# Objectives 目标

• Understand Bernoulli Distribution and Binoimal Distribution
  了解伯努利分布和二项分布
  Bernoulli and Binomial distributions are related to experiments with binary outcomes (success or failure).
  伯努利分布和二项分布与二元结果（成功或失败）的实验有关。
• Understand the interpretation of these distributions
  理解这些分布的解释
• Know how to do the simulations of these two experiments and barplot
  知道如何做这两个实验的模拟和 barplot
• Know how to use sample to generate training and testing data set
  知道如何使用 sample 生成训练和测试数据集

# Bernoulli Distribution 伯努利分布

Bernoulli Experiment 伯努利实验

A single random experiment with outcome either success or failure, and the probability of the success is $p$.
一个随机实验，结果要么成功要么失败，成功的概率是 $p$。
Examples: Flip a single coin; Attempt a free throw
例子： 抛一枚硬币；尝试罚球

Bernoulli Random Variable 伯努利随机变量

A random variable that takes either $1$ or $0$, where it takes 1 when the outcome of the Bernoulli trial is a success and 0 when the outcome of the Bernoulli trial is a failure.
一个随机变量 $1$ 或者 $0$，当伯努利试验的结果成功时取 1，当伯努利试验的结果失败时取 0。

Bernoulli Distribution 伯努利分布

The distribution of a Bernoulli random variable is called the Bernoulli distribution.
伯努利随机变量的分布称为伯努利分布。

• Probability Mass Function of Bernoulli Distribution/Random Variable:
  伯努利分布 / 随机变量的概率质量函数：

• Expectation and Variance 期望值和方差值

$$\mu = E(X) = (0)(1-p)+(1)p = p, \\ \sigma^2 = var(X) = (0)^2(1-p)+(1)^2p -p^2 = p(1-p)$$

# A Simple Example: We toss a fair coin. 扔一枚公平的硬币

• Each time of the tossing coin is a Bernoulli Experiment. Consider head is success (1).
  每次抛硬币都是伯努利实验。考虑头像面是成功（1）。

• What is the probability $p$?
  概率$p$ 是多少？

  	n <- 10 # number of random experiments
  	x <- c(0,1) # sample space for tossing a coin, 0--Tail, 1--Head
  	coinToss <- sample(x, size=n, replace=TRUE)
  	# generate n outcomes of the experiments
  	mean(coinToss)

  [1] 0.4
  

• What about 100 times? 100 次呢？

  	n <- 100 # number of random experiments
  	x <- c(0,1) # sample space for tossing a coin, 0--Tail, 1--Head
  	coinToss <- sample(x, size=n, replace=TRUE)
  	# generate n outcomes of the experiments
  	mean(coinToss)

  [1] 0.5
  

• Seems close to $p$. Let's try $n = 1000000$.
  似乎接近$p$ 了，再试试 $n = 1000000$。

  	n <- 1000000 # number of random experiments
  	x <- c(0,1) # sample space for tossing a coin, 0--Tail, 1--Head
  	coinToss <- sample(x, size=n, replace=TRUE)
  	# generate n outcomes of the experiments
  	mean(coinToss)

  [1] 0.5
  

• We can use a barplot to generate the frequence for coinToss
  我们可以使用 abarplot 来生成频率 coinToss

  barplot(table(coinToss)) #

# the sample function is also used to split a data set

sample 函数还用于拆分数据集

例如机器学习数据挖掘部分，将考虑 80% 作为训练数据，20% 作为测试数据。

i <- sample(2, nrow(cars), replace=TRUE, prob=c(0.8, 0.2)) # prob 后面跟着的是两个概率

carsTrainingData <- cars[i==1,] # 等于 1 的数据归于训练集

carsTestingData <- cars[i==2,] # 等于 2 的数据归于测试集

summary(cars)

    speed           dist       
Min.   : 4.0   Min.   :  2.00  
1st Qu.:12.0   1st Qu.: 26.00  
Median :15.0   Median : 36.00  
Mean   :15.4   Mean   : 42.98  
3rd Qu.:19.0   3rd Qu.: 56.00  
Max.   :25.0   Max.   :120.00  

summary(carsTrainingData)

    speed            dist      
Min.   : 4.00   Min.   :  2.0  
1st Qu.:12.00   1st Qu.: 26.0  
Median :15.00   Median : 35.0  
Mean   :15.19   Mean   : 41.9  
3rd Qu.:18.75   3rd Qu.: 54.0  
Max.   :24.00   Max.   :120.0  

summary(carsTestingData)

    speed           dist      
Min.   : 7.0   Min.   :16.00  
1st Qu.:11.0   1st Qu.:23.50  
Median :18.0   Median :52.00  
Mean   :16.5   Mean   :48.62  
3rd Qu.:21.0   3rd Qu.:68.50  
Max.   :25.0   Max.   :85.00  

# In-class Exercise

1. Please use the above the code to divide the iris data set into two parts, $80\%$ in training data set and $20\%$ in testing data set

   请使用上面的代码将 iris 数据集分成两部分，$80\%$ 在训练数据集和 $20\%$ 在测试数据集

   	i <- sample(2, nrow(iris), replace=TRUE, prob=c(0.8, 0.2)) # prob 后面跟着的是两个概率
   	irisTrainingData <- iris[i==1,] # 等于 1 的数据归于训练集
   	irisTestingData <- iris[i==2,] # 等于 2 的数据归于测试集
   	summary(iris)

