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Maths

基础公式

希腊字母表

音标 希腊字母大写 希腊字母小写 LaTeX 大写 LaTeX 小写
/ˈælfə/ \(\Alpha\) \(\alpha\) \Alpha \alpha
/ˈbeɪtə/ \(\Beta\) \(\beta\) \Beta \beta
/ˈɡæmə/ \(\Gamma\) \(\gamma\) \Gamma \gamma
/ˈdeltə/ \(\Delta\) \(\delta\) \Delta \delta
/ˈepsɪlɒn/ \(\Epsilon\) \(\epsilon\) \Epsilon \epsilon
/ˈepsɪlɒn/ \(\Epsilon\) \(\varepsilon\) \Epsilon \varepsilon
/ˈzeɪtə/ \(\Zeta\) \(\zeta\) \Zeta \zeta
/ˈiːtə/ \(\Eta\) \(\eta\) \Eta \eta
/ˈθiːtə/ \(\Theta\) \(\theta\) \Theta \theta
/ˈθiːtə/ \(\Theta\) \(\vartheta\) \Theta \vartheta
/aɪˈəʊtə/ \(\Iota\) \(\iota\) \Iota \iota
/ˈkæpə/ \(\Kappa\) \(\kappa\) \Kappa \kappa
/ˈkæpə/ \(\Kappa\) \(\varkappa\) \Kappa \varkappa
/ˈlæmdə/ \(\Lambda\) \(\lambda\) \Lambda \lambda
/mjuː/ \(\Mu\) \(\mu\) \Mu \mu
/njuː/ \(\Nu\) \(\nu\) \Nu \nu
/zaɪ/ \(\Xi\) \(\xi\) \Xi \xi
/əʊˈmaɪkrɒn/ \(\Omicron\) \(\omicron\) \Omicron \omicron
/paɪ/ \(\Pi\) \(\pi\) \Pi \pi
/paɪ/ \(\Pi\) \(\varpi\) \Pi \varpi
/rəʊ/ \(\Rho\) \(\rho\) \Rho \rho
/rəʊ/ \(\Rho\) \(\varrho\) \Rho \varrho
/ˈsɪɡmə/ \(\Sigma\) \(\sigma\) \Sigma \sigma
/ˈsɪɡmə/ \(\Sigma\) \(\varsigma\) \Sigma \varsigma
/tɔː/ \(\Tau\) \(\tau\) \Tau \tau
/ˈjuːpsɪlɒn/ \(\Upsilon\) \(\upsilon\) \Upsilon \upsilon
/faɪ/ \(\Phi\) \(\phi\) \Phi \phi
/faɪ/ \(\Phi\) \(\varphi\) \Phi \varphi
/kaɪ/ \(\Chi\) \(\chi\) \Chi \chi
/psaɪ/ \(\Psi\) \(\psi\) \Psi \psi
/ˈəʊmɪɡə/ \(\Omega\) \(\omega\) \Omega \omega

扩展欧几里得算法

\(q\) \(a\) \(b\) \(s_0\) \(s_1\) \(t_0\) \(t_1\)
\(-\) \(25\) \(21\) \(1\) \(0\) \(0\) \(1\)
\(1\) \(21\) \(4\) \(0\) \(1\) \(1\) \(-1\)
\(5\) \(4\) \(1\) \(1\) \(-5\) \(-1\) \(6\)
\(4\) \(1\) \(0\) \(-5\) \(21\) \(6\) \(-25\)
\[ (-5) * 25 + 6 * 21 = 1 \]

和差化积与积化和差

\[ \begin{aligned} \sin\alpha+\sin\beta &= 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}\\ \cos\alpha+\cos\beta &= 2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}\\ \sin\alpha-\sin\beta &= 2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}\\ \cos\alpha-\cos\beta &= -2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}\\ \end{aligned} \]
\[ \begin{aligned} \sin\alpha\cos\beta &= \frac{1}{2}[\sin(\alpha+\beta)+\sin(\alpha-\beta)]\\ \cos\alpha\cos\beta &= \frac{1}{2}[\cos(\alpha+\beta)+\cos(\alpha-\beta)]\\ \cos\alpha\sin\beta &= \frac{1}{2}[\sin(\alpha+\beta)-\sin(\alpha-\beta)]\\ \sin\alpha\sin\beta &= -\frac{1}{2}[\cos(\alpha+\beta)-\cos(\alpha-\beta)]\\ \end{aligned} \]

