UNIT06A：離散機率理論驗證

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pacman::p_load(dplyr)

【1】\(n\)很大時，二項分佈(Binominal Dist.)會趨近於常態(Normal)分佈

• When \(n\) is large, \(Binom[n, p]\) approaches \(Norm[\mu = n p, \sigma=\sqrt{n p (1-p)}]\)

• \(X \sim Binom[n, p] \, \Rightarrow \, Exp(X) = n \cdot p \, , \, Var(X) = n \cdot p \cdot (1-p)\)

par(mfrow=c(1,1), mar=c(3,4,3,1), cex=0.7)
n = 1000; p = 0.2
rbinom(500000, n, p) %>% hist(breaks=80, freq=F, main="")
curve(dnorm(x, mean=n*p, sd=sqrt(n*p*(1-p))), col='red', lwd=2, add=T)

par(mfrow=c(1,2), cex=0.7)
n = 10; p = 0.2
rbinom(100000, n, p) %>% hist(freq=F, breaks=(0:n)-0.01)
rnorm(100000, n*p, sqrt(n*p*(1-p))) %>% hist(freq=F)

💡 : 當期望值夠大的時候， 二項分佈會以期望值為中心向兩邊對稱的伸展，但是如果期望值不夠大的話，這個分佈的左尾就會受到擠壓，變成一個不對稱的分佈。

【2】\(n \times p\)不大時，二項分佈(Binominal Dist.)會趨近於Poisson分佈

• When \(n\) is large and \(p\) is small, \(Binom[n, p]\) approaches \(Pois[\lambda = n p]\)

par(mfrow=c(1,2), cex=0.7)
rbinom(100000, 1000, 0.002) %>% table %>% barplot(main="Boinomial")
rpois(100000, 2)  %>% table %>% barplot(main="Poisson")

【3】Poisson分佈的特性：(1)期望值等於標準差 & (2)期望值對加法有封閉性

• \(X \sim Pois[\lambda] \, \Rightarrow \, E(X) = Var(X) = \lambda\)

sapply(1:10, function(lambda) {
  x = rpois(1000000, lambda)
  c(mean(x), var(x))
  })

        [,1]   [,2]   [,3]   [,4]   [,5]   [,6]   [,7]   [,8]   [,9]  [,10]
[1,] 0.99999 1.9986 3.0031 3.9977 5.0017 6.0002 7.0010 7.9968 9.0017 9.9998
[2,] 1.00139 2.0010 3.0031 3.9900 5.0201 5.9911 7.0112 7.9797 9.0152 9.9987

• \(X \sim Pois[\lambda_1], \, Y \sim Pois[\lambda_2] \, \Rightarrow \, X+Y \sim Pois[\lambda_1 + \lambda_2]\)

par(mfrow=c(1,2), cex=0.7)
(rpois(100000, 1) + rpois(100000, 2)) %>% table %>% barplot(main="Pois[1] + Pois[2]")
rpois(100000, 3)  %>% table %>% barplot(main="Pois[3]")

dpois(0:10, 3) + dpois(0:10, 4) - dpois(0:10, 7)

 [1]  0.0671908  0.2162406  0.3482258  0.3672794  0.2721720  0.1293956
 [7]  0.0056023 -0.0678584 -0.0925057 -0.0854730 -0.0648806

【4】Geometric Dist.基本上是等待時間的分佈

我們可以用二項分佈來模擬Geometric Dist.

par(mfrow=c(1,2), mar=c(3,3,3,1), cex=0.7)
replicate(100000, which(rbinom(100, 1, .3) == 1)[1] - 1) %>% 
  table %>% barplot(main="Binomial Simulation")
rgeom(100000, 0.3) %>% table %>% barplot(main="Geometric")

🗿 : 如果有一台機器每一天壞掉的機率是0.05，那麼在20天之內，它還能正常工作的機率分別是多少呢？

p = 0.05
data.frame(end.of.day=1:20) %>% mutate(
  brokenProb = pgeom(end.of.day-1, p),
  workingProb = 1 - brokenProb
  )

   end.of.day brokenProb workingProb
1           1    0.05000     0.95000
2           2    0.09750     0.90250
3           3    0.14263     0.85737
4           4    0.18549     0.81451
5           5    0.22622     0.77378
6           6    0.26491     0.73509
7           7    0.30166     0.69834
8           8    0.33658     0.66342
9           9    0.36975     0.63025
10         10    0.40126     0.59874
11         11    0.43120     0.56880
12         12    0.45964     0.54036
13         13    0.48666     0.51334
14         14    0.51233     0.48767
15         15    0.53671     0.46329
16         16    0.55987     0.44013
17         17    0.58188     0.41812
18         18    0.60279     0.39721
19         19    0.62265     0.37735
20         20    0.64151     0.35849

【5】Geometric Dist.的期望值

• \(X \sim Geom[p] \, \Rightarrow \, E[X] = \frac{1}{p}-1\)
  

🗿 : 如果平均而言每一個捐贈者有我需要的器官的機率是5%，那麼平均我要等多少個捐贈者才能等到我想要用的器官呢？

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