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---
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title: 信息论角度推导高斯分布的概率密度函数
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mathjax: true
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comments: true
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max_depth: 4
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abbrlink: '741621e3'
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date: 2025-07-02 09:20:07
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tags:
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- 高斯分布
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- 信息论
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categories:
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- [数学, 概率论]
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---
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$\newcommand\diff{\mathop{}\!\mathrm{d}}$
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# 背景与目标
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在统计建模中，我们常常面临这样一个问题：**已知某个连续随机变量的期望和方差，如何合理地确定它的概率分布？** 最大熵原理（Maximum Entropy Principle）提供了一个具有信息论意义的答案：
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> 在已知约束（如期望与方差）下，熵最大的概率密度函数是最“中立”或最“无偏”的选择。
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熵在此表示不确定性或信息量。连续型随机变量的微分熵定义为：
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$$
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\begin{equation*}
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H[p] = -\int_{-\infty}^{\infty} p(x) \ln p(x) \, \diff x
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\end{equation*}
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$$
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我们的目标是：**在已知期望和方差的约束下，最大化熵 $H[p]$，求出最优概率密度函数 $p(x)$**。
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<!-- more -->
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# 约束条件
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我们假设随机变量 $X$ 满足：
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- 归一化约束：
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$$
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\begin{equation*}
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\int_{-\infty}^{\infty} p(x) \, \diff x = 1
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\end{equation*}
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$$
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- 期望约束：
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$$
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\begin{equation*}
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\int_{-\infty}^{\infty} x \, p(x) \, \diff x = \mu
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\end{equation*}
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$$
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- 方差约束：
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$$
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\begin{equation*}
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\int_{-\infty}^{\infty} (x - \mu)^2 \, p(x) \, \diff x = \sigma^2
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\end{equation*}
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$$
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# 构造拉格朗日泛函
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引入拉格朗日乘子 $\lambda_0, \lambda_1, \lambda_2$，构造带约束的目标泛函：
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$$
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\begin{equation*}
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\mathcal{L}[p] = -\int p(x) \ln p(x) \diff x
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+ \lambda_0 \left( \int p(x) \diff x - 1 \right)
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+ \lambda_1 \left( \int x p(x) \diff x - \mu \right)
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+ \lambda_2 \left( \int (x - \mu)^2 p(x) \diff x - \sigma^2 \right)
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\end{equation*}
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$$
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将所有积分项合并为：
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$$
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\begin{equation*}
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\mathcal{L}[p] = \int \left[
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- p(x) \ln p(x)
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+ \lambda_0 p(x)
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+ \lambda_1 x p(x)
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+ \lambda_2 (x - \mu)^2 p(x)
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\right] \diff x
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\end{equation*}
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$$
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# 变分法求极值
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我们考虑一个扰动：
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$$
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\begin{equation*}
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p(x) \rightarrow p(x) + \epsilon \eta(x)
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\end{equation*}
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$$
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其中 $\epsilon$ 是足够小的实数，$\eta(x)$ 是任意光滑、在积分区间内有界且满足边界条件的函数（可理解为“任意小的扰动”）。为了求出 $\dfrac{\delta \mathcal{L}}{\delta p(x)}$，我们代入扰动项进行展开：
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$$
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\begin{equation*}
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\mathcal{L}[p + \epsilon \eta]
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= -\int (p(x) + \epsilon \eta(x)) \ln (p(x) + \epsilon \eta(x)) \, \diff x + \text{其余约束项}
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\end{equation*}
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$$
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> 注意：这里只展开了主熵项 $-\displaystyle\int p \ln p$，约束项形式类似，将在后面统一处理。
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---
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对主项使用泰勒展开（忽略高阶小量）：
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> **泰勒展开回顾：**
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>
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> 设 $f(p) = p \ln p$，我们对 $p(x) + \epsilon \eta(x)$ 展开，有：
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> $$
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> f(p + \epsilon \eta) = p \ln p + \epsilon \eta (1 + \ln p) + o(\epsilon)
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> $$
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> 其中：
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> $$
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> \frac{\diff }{\diff p}(p \ln p) = \ln p + 1
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> $$
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将其代入：
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$$
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\begin{align*}
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\mathcal{L}[p + \epsilon \eta]
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&= -\int \left[ p(x) \ln p(x) + \epsilon \eta(x)(1 + \ln p(x)) \right] \diff x + o(\epsilon) + \text{约束项} \\
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&= \mathcal{L}[p] - \epsilon \int \eta(x) (1 + \ln p(x)) \diff x + o(\epsilon) + \text{约束项}
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\end{align*}
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$$
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对约束项也类似展开：
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- 对归一化项：
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$$
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\begin{equation*}
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\lambda_0 \left( \int (p + \epsilon \eta) \diff x - 1 \right)