微积分

积分

\[ \lim_{n\to\infty}\sum_{k = 1}^n{\dfrac{1}{n+k}} = \lim_{n\to\infty}\sum_{k = 1}^n{\dfrac{1}{1+\frac{k}{n}}}=\int_0^1{\dfrac{1}{1+x}}dx =\ln2. \]

常微分方程

  • \(\dfrac{dy}{dx}=f(x)g(y)\)
\[ \int{\dfrac{dy}{g(y)}=\int{f(x)dx}+C} \]
  • \(\dfrac{dy}{dx}=f(x,y)\)
\[ \begin{aligned} &\text{let}\ y = ux\\ f(x, y) &= f(x, ux)=\varphi(u)\\ \varphi(u) &= u+x\dfrac{du}{dx}\\ \int \dfrac{du}{\varphi(u)-u}&=\int \dfrac{dx}{x} + C =\ln|x| + C\\ \Phi(u)&=\Phi(\dfrac{y}{x}) = \ln|x| + C\\ \end{aligned} \]
  • \(y'+p(x)y=q(x)\)
\[ y = \dfrac{\displaystyle\int e^{\int p(x)dx} q(x)dx+C}{e^{\int p(x)dx}} \]
  • \(y'+p(x)y=q(x)y^n\)
\[ \begin{aligned} &y'y^{-n}+p(x)y^{1-n} = q(x)\\ \text{let}\ &z = y^{1-n},\dfrac{dz}{dx}=(1-n)\dfrac{dy}{dx}y^{-n}\\ &\dfrac{1}{1-n}\dfrac{dz}{dx}+p(x)z = q(x) \end{aligned} \]

多元函数微积分

  • 方向导数

对于多元函数 \(f(x_1, x_2, \cdots, x_n)\)，在点 \(\mathbf{x}_0=(x_{01}, x_{02}, \cdots, x_{0n})\) 处，沿单位向量 \(\mathbf{v} = (v_1, v_2, \cdots, v_n)\) 的方向导数记为

\[ D_{\mathbf{v}} f(\mathbf{x}_0) = \lim_{t \to 0^+} = \dfrac{f(\mathbf{x}_0 + t\mathbf{v}) - f(\mathbf{x}_0)}{t} \]

如果函数在 \(\mathbf{x}_0\) 处可微，则

\[ D_{\mathbf{v}} f(\mathbf{x}_0) = \nabla f(\mathbf{x}_0) \cdot \mathbf{v} \]

概率论与数理统计

基本概念

常用统计量

  • 样本均值 \(\overline{X} = \dfrac{1}{n}\displaystyle\sum^{n}_{i=1}\)，观测值 \(\overline{x} = \dfrac{1}{n}\displaystyle\sum_{i=1}^{n}x_i\).

  • 样本方差 \(S^2 = \dfrac{1}{n-1}\displaystyle\sum_{i=1}^{n}(X_i - \overline{X})^2 = \dfrac{1}{n-1}(\displaystyle\sum_{i=1}^nX_i^2 - n \overline{X}^2)\)

  • 样本标准差 \(S = \sqrt{S^2}\)

  • 样本 \(k\) 阶原点矩 \(A_k = \dfrac{1}{n}\displaystyle\sum_{i=1}^nX_i^k\)

  • 样本 \(k\) 阶中心距 \(B_k = \dfrac{1}{n}\displaystyle\sum_{i=1}^n(X_i - \overline{X})^k\)

定理：对总体 \(X\) 有 \(E(X) = \mu, D(X) = \sigma^2\)；从总体中取出样本，\(E(\overline{X}) = \mu, D(\overline{X}) = \dfrac{\sigma^2}{n},E(S^2) = \sigma^2\)

三种分布

\(\chi^2\) 分布

若 \(X_1,X_2\cdots X_n\) 独立同分布，且满足 \(X\sim N(0,1)\)，则随机变量 \(X^2\) 满足 \(X\sim \chi^2(n)\)，其中 \(n\) 为自由度。

  • 若 \(X\sim N(\mu, \sigma)\)，则 \(\dfrac{1}{\sigma^2}\displaystyle\sum_{i=1}^n(X_i-\mu)\sim \chi^2(n)\)

  • \(E(x) = n, D(X) = 2n\)

  • 若 \(X\sim \chi^2(m),Y\sim \chi^x(n)\)，则 \(X+Y\sim \chi^2(m+n)\)

