MCS: discrete random variable Poisson distribution

Poisson

When the event is within a specified time interval ( Unit time ), At a fixed average instantaneous rate ( The average number of occurrences )$\theta$ happen , Then the variable describing the number of events per unit time is Poisson variable . Poisson distribution is suitable for describing the number of random events per unit time , for example ： The number of vehicles passing the intersection per minute ; The number of passengers waiting at the station every hour ;

$$P(x)=\frac{{\theta }^x{e}^{-\theta }}{x!},x=0,1,2,...$$

Expectation and variance ：

$$E(x)=\theta$$
$$V(x)=\theta$$

Poisson variable & Exponential variable ：$t$

$$E(t)=\frac{1}{\theta }$$

Generate random Poisson variables

Generate random Poisson variables $x$, And $E(x)=V(x)=\theta$

1. Make $x=0,i=0,S_t=0$
2. $i=i+1$, Generate random exponential variables ：$t$, And $E(t)=1/\theta$,$S_t=S_t+t$
3. if $S_t>1$,$\to step5$
4. if $S_t<=1,x=x+1$,$\to step2$
5. Return $x$

example ： hypothesis $x$ Is a random variable subject to Poisson distribution , The expected number of events per unit time is $\theta =2.4$, Let's use the exponential variable $t$,($E(t)=1/2.4$) To generate Poisson variables ：

1. Make $x=0,S_t=0$
2. $i=1,t=0.21,S_t=0.21,x=1$
3. $i=2,t=0.43,S_t=0.64,x=2$
4. $i=3,t=0.09,S_t=0.73,x=3$
5. $i=4,t=0.31,S_t=1.04,x=4$
6. $x=3$

example ： One 24 A gas station that is open 24 hours , Every day 200-300 A car came here to refuel , among 80% It's a car ,15% It's a truck ,5% It's a motorcycle . The minimum gasoline consumption per car is 3 Gallon , The average is 11 Gallon . Trucks consume at least 8 Gallon , The average is 20 Gallon . Motorcycles consume at least 2 Gallon , The average is 4 Gallon . Analysts want to figure out the distribution of the total consumption of gas stations in a day .

Simulate the future with random numbers 1000 God , Gasoline consumption in gas stations , The type of vehicle and the amount of fuel consumed are generated by random numbers .

1. Record analog data , Total fuel consumption per day $G$
2. Sort from small to large ,$G(1)<=G(2)<=G(3)<=...G(1000)$
3. Estimated quantile ：$G(p\times 1000)$,$p=0.01,G(0.01\times 1000)=G(10)$

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt

def gas_consum_oneday():
    num_vehicle = np.clip(np.random.poisson(300), 200, 500)
    num_cars = int(num_vehicle * 0.8)
    num_trucks = int(num_vehicle * 0.15)
    num_motos = num_vehicle - (num_cars + num_trucks)
    car_consume = [np.random.randint(3, 19) for i in range(num_cars)]
    truck_consume = [np.random.randint( 8, 32) for i in range(num_trucks)]
    moto_consume = [np.random.randint( 2, 6) for i in range(num_motos)]
    sum_gas = np.sum(car_consume + truck_consume + moto_consume)
    return num_vehicle, num_cars, num_trucks, num_motos, sum_gas

	vehicle_record	cars_record	trucks_record	motos_record	gas_record	P
9	201	160	30	11	2362	0.01
49	202	161	30	11	2447	0.05
99	213	170	31	12	2498	0.10
199	215	172	32	11	2617	0.20
299	228	182	34	12	2723	0.30
399	245	196	36	13	2841	0.40
499	254	203	38	13	2965	0.50
599	259	207	38	14	3089	0.60
699	262	209	39	14	3199	0.70
799	276	220	41	15	3332	0.80
899	288	230	43	15	3468	0.90
949	297	237	44	16	3536	0.95
989	299	239	44	16	3616	0.99

Simulation results ：

• The daily oil consumption exceeds 3536 The probability of gallons is ：5%
• The daily fuel consumption exceeds 3616 The probability of gallons is ：1%
• Daily fuel consumption 90% The interval is ：$P(2447<=G<=3536)=0.90$