Mathematical Foundations of Artificial Intelligence--Probability and Statistics 14: Exponential Distribution, Weibull Distribution and Uniform Distribution of Continuous Random Variables

I. Introduction

In " Mathematics of Artificial Intelligence - Probability and Statistics 12: Probability Density Function and Normal Distribution of Continuous Random Variables ", the probability distribution, probability density function and normal distribution of continuous random variables are introduced, " Mathematics of Artificial Intelligence - Probability and Statistics 13 : The Standard Normal Distribution of Continuous Random Variables " introduces the standard normal distribution, and this article will continue to introduce the distribution functions of several continuous random variables.

2. Exponential distribution

2.1. Definition

If the random variable X has a probability density function: f ( x ) = { 0 when x ≤ 0 λ e − λ x when x > 0 f(x) = {\Huge \{}{\huge^{λe^{ -λx}\;\;\;\;When x>0}_{0\;\;\;\;\;\;\;\;\;\;\;\;When x≤0} }f(x)={ 0when x ≤ 0 _λe−λxwhen x > 0 _
It is said that X obeys an exponential distribution , where λ is a parameter, and its value is greater than 0. When x is greater than 0, -λx is a negative number, so this distribution is also called a negative exponential distribution .

For the exponential distribution, when x≤0, f(x) = 0, which means that the probability of the random variable taking a negative value is 0, so X only takes a positive value, and the function f(x) is discontinuous at x=0.

2.2. Distribution function of exponential distribution

Distribution function of exponential distribution F ( x ) = ∫ − ∞ xf ( t ) d ( t ) = { 0 When x ≤ 0 1 − e − λ x When x > 0 F(x)=∫_{-∞ }^xf(t)d(t)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;= {\Huge \{}{\Large^{1-e^{-λx}\;\;\;\;When x>0}_{0\;\;\;\ ;\;\;\;\;\;\;\;\;When x≤0}}F(x)=∫−∞x​f(t)d(t)={ 0when x ≤ 0 _1−e−λxwhen x > 0 _

2.3. Applicable scenarios and derivation of exponential distribution

One of the most common scenarios for the exponential distribution is the lifetime distribution.

Imagine a mass-produced electronic component whose lifetime X is a random variable, and the distribution function of X is recorded as F(x). Proof: Under certain conditions, F(x) is the distribution function of the exponential distribution.

In order to prove the technical "no aging" assumption, that is to say, "under the condition that the component is still working normally at time x, its failure rate always remains a certain constant λ>0, which has nothing to do with x. The failure rate is the unit The probability of failure within a length of time. In the form of conditional probability, the above assumption can be expressed as:
$$P(x\le X\le x+h|X>x)/h=l(h\to 0)$$
This formula is explained as follows:

1. The component works normally at time x, which means its life is greater than x, that is, X>x;
2. At x, it fails within a period of time h, that is, x≤X≤x+h;
3. Divide this conditional probability by the length h of the time period to get the average failure rate at time x;
4. Let h→0, the instantaneous failure rate is obtained, which should be a constant λ according to the assumption.

According to the definition of conditional probability, notice that P(X>x)=1-F(x), and { X > x } { x ≤ X ≤ x + h } = { x < X ≤ x + h } \{X >x\}\{x≤X≤x+h\}=\{x<X≤x+h\}{ X>x}{ x≤X≤x+h}={ x<X≤x+h}
有$$\begin{gathered} P(x\le X\le x+h|X>x)/h=P(x<X\le x+h)/(h(1-F(x))) \\ =(F(x+h)-F(x))/h]/(1-F(x)) \\ \to F^{\prime }(x)/(1-F(x\left)\right)= \end{gathered}$$
The general solution of the differential equation$$\lambda$$$F(x)=1-C{e}^{-\lambda x}$ (x>0), when x≤0, F(x) is 0. The constant C can be obtained as 1 with the initial condition F(0)=0.