  • 对于 \(0 < \alpha < 1\)，当 \(P(X^2 \geq \chi^2_{\alpha}(n)) = \alpha\) 称为 \(\chi_{\alpha}^2(n)\) 的 \(\alpha\) 的上侧分位数

\(t\) 分布

若 \(X,Y\) 独立分布，\(X\sim N(0,1), Y\sim \chi^(n)\)，则 \(T = \dfrac{X}{\sqrt{Y/n}}\) 称为 \(t(n)\) 分布（密度函数关于 \(y\) 轴对称）

  • 对于 \(0 < \alpha < 1\)，当 \(P(t \geq t_{\alpha}(n)) = \alpha\) 称为 \(t_{\alpha}(n)\) 的 \(\alpha\) 的上侧分位数

\(F\) 分布

若 \(X\sim \chi^2(m), Y\sim \chi^2(n)\) 相互独立，则 \(F = \dfrac{X/m}{Y/n}\) 称为 \(F(m,n)\) 分布

  • 设 \(X\sim F(m,n)\)，则 \(\dfrac{1}{X}\sim F(n,m)\)

  • 对于 \(0< \alpha < 1\)，当 \(P(F \geq F_{\alpha}(m,n)) = \alpha\) 称为 \(F_{\alpha}(m,n)\) 的上侧分位数

正态总体统计量的分布

单变量

若 \(X\sim N(\mu, \sigma^2)\)，有

  • \(\overline{X} \sim N(\mu, \dfrac{\sigma^2}{n})\)，\(\dfrac{\overline{X} - \mu}{\sigma / \sqrt{n}}\sim N(0,1)\)
  • \(\dfrac{1}{\sigma^2}\displaystyle\sum_{i=1}^n(X_i - \mu)^2\sim \chi^2(n)\)
  • \(\overline{X}\) 与 \(S^2\) 相互独立，且 \(\dfrac{(n-1)S^2}{\sigma^2}\sim \chi^2(n-1)\)

\(X^2 = \displaystyle\sum_{i=1}^n (\dfrac{X_i-\overline{X}}{\sigma})^2\)，但是 \(\displaystyle\sum _{i=1}^n(\dfrac{X_i-\overline{X}}{\sigma}) = 0\) 是一个约束，故自由度从 \(n\) 减为 \(n-1\).

  • \(\dfrac{\overline{X} - \mu}{S/\sqrt{n}} \sim t(n-1)\)

\(z = \dfrac{\overline{X} - \mu}{\sigma / \sqrt{n}}\sim N(0,1)\)，\(\chi^2 = \dfrac{(n-1)S^2}{\sigma^2}\sim \chi^2(n-1)\) 满足相互独立，则 \(t = \dfrac{z}{\sqrt{\frac{\chi^2}{n-1}}}\) 是自由度为 \(n-1\) 的 \(t\) 分布

双变量

若 \(X\sim N(\mu, \sigma^2),Y\sim N(0,1)\)，有

  • \(\dfrac{\overline{X} - \overline{Y} - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{m}+\frac{\sigma_2^2}{n}}}\sim N(0,1)\)

  • \(\dfrac{\overline{X} - \overline{Y} - (\mu_1 - \mu_2)}{S_w\sqrt{\frac{1}{m} + \frac{1}{n}}}\sim t(m+n-2),其中S_w^2 = \sqrt{\dfrac{(m-1)S_1^2+(n-1)S_2^2}{m+n-2}}\)

  • \(\dfrac{S_1^2/S^2_2}{\sigma_1^2/\sigma_2^2}\sim F(m-1,n-1)\)

\(\dfrac{(n-1)S_1^2}{\sigma_1^2}\sim \chi^2(m-1)\), \(\dfrac{(n-1)S_2^2}{\sigma_2^2}\sim \chi^2(n-1)\) 两者相互独立

参数估计

点估计

矩估计

用样本均值 \(\overline{X}\) 代替总体均值 \(\mu\)，用 样本的二阶中心距 代替总体方差 \(\sigma^2\).

