $p$. We find, applying Eq. (3.14) to Eq. (3.39):
$$\ln f(x) \approx N \ln \left(\frac{N q}{N-x}\right)+x \ln \left(\frac{N-x}{x} \frac{p}{q}\right) \equiv g(x) .$$
The first and second derivatives of $g(x)$ are:
$$g^{\prime}(x)=\ln \left(\frac{(N-x)}{x} \frac{p}{q}\right) \quad g^{\prime \prime}(x)=-\frac{N}{x(N-x)} .$$
Thus, $g(x)$ has an extremum (a maximum) at $x=N p=\mu$, i.e., $g^{\prime}(x=N p)=0$ (show this). Moreover, the function itself has the value zero at $x=N p, g(x=N p)=0$. The Taylor series of $g(x)$ about $x=\mu$ is therefore at lowest order
$$g(x) \approx-\frac{1}{2 N p q}(x-\mu)^{2}=-\frac{1}{2 \sigma^{2}}(x-\mu)^{2},$$
where $\sigma^{2} \equiv N p q$. Combining Eqs. (3.47) and (3.45),
$$f(x)=\mathrm{e}^{g(x)} \approx \exp \left(-\frac{1}{2 \sigma^{2}}(x-\mu)^{2}\right) .$$
Distributions in the form of Eq. (3.48) are known as the Gaussian or normal distribution. Its shape is shown in Fig. 3.6.

We would expect that Eq. (3.48), an approximate expression derived from a second-order laylor expansion, would provide accurate predictions only for sufficiently small values of $x-\mu$, say for $|x-\mu| \lesssim \sqrt{N}$. A theorem, the DeMoivre-Laplace theorem states that Eq. (3.48) holds for all $x$ in the limit $N \rightarrow \infty$. For $A \sqrt{N} \leq x-\mu \leq B \sqrt{N}$, where $A, B$ are arbitrary numbers with $A<B$, we have the normalized probability distribution.

## 物理代写|统计力学代写Statistical mechanics代考|The law of large numbers

Consider, in the sequence of experiments underlying the binomial distribution, that we define a new random variable $x_{i}$ that indicates whether success is achieved on the $i^{\text {th }}$-trial, with $x_{i}=1$ for a successful outcome and $x_{i}=0$ if not. Each variable $x_{i}$ is independent. The quantity $S_{N}=\sum_{i=1}^{N} x_{i}$ records the number of successful outcomes achieved in $N$ trials. We would expect that after many trials the ratio $S_{N} / N \rightarrow p$, the probability for success on a single trial. This expectation is made precise by the law of large numbers.

Theorem. Law of large numbers. For $\left{x_{i}\right}$ a sequence of mutually independent random variables having a common distribution, then for every $\epsilon>0$,
$$P\left(\left|\frac{x_{1}+\cdots+x_{N}}{N}-p\right|>\epsilon\right) \stackrel{N \rightarrow \infty}{\longrightarrow} 0 .$$
Said differently, the probability that $\left(x_{1}+\cdots+x_{N}\right) / N$ differs from $p$ by less than an arbitrarily prescribed number $\epsilon$ tends to unity as $N \rightarrow \infty$.
Proof. Using Chebyshev’s inequality, we have, for any $t>0$,
$$P\left(\left|S_{N}-\mu\right| \geq t\right) \leq \frac{N \sigma^{2}}{t^{2}}$$
where the variance of $S_{N}=\sum_{i=1}^{N} x_{i}$ is $N \sigma^{2}$, with $\sigma^{2}$ the variance of $x_{i}$. For $t=N \epsilon$, the right side becomes $\sigma^{2} /\left(N \epsilon^{2}\right)$, which tends to zero as $N \rightarrow \infty$. This accomplishes the proof of Eq. (3.52).

# 统计力学代考

## 物理代写|统计力学代写Statistical mechanics代考|Gaussian distribution

$$\ln f(x) \approx N \ln \left(\frac{N q}{N-x}\right)+x \ln \left(\frac{N-x}{x} \frac{p}{q}\right) \equiv g(x) .$$

$$g^{\prime}(x)=\ln \left(\frac{(N-x)}{x} \frac{p}{q}\right) \quad g^{\prime \prime}(x)=-\frac{N}{x(N-x)} .$$

$$g(x) \approx-\frac{1}{2 N p q}(x-\mu)^{2}=-\frac{1}{2 \sigma^{2}}(x-\mu)^{2},$$

$$f(x)=\mathrm{e}^{g(x)} \approx \exp \left(-\frac{1}{2 \sigma^{2}}(x-\mu)^{2}\right) .$$

## 物理代写|统计力学代写Statistical mechanics代考|The law of large numbers

$$P\left(\left|\frac{x_{1}+\cdots+x_{N}}{N}-p\right|>\epsilon\right) \stackrel{N \rightarrow \infty}{\longrightarrow} 0 .$$

$$P\left(\left|S_{N}-\mu\right| \geq t\right) \leq \frac{N \sigma^{2}}{t^{2}}$$

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