Old Ape Note:

1. The above derivation process uses the definition of limit, conditional probability formula, conditional probability definition, definition and properties of distribution function, which is quite an interesting derivation process;
2. $F\left(x\right)=1-C{e}^{-\lambda x}$ (x>0) is obtained by solving the differential equation. This process is demonstrated as follows:
   Let F(x)=y, then F ′ (x)/(1−F(x))=λ can be simplified is dy/(dx(1-y))=λ, then we can get:
   dy/(1-y)=λdx, and integrate both sides of this formula: ∫dy/(1-y)=∫λdx, then we get:
   -ln(1-y)+c1 = λx+c2
   ∴ ln(1-y)=-λx+c3
   ∴$1-y=e^{-\lambda x+c3}=e^{c3}{e}^{-\lambda x}=C{e}^{-\lambda x}$
   ∴y=1-$C{e}^{-\lambda x}$

Note that in the whole derivation process, the result of the constant and difference power operation is still a constant, so there are c1, c2, c3 and C.

2.4. Supplementary instructions

From the above derivation process, we can know that the meaning of λ is the failure rate, the higher the failure rate, the smaller the average life, so the exponential distribution describes the life distribution without aging, but in practice, "no aging" is impossible, so The exponential distribution is only an approximate lifetime distribution. For some long-lived components, in the initial stage, the aging phenomenon is very small, and the exponential distribution at this stage is more accurate and quite describes the life distribution.

3. Weibull distribution

If aging is considered during the derivation of exponential distribution, the failure rate will increase with time, so it should be taken as an increasing function of x λ xm λx $\lambda x^m$ , where both λ and m are constants greater than 0. Under this condition, according to the reasoning of exponential distribution, it will be concluded that the life distribution F(x) satisfies the differential equation$$F^{\prime }（x）/[1-F(x)]=\lambda x^m$$
combined with F(0)=0, we get:$$F(x)=1-e^{-(\lambda /(m+1\left)\right)x^{m+1take}}$$
α = m+1(α>1)，以把λ/(m+1)记的λ，电影：$$F(x)=1-e^{-\lambda x^a}(x>0)$$
and when x≤0, F(x)=0, the function F(x) at this time is called Weibull distribution function. The density function of this distribution is:
f ( x ) = { 0 when x ≤ 0 λ α x α − 1 e − λ x α when x > 0 f(x) = {\Huge \{}{\huge^ {λαx^{α -1}e^{-λx^α }\;\;\;\;When x>0}_{0\;\;\;\;\;\;\;\;\ ;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;When x≤0}}f(x)={ 0when x ≤ 0 _λαxa − 1 e−λxawhen x > 0 _

Like the exponential distribution, the Weibull distribution plays an important role in reliability statistical analysis. In fact, the exponential distribution is a special case of α=1 of the Weibull distribution.

3. Uniform distribution

3.1. Definition

Let the random variable X have a probability density function:
f ( x ) = { 0 other 1 b − a when a ≤ x ≤ b f(x) = {\Huge \{} ^{ {\huge\frac {1}{ ba}}{\large\;\;\;\;\;\;\;\;\;\;\;\;when a≤x≤b}}_{\large0\;\;\;\ ;\;\;\;\;\;\;\;\;\;\;\;\;\;\;other}f(x)={ 0otherb−a1​this a≤x≤when b
Then it is said that X obeys the uniform distribution on the interval [a,b] , which is denoted as X~R(a,b), where a and b are constants and satisfy -∞<a<b<∞.

The distribution function for a uniform distribution is:

3.2 An application of uniform distribution

For the general distribution function F(x), if X~R(0,1), and F(x) is continuously and strictly monotonically increasing everywhere, its inverse function G exists, and G(x) ~ F.

Proof :
∵{G(X)≤x}
∴{F(G(X))≤F(x)}
∴{X≤F(x)}
∵X ~ R(0,1)
∴R(0,1 ) distribution function is F(x)=x（0<x<1）
∴P(G(X)≤x)=P(X≤F(x))=F(x)
∴G(X) ~ F

Therefore, a uniform distribution can be used to simulate the general distribution F satisfying the above conditions.

Four. Summary

This article is the summary and thinking of the old ape learning Mr. Chen Xiru's "Probability Theory and Mathematical Statistics" published by the University of Science and Technology of China Press. In the article, it introduces the concepts of exponential distribution, Weibull distribution and uniform distribution, as well as some of the derivation process , in the text, some derivation processes are supplemented according to Lao Yuan's own understanding.

For more mathematical foundations of artificial intelligence, please refer to the column " Mathematical Foundations of Artificial Intelligence ".

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