极大似然估计

选择能够使概率取到极值的参数

\[ \begin{aligned} &P(X = x) = f(x;\theta)\\ 极大似然函数&L_n(\theta) = \displaystyle\prod_{i = 1}^{n}f(x_i;\theta)\\ 解方程&\dfrac{d}{d\theta}\ln {L_n(\theta)} = 0\\ \end{aligned} \]

衡量点估计的标准

  • 无偏性：\(E(\hat{\theta}) = \theta\)
  • 有效性：若 \(D(\hat{\theta_1}) < D(\hat{\theta_2})\)，则 \(\theta_1\) 比 \(\theta_2\) 更有效
  • 一致性：\(\displaystyle\lim_{n \to \infty}P(\lvert \hat{\theta_n} - \theta\rvert < \varepsilon) = 1\)

样本均值是总体均值的无偏且一致的估计量；

样本方差是总体方差的无偏且一致估计量

区间估计

单正态总体下的参数区间估计

用枢轴量法求参数参数 \(\mu\) 的置信水平为 \(1-\alpha\) 的置信区间

  • \(\sigma\) 已知

构造 \(\overline{X}\) 和 \(\mu\) 的函数

\[ \begin{aligned} &Z = \dfrac{\overline{X}-\mu}{\sigma/\sqrt{n}}\sim N(0,1)\\ 对给定\alpha，取 a < b，满足&P(a\leq \dfrac{\overline{X}-\mu}{\sigma/\sqrt{n}} \leq b) = 1-\alpha\\ 为使置信区间最短，取&a = -z_{\frac{\alpha}{2}}, b = z_{\frac{\alpha}{2}}\\ 则\mu 的置信水平为 1-\alpha 的置信区间为&[\overline{X}-\dfrac{\sigma}{\sqrt{n}}z_{\frac{\alpha}{2}},\overline{X}+\dfrac{\sigma}{\sqrt{n}}z_{\frac{\alpha}{2}}]\\ \end{aligned} \]
  • \(\sigma\) 未知

用样本标准差代替总体标准差，构造 \(\overline{X}\) 和 \(\mu\) 的函数

\[ \begin{aligned} &T = \dfrac{\overline{X}-\mu}{S/\sqrt{n}} \sim t(n-1)\\ 对给定\alpha，取 a < b，满足&P(a\leq \dfrac{\overline{X}-\mu}{S/\sqrt{n}} \leq b) = 1-\alpha\\ t 分布也是对称分布，为使区间长度最短，取&a = -t_{\frac{\alpha}{2}}(n-1), b = t_{\frac{\alpha}{2}}(n-1)\\ 则\mu 的置信水平为 1-\alpha 的置信区间为&[\overline{X} - \dfrac{S}{\sqrt{n}}t_{\frac{\alpha}{2}}(n-1),\overline{X} + \dfrac{S}{\sqrt{n}}t_{\frac{\alpha}{2}}(n-1)]\\ \end{aligned} \]

用枢轴量法求参数参数 \(\sigma^2\) 的置信水平为 \(1-\alpha\) 的置信区间

  • \(\mu\) 已知

参数 \(\sigma^2\) 的无偏估计为 \(\hat{\sigma^2} = \dfrac{1}{n}\displaystyle\sum_{i = 1}^n(X_i - \mu)^2\) 构造 \(\dfrac{1}{n}\displaystyle\sum_{i=1}^n(X_i - \mu)^2\) 和参数 \(\sigma^2\) 的函数

\[ \begin{aligned} &G = \dfrac{n\hat{\sigma^2}}{\sigma^2} = \dfrac{\displaystyle\sum_{i = 1}^n (X_i - \mu)^2}{\sigma^2}\sim \chi^2(n)\\ 对给定\alpha，取 a < b，满足&P(a\leq \dfrac{\displaystyle\sum_{i = 1}^n (X_i - \mu)^2}{\sigma^2} \leq b) = 1-\alpha\\ 通常选取&P(G < a) = P(G > b) = \dfrac{\alpha}{2}, 即 a = \chi_{1-\alpha /2}^2(n), b = \chi_{\alpha / 2}^2(n)\\ 由&\chi_{1-\alpha /2}^2(n) \leq \dfrac{\displaystyle\sum_{i = 1}^n (X_i - \mu)^2}{\sigma^2} \leq \chi_{\alpha / 2}^2(n)\\ 求得置信区间为&\bigg [\dfrac{\displaystyle\sum_{i = 1}^n (X_i-\mu)^2}{\chi_{\alpha / 2}^2(n)}, \dfrac{\displaystyle\sum_{i = 1}^n (X_i-\mu)^2}{\chi^2_{1-\alpha/ 2}(n)}\bigg]\\ \end{aligned} \]
  • \(\mu\) 未知

