22. Truncated Normal Distribution

Jun He May 31, 2026
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Note

本章主题：截断正态分布。 设 \(X\sim\mathcal N(\mu,\sigma^2)\)，则 \(X\mid a截断正态分布。本章推导其四个性质（记 \(\alpha\equiv\frac{a-\mu}{\sigma}\)、\(\beta\equiv\frac{b-\mu}{\sigma}\)）：p.d.f（22.1） \(f=\frac{\phi\left(\frac{x-\mu}{\sigma}\right)}{\sigma\left(\Phi(\beta)-\Phi(\alpha)\right)}\)（分子不被截断改变、分母重标定使积分为 1）；c.d.f（22.2） \(F=\frac{\Phi\left(\frac{x-\mu}{\sigma}\right)-\Phi(\alpha)}{\Phi(\beta)-\Phi(\alpha)}\)（换元 \(z=\frac{x-\mu}{\sigma}\)）；均值（22.3） \(\mathbb E[X\mid a方差（22.4） \(\text{Var}=\sigma^2\left[1+\frac{\alpha\phi(\alpha)-\beta\phi(\beta)}{\Phi(\beta)-\Phi(\alpha)}-\left(\frac{\phi(\alpha)-\phi(\beta)}{\Phi(\beta)-\Phi(\alpha)}\right)^2\right]\)（经 \(\mathbb E[X^2]\)（22.5）（22.6）分部积分）。单边截断：令 \(a=-\infty\) 或 \(b=\infty\) 即得，同样结果成立。

Note

Chapter theme: the truncated normal distribution. Let \(X\sim\mathcal N(\mu,\sigma^2)\); then \(X\mid atruncated normal distribution. This chapter derives its four properties (write \(\alpha\equiv\frac{a-\mu}{\sigma}\), \(\beta\equiv\frac{b-\mu}{\sigma}\)): p.d.f (22.1) \(f=\frac{\phi\left(\frac{x-\mu}{\sigma}\right)}{\sigma\left(\Phi(\beta)-\Phi(\alpha)\right)}\) (the numerator is unchanged by truncation, the denominator rescales it to integrate to 1); c.d.f (22.2) \(F=\frac{\Phi\left(\frac{x-\mu}{\sigma}\right)-\Phi(\alpha)}{\Phi(\beta)-\Phi(\alpha)}\) (substitution \(z=\frac{x-\mu}{\sigma}\)); mean (22.3) \(\mathbb E[X\mid avariance (22.4) \(\text{Var}=\sigma^2\left[1+\frac{\alpha\phi(\alpha)-\beta\phi(\beta)}{\Phi(\beta)-\Phi(\alpha)}-\left(\frac{\phi(\alpha)-\phi(\beta)}{\Phi(\beta)-\Phi(\alpha)}\right)^2\right]\) (via \(\mathbb E[X^2]\) (22.5) (22.6) and integration by parts). One-sided truncation: plug in \(a=-\infty\) or \(b=\infty\) and the same results hold.

设 \(X\) 服从均值为 \(\mu\)、方差为 \(\sigma^2\) 的正态分布。则变量 \(X\mid a截断正态分布，具有如下性质。本章统一记 $$\alpha\equiv\frac{a-\mu}{\sigma},\qquad\beta\equiv\frac{b-\mu}{\sigma}$$ 其中 \(\phi(\cdot)\) 是标准正态分布的 p.d.f、\(\Phi(\cdot)\) 是标准正态分布的 c.d.f。

  • \(X\mid a
  • \(X\mid a
Important

p.d.f（22.1） $$f(x;\mu,\sigma,a,b)=\frac{\phi\left(\frac{x-\mu}{\sigma}\right)}{\sigma\left(\Phi\left(\frac{b-\mu}{\sigma}\right)-\Phi\left(\frac{a-\mu}{\sigma}\right)\right)}\tag{22.1}$$

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证明 为证 \(X\mid a

Suppose \(X\) has a normal distribution with mean \(\mu\) and variance \(\sigma^2\). Then, the variable \(X\mid a

  • The support of \(X\mid a
  • The p.d.f. of \(X\mid a
Important

p.d.f (22.1) $$f(x;\mu,\sigma,a,b)=\frac{\phi\left(\frac{x-\mu}{\sigma}\right)}{\sigma\left(\Phi\left(\frac{b-\mu}{\sigma}\right)-\Phi\left(\frac{a-\mu}{\sigma}\right)\right)}\tag{22.1}$$

Note

Proof To prove that the p.d.f. of \(X\mid a

  • \(X\mid a
Important

c.d.f（22.2） $$F(x;\mu,\sigma,a,b)=\frac{\Phi\left(\frac{x-\mu}{\sigma}\right)-\Phi\left(\frac{a-\mu}{\sigma}\right)}{\Phi\left(\frac{b-\mu}{\sigma}\right)-\Phi\left(\frac{a-\mu}{\sigma}\right)}\tag{22.2}$$

Note

证明 为证 \(X\mid a

  • The c.d.f. of \(X\mid a
Important

c.d.f (22.2) $$F(x;\mu,\sigma,a,b)=\frac{\Phi\left(\frac{x-\mu}{\sigma}\right)-\Phi\left(\frac{a-\mu}{\sigma}\right)}{\Phi\left(\frac{b-\mu}{\sigma}\right)-\Phi\left(\frac{a-\mu}{\sigma}\right)}\tag{22.2}$$

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Proof To prove that the c.d.f. of \(X\mid a

  • \(X\mid a
Important

均值（22.3） $$\mathbb E[X\mid a

Note

证明 为证 \(X\mid a

  • The mean of \(X\mid a
Important

Mean (22.3) $$\mathbb E[X\mid a

Note

Proof To prove that the mean of \(X\mid a

  • \(X\mid a
Important

方差（22.4） $$\text{Var}[X\mid a

Note

证明 为证方差为 (22.4)，考虑（其中用到均值 (22.3)）： $$\begin{aligned}\text{Var}[X\mid a

把 (22.6) 代入 (22.5)： $$\begin{aligned}\text{Var}[X\mid a

  • 对单边截断，只需令 \(a=-\infty\) 或 \(b=\infty\) 代入，上述同样结果仍成立。
  • The variance of \(X\mid a
Important

Variance (22.4) $$\text{Var}[X\mid a

Note

Proof To prove that the variance is (22.4), consider the following (using the mean (22.3)): $$\begin{aligned}\text{Var}[X\mid a

Plug (22.6) into (22.5): $$\begin{aligned}\text{Var}[X\mid a

  • For one-sided truncation, simply plug in \(a=-\infty\) or \(b=\infty\), and the same results from above hold.