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= \lambda_0 \left( \int p \, \diff x - 1 + \epsilon \int \eta \, \diff x \right)
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\end{equation*}
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$$
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- 对期望项：
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$$
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\begin{equation*}
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\lambda_1 \left( \int x (p + \epsilon \eta) \diff x - \mu \right)
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= \lambda_1 \left( \int x p \, \diff x - \mu + \epsilon \int x \eta \, \diff x \right)
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\end{equation*}
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$$
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- 对方差项：
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$$
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\begin{equation*}
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\lambda_2 \left( \int (x - \mu)^2 (p + \epsilon \eta) \diff x - \sigma^2 \right)
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= \lambda_2 \left( \int (x - \mu)^2 p \, \diff x - \sigma^2 + \epsilon \int (x - \mu)^2 \eta(x) \diff x \right)
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\end{equation*}
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$$
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---
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将所有一阶 $\epsilon$ 项收集起来：
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$$
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\begin{equation*}
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\mathcal{L}[p + \epsilon \eta] = \mathcal{L}[p]
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+ \epsilon \int \eta(x) \left[ - (1 + \ln p(x)) + \lambda_0 + \lambda_1 x + \lambda_2 (x - \mu)^2 \right] \diff x + o(\epsilon)
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\end{equation*}
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$$
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对 $\epsilon$ 求导并令其在 $\epsilon = 0$ 处为零：
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$$
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\begin{equation*}
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\left. \frac{\diff}{\diff \epsilon} \mathcal{L}[p + \epsilon \eta] \right|_{\epsilon=0}
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= \int \eta(x) \left[ -\ln p(x) - 1 + \lambda_0 + \lambda_1 x + \lambda_2 (x - \mu)^2 \right] \diff x = 0
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\end{equation*}
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$$
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根据变分法基本引理，这个积分当且仅当括号内为 $0$ 时对所有扰动函数 $\eta(x)$ 为零：
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> 变分法基本引理：
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>
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> 若函数 $F(x)$ 满足
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> $$
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> \int_{a}^{b} F(x) \cdot \eta(x) dx = 0
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> $$
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> 对所有合适扰动 $\eta(x)$ 成立，则必有
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> $$
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> F(x) = 0 \quad \forall x \in (a, b)
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> $$
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>
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> 这是因为若 $F(x_0) \ne 0$，则可构造在 $x_0$ 附近支持的 $\eta(x)$ 使得积分不为零，导致矛盾。
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$$
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\begin{equation*}
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\frac{\delta \mathcal{L}}{\delta p(x)} = -\ln p(x) - 1 + \lambda_0 + \lambda_1 x + \lambda_2 (x - \mu)^2 = 0
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\end{equation*}
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$$
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也即求得了变分导数。移项得：
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$$
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\begin{equation*}
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\ln p(x) = \lambda_0' + \lambda_1 x + \lambda_2 (x - \mu)^2
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\end{equation*}
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$$
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两边取指数：
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$$
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\begin{equation*}
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p(x) = \exp\left( \lambda_0' + \lambda_1 x + \lambda_2 (x - \mu)^2 \right)
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\end{equation*}
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$$
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要满足可积性，需 $\lambda_2 < 0$。设
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$$
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\begin{equation*}
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p(x) = A \cdot \exp\left( -\alpha (x - \mu)^2 \right)
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\end{equation*}
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$$
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# 归一化与确定参数
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根据归一化条件：
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$$
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\begin{equation*}
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A \cdot \int_{-\infty}^{\infty} \exp\left( -\alpha (x - \mu)^2 \right) \diff x = 1
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\end{equation*}
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$$
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已知高斯积分结果：
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$$
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\begin{equation*}
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\int_{-\infty}^{\infty} \exp\left( -\alpha (x - \mu)^2 \right) \diff x = \sqrt{\frac{\pi}{\alpha}}
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\end{equation*}
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$$
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得到：
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$$
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\begin{equation*}
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A = \sqrt{\frac{\alpha}{\pi}}
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\end{equation*}
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$$
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若令 $\alpha = \dfrac{1}{2\sigma^2}$，则最终得：
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$$
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\begin{equation*}
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p(x) = \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left( -\frac{(x - \mu)^2}{2\sigma^2} \right)
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\end{equation*}
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$$
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# 总结
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从最大熵原理出发，在已知均值与方差的约束下，最“无偏”或熵最大的概率密度函数是：
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$$
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\begin{equation*}
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p(x) = \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left( -\frac{(x - \mu)^2}{2\sigma^2} \right)
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\end{equation*}
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$$
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这正是**高斯分布**的标准形式，说明它不仅数学上方便，**信息论上也是最合理的选择**。
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