由于 \(\mu\) 未知，参数 \(\sigma^2\) 的无偏估计为 \(S^2 = \dfrac{1}{n-1}\displaystyle\sum_{i = 1}^n(X_i - \overline{X})^2\)，构造 \(S^2\) 和 \(\sigma^2\) 的函数

\[ \begin{aligned} &\chi^2 = \dfrac{(n-1)S^2}{\sigma^2}\sim \chi^2(n-1)\\ 类似地，可以得到置信区间为&\bigg [\dfrac{\displaystyle\sum_{i = 1}^n (X_i-\mu)^2}{\chi_{\alpha / 2}^2(n-1)}, \dfrac{\displaystyle\sum_{i = 1}^n (X_i-\mu)^2}{\chi^2_{1-\alpha/ 2}(n-1)}\bigg] \end{aligned} \]
待估参数 待估参数 \(G(\hat{\theta}, \theta)\) 双侧置信区间
均值 \(\mu\) \(\sigma\) 已知 \(G = \dfrac{(\overline{X}-\mu)}{\sigma/\sqrt{n}}\sim N(0,1)\) \([\overline{X}-\dfrac{\sigma}{\sqrt{n}}z_{\frac{\alpha}{2}},\overline{X}+\dfrac{\sigma}{\sqrt{n}}z_{\frac{\alpha}{2}}]\)
均值 \(\mu\) \(\sigma\) 未知 \(G = \dfrac{(\overline{X}-\mu)}{S/\sqrt{n}}\sim t(n-1)\) \([\overline{X} - \dfrac{S}{\sqrt{n}}t_{\frac{\alpha}{2}}(n-1),\overline{X} + \dfrac{S}{\sqrt{n}}t_{\frac{\alpha}{2}}(n-1)]\)
方差 \(\sigma\) \(\mu\) 已知 \(G = \dfrac{1}{\sigma^2}\displaystyle\sum_{i=1}^n(X_i-\mu)^2\sim \chi^2(n)\) \(\bigg[\dfrac{\displaystyle\sum_{i=1}^n (X_i-\mu)^2}{\chi_{\alpha / 2}^2(n)}, \dfrac{\displaystyle\sum_{i=1}^n (X_i-\mu)^2}{\chi^2_{1-\alpha/ 2}(n)}\bigg]\)
方差 \(\sigma\) \(\mu\) 未知 \(G = \dfrac{(n-1)S^2}{\sigma^2}\sim \chi^2(n-1)\) \(\bigg[\dfrac{\displaystyle\sum_{i=1}^n (X_i-\mu)^2}{\chi_{\alpha / 2}^2(n-1)}, \dfrac{\displaystyle\sum_{i=1}^n (X_i-\mu)^2}{\chi^2_{1-\alpha/ 2}(n-1)}\bigg]\)

两个正态总体下的参数区间估计

待估参数 待估参数 \(G(\hat{\theta}, \theta)\) 双侧置信区间
均值差 \(\mu_1-\mu_2\) \(\sigma_1,\sigma_2\) 已知 \(G = \dfrac{\overline{X} - \overline{Y}-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{m} + \frac{\sigma_2^2}{n}}}\sim N(0,1)\) 不背， 现推:sleepy:
均值差 \(\mu_1 - \mu_2\) \(\sigma_1=\sigma_2=\sigma^2\)，但 \(\sigma^2\) 未知 \(G = \dfrac{\overline{X} - \overline{Y}-(\mu_1 - \mu_2)}{S_w\sqrt{\frac{1}{m} + \frac{1}{n}}}\sim t(m+n-2)\) \(\cdots\)
方差比 \(\dfrac{\sigma_1^2}{\sigma_2^2}\) \(\mu_1, \mu_2\) 已知 \(\dfrac{\hat{\sigma^2_1}/\hat{\sigma_2^2}}{\sigma_1^2/\sigma_2^2}\sim F(m.n)\) \(\cdots\)
方差比 \(\dfrac{\sigma_1^2}{\sigma_2^2}\) \(\mu_1, \mu_2\) 未知 \(G = \dfrac{S_1^2/S_2^2}{\sigma_1^2/\sigma_2^2}\sim F(m-1,n-1)\) \(\cdots\)

假设检验

  1. 根据问题提出原假设 \(H_0\) 和备择假设 \(H_1\)；
  2. 选择统计量，在原假设成立的条件下确定统计量的分布；
  3. 根据 \(\alpha\) 确定统计量对应的临界值；
  4. 根据样本观测值计算统计量的观测值与临界值比较，选择接受或拒绝原假设 \(H_0\)；

