広島大学 先進理工系科学研究科 情報科学プログラム 2022年8月実施 専門科目II 問題6

标签：

• Hiroshima-University
• Sigmoid-Function

Author

祭音Myyura

Description

Let the sigmoid function be

$$\sigma(x) = \frac{1}{1 + \exp(-x)}.$$

(1) Show that the following equation holds:

$$1 - \sigma(x) = \sigma(-x).$$

(2) Show that the derivative of the sigmoid function can be written as

$$\frac{d}{dx}\sigma(x) = \sigma(x)(1-\sigma(x)).$$

(3) The inverse function of the sigmoid function $\sigma$ is called the logit function. Show that the logit function $\sigma^{-1}$ can be written as

$$\sigma^{-1}(p) = \log\!\left(\frac{p}{1-p}\right),$$

where $0 \le p \le 1$.

题目描述

定义 sigmoid 函数

$$\sigma(x)=\frac{1}{1+\exp(-x)}.$$

1. 证明

   $$1-\sigma(x)=\sigma(-x).$$

2. 证明其导数可写为

   $$\frac{d}{dx}\sigma(x)=\sigma(x)\bigl(1-\sigma(x)\bigr).$$

3. sigmoid 函数 $\sigma$ 的反函数称为 logit 函数。证明

   $$\sigma^{-1}(p)=\log\!\left(\frac{p}{1-p}\right),$$

   其中题面给定 $0\le p\le1$。

考点

• Sigmoid 函数：由定义推导其对称恒等式、导数的自表达形式，并通过代数反解得到 logit 反函数。

Kai

Throughout, $\log$ denotes the natural logarithm.

(1)

Start from the definition:

$$\sigma(x) = \frac{1}{1+e^{-x}}.$$

Then

$$1 - \sigma(x) = 1 - \frac{1}{1+e^{-x}} = \frac{1+e^{-x} - 1}{1+e^{-x}} = \frac{e^{-x}}{1+e^{-x}}.$$

Multiply numerator and denominator by $e^{x}$:

$$1 - \sigma(x) = \frac{e^{-x}e^{x}}{(1+e^{-x})e^{x}} = \frac{1}{1+e^{x}} = \sigma(-x).$$

(2)

Differentiate the definition:

$$\sigma(x) = \frac{1}{1+e^{-x}}.$$

Using the chain rule:

$$\begin{aligned} \frac{d}{dx}\sigma(x) &= \frac{d}{dx}\left(1+e^{-x}\right)^{-1} \\ &= -1\cdot\left(1+e^{-x}\right)^{-2}\cdot\frac{d}{dx}(1+e^{-x}) \\ &= -\left(1+e^{-x}\right)^{-2}\cdot(-e^{-x}) \\ &= \frac{e^{-x}}{\left(1+e^{-x}\right)^2}. \end{aligned}$$

Now compute $\sigma(x)(1-\sigma(x))$, from (1), we have

$$1-\sigma(x) = \frac{e^{-x}}{1+e^{-x}}.$$

hence

$$\sigma(x)(1-\sigma(x)) = \frac{1}{1+e^{-x}}\cdot\frac{e^{-x}}{1+e^{-x}} = \frac{e^{-x}}{\left(1+e^{-x}\right)^2}.$$

(3)

Let $p = \sigma(x)$. Then

$$p = \frac{1}{1+e^{-x}}.$$

We want to solve for $x$ in terms of $p$.

First invert the fraction:

$$\frac{1}{p} = 1 + e^{-x}.$$

So

$$e^{-x} = \frac{1}{p} - 1 = \frac{1-p}{p}.$$

Take the natural logarithm of both sides:

$$-x = \log\left(\frac{1-p}{p}\right),$$

so

$$x = -\log\left(\frac{1-p}{p}\right) = \log\left(\frac{p}{1-p}\right).$$

Therefore the inverse function of $\sigma$ is

$$\sigma^{-1}(p) = \log\left(\frac{p}{1-p}\right),$$

which is defined for $0 < p < 1$ (and extends to $\pm\infty$ at the endpoints $p=0,1$).