MCS: discrete random variables - geometric distribution

Geometric

Geometric distribution （Geometric distribution） It's a discrete probability distribution . One of them is defined as ： stay $n$ In the thibernoulli experiment , test $k$ The probability of first success . In detail , yes ： front $k-1$ I failed every time , The first $k$ Probability of secondary success , Probability of success in each experiment $p$ remain unchanged . The geometric distribution is Pascal distribution $r=1$ The special case of time .

$$P(k)=p(1-p{)}^{k-1},k=1,2,...$$

$$F(k)=1-(1-p{)}^k,k=1,2,...$$

$$E(k)=\frac{1}{p}$$

$$V(k)=\frac{1-p}{p^2}$$

When variables $x^{\prime }$ It is defined as the number of failures when the experiment succeeds for the first time ,$x^{\prime }=k-1$:

$$P(x^{\prime })=p(1-p{)}^{x^{\prime }}$$

$$F(x^{\prime })=1-(1-p{)}^{x^{\prime }+1}$$

$$E(x^{\prime })=E(x)-1=(1-p)/p$$

$$V(x^{\prime })=V(x)=(1-p)/p^2$$

Generate random variables of geometric distribution $k$:

1. Generate random continuous uniform variables ：$u\sim U(0,1)$
2. $x=int[\frac{ln(1-u)}{ln(1-p)}]+1$

example ： Suppose the probability of a successful experiment is $p=0.2$, Random geometric variable $x$ The first success for this experiment is the number of attempts , Generate a random geometric variable ：

1. Generate random uniform variables ：$u\sim U(0,1),u=0.27$
2. $x=int(ln(1-0.27)/ln(1-0.2))+1=2$

Simulation generates random geometric variables

import numpy as np
import matplotlib.pyplot as plt

def generate_geometric(p=0.1):
    u = np.random.uniform(0, 1)
    x = int(np.log(1 - u)/ np.log(1 - p)) + 1
    return x

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