两类错误

  • 弃真：当且仅当小概率事件 \(A\) 发生时才拒绝 \(H_0\)，\(P(A|H_0)\leq \alpha\)
  • 取伪：当 \(\alpha\) 确定后，可以通过增加样本容量来减小取伪概率

显著性检验 只对第一类错误的概率加以控制，而不考虑犯第二类错误的概率，称犯第一类错误的概率 \(\alpha\) 为 显著性水平

后果严重的作为原假设

正态总体的假设检验

单边检验：拒绝域与备择假设的不等号方向一致

渐进无偏估计

贝塞尔校正

设 \(X_1, X_2, \cdots, X_n\) 是来自总体 \(X\) 的样本，且 \(E(X) = \mu\) 已知，\(D(X)=\sigma^2\) 未知，则 \(\dfrac{1}{n}\displaystyle\sum_{i=1}^n(X_i-\mu)^2\) 是 \(\sigma^2\) 的无偏估计。

\(\forall i=1,2,,\cdots,n\)，有 \(E[(X_i-\mu)^2]=\sigma^2\)，则

\[ E\left(\dfrac{1}{n}\sum_{i = 1}^n(X_i-\mu)^2\right) = \dfrac{1}{n}E\left(\sum_{i = 1}^n(X_i-\mu)^2\right)=\dfrac{1}{n}n\sigma^2 \]

设 \(X_1, X_2, \cdots, X_n\) 是来自总体 \(X\) 的样本，\(E(X) = \mu\) 未知，\(D(X)=\sigma^2\) 未知，则 \(\dfrac{1}{n}\displaystyle\sum_{i=1}^n(X_i-\overline{X})^2\) 不是 \(\sigma^2\) 的无偏估计。

\[ \begin{aligned} \dfrac{1}{n}\displaystyle\sum_{i = 1}^n(X_i-\overline{X})^2 &= \dfrac{1}{n}\sum_{i = 1}^n\left((X_i-\mu) + (\mu - \overline{X})\right)^2\\ &= \dfrac{1}{n}\displaystyle\sum_{i = 1}^n(X_i-\mu)^2 + 2(\overline{X}-\mu)(\mu-\overline{X}) + (\mu-\overline{X})^2\\ &= \dfrac{1}{n}\displaystyle\sum_{i = 1}^n(X_i-\mu)^2 -(\mu-\overline{X})^2\\ &\leq \sigma^2 \end{aligned} \]

当 \(\overline{X}\not=\mu\) 时，\(\dfrac{1}{n}\displaystyle\sum_{i=1}^n(X_i-\overline{X})^2\) 会造成对方差的低估，因此需要缩小分母。

已知

\[ \begin{aligned} E(\overline{X}) &= E(\dfrac{1}{n}\sum_{i = 1}^nX_i) = \dfrac{1}{n}\sum_{i = 1}^nE(X_i) = \dfrac{1}{n}n\mu = \mu\\ D(\overline{X}) &= D(\dfrac{1}{n}\sum_{i = 1}^nX_i) = \dfrac{1}{n^2}D(\sum_{i = 1}^nX_i) = \dfrac{1}{n^2}n\sigma^2 = \dfrac{1}{n}\sigma^2\\ E(\overline{X}^2) &= E(\overline{X})^2 + D(\overline{X}) = \mu^2 + \dfrac{1}{n}\sigma^2\\ \end{aligned} \]

于是有

\[ \begin{aligned} E\left(\sum_{i = 1}^n(X_i-\overline{X})^2\right) &= E(\sum_{i = 1}^nX_i^2-n\overline{X}^2)\\ &= \sum_{i = 1}^nE(X_i^2) - nE(\overline{X}^2)\\ &= n(\mu^2 + \sigma^2) - n(\mu^2 + \dfrac{\sigma^2}{n})\\ &= (n - 1)\sigma^2\\ \end{aligned} \]

最终得到 \(\dfrac{1}{n-1}\displaystyle\sum_{i=1}^n(X_i-\overline{X})^2\) 是 \(\sigma^2\) 的无偏估计。

由于 \(\dfrac{1}{n}\displaystyle\sum_{i=1}^n(X_i-\overline{X})^2=\dfrac{n}{n-1}\sigma^2\)，且 \(\displaystyle\lim_{n\to\infty}\dfrac{n-1}{n}\sigma^2=\sigma^2\)，则称 \(\dfrac{1}{n}\displaystyle\sum_{i=1}^n(X_i-\overline{X})^2\) 为 \(\sigma^2\) 的渐进无偏估计。