    Sepal.Length    Sepal.Width     Petal.Length    Petal.Width          Species  
   Min.   :4.300   Min.   :2.000   Min.   :1.000   Min.   :0.100   setosa    :50  
   1st Qu.:5.100   1st Qu.:2.800   1st Qu.:1.600   1st Qu.:0.300   versicolor:50  
   Median :5.800   Median :3.000   Median :4.350   Median :1.300   virginica :50  
   Mean   :5.843   Mean   :3.057   Mean   :3.758   Mean   :1.199                  
   3rd Qu.:6.400   3rd Qu.:3.300   3rd Qu.:5.100   3rd Qu.:1.800                  
   Max.   :7.900   Max.   :4.400   Max.   :6.900   Max.   :2.500             
   

2. Do some data exploration in these two data sets. E.g. summary , histogram , boxplot , or barplot

   在这两个数据集中做一些数据探索。

   summary(irisTrainingData)

    Sepal.Length    Sepal.Width     Petal.Length    Petal.Width          Species  
   Min.   :4.300   Min.   :2.000   Min.   :1.000   Min.   :0.100   setosa    :40  
   1st Qu.:5.100   1st Qu.:2.800   1st Qu.:1.600   1st Qu.:0.300   versicolor:44  
   Median :5.800   Median :3.000   Median :4.400   Median :1.400   virginica :41  
   Mean   :5.853   Mean   :3.074   Mean   :3.783   Mean   :1.217                  
   3rd Qu.:6.400   3rd Qu.:3.400   3rd Qu.:5.100   3rd Qu.:1.800                  
   Max.   :7.900   Max.   :4.400   Max.   :6.900   Max.   :2.500          
   

   summary(irisTestingData)

    Sepal.Length    Sepal.Width     Petal.Length    Petal.Width          Species  
   Min.   :4.400   Min.   :2.200   Min.   :1.200   Min.   :0.200   setosa    :10  
   1st Qu.:5.000   1st Qu.:2.600   1st Qu.:1.500   1st Qu.:0.200   versicolor: 6  
   Median :5.700   Median :3.000   Median :4.000   Median :1.300   virginica : 9  
   Mean   :5.796   Mean   :2.976   Mean   :3.632   Mean   :1.112                  
   3rd Qu.:6.300   3rd Qu.:3.200   3rd Qu.:5.200   3rd Qu.:1.800                  
   Max.   :7.700   Max.   :3.900   Max.   :6.700   Max.   :2.300  
   

   	par(mfrow=c(1,3))
   	hist(iris$Sepal.Length)
   	hist(irisTrainingData$Sepal.Length)
   	hist(irisTestingData$Sepal.Length)

   	par(mfrow=c(1,3))
   	boxplot(iris$Petal.Width)
   	boxplot(irisTrainingData$Petal.Width)
   	boxplot(irisTestingData$Petal.Width)

   	par(mfrow=c(1,3))
   	barplot(table(iris$Species))
   	barplot(table(irisTrainingData$Species))
   	barplot(table(irisTestingData$Species))

# If we have an unfair coin 如果扔一个不公平的硬币

what should we do? 我们应该怎么做？

• Suppose we have one unfair coin, whose probabilities of heads are $p = 0.25$.
  假设我们有一枚不公平的硬币，其正面概率为$p = 0.25$.

• How many heads will we get, if we flip the coin 10 times? 100 times? 1000 times?
  如果我们抛硬币 10 次，我们会得到多少个正面？抛 100 次？抛 1000 次？

  	p = 0.25;
  	n = 10;
  	x = c(0,1)
  	unfairCoinToss <- sample(x, size=n, replace=TRUE, prob = c(1-p,p))
  	sum(unfairCoinToss)

  [1] 2
  

  	n = 100;
  	unfairCoinToss <- sample(x, size=n, replace=TRUE, prob = c(1-p,p))
  	sum(unfairCoinToss)

  [1] 30
  

  	n = 1000;
  	unfairCoinToss <- sample(x, size=n, replace=TRUE, prob = c(1-p,p))
  	sum(unfairCoinToss)

  [1] 241
  

• Question: Do these results make sense?
  问题：这些结果有意义吗？

# In-class Exercise

3. Suppose I have an unfair coin, whose probabilities of heads are $p = 0.6$. Can you modify the code above to find the number of heads when flipping the code $1e6$ times?

   假设我有一枚不公平的硬币，其正面概率为 $p = 0.6$. 能不能修改上面的代码，求抛 $1e6$ 次正面朝上的次数？

   	p = 0.6
   	n = 1e6
   	x = c(0, 1)
   	unfairCoinToss <- sample(x, size=n, replace=TRUE, prob = c(1-p,p))
   	sum(unfairCoinToss)

# Binomial Distribution 二项分布

Binomial Experiment 二项式实验

Repeat Bernoulli experiment independently for a certain number ($n$) of times, each repetition has the same possible outcomes (success or failure), and the probability of each success ($p$) is consistent for all trials. This experiment is called Binomial trial.
独立重复伯努利实验一定数量（$n$) 次，每次重复都有相同的可能结果（成功或失败），以及每次成功的概率（$p$) 对于所有试验都是一致的。这个实验叫做二项式试验。

Binomial Random Variable 二项式随机变量

The number of $X$ of a binomial experiment is called a Binomial random variable.
二项式实验结果的发生次数 X 称为二项式随机变量。

Binomial Distribution 二项式分布

The distribution of a Binomial random variable is called the binomial distribution
二项式随机变量的分布称为二项式分布

# Examples: Number of heads in 40 coin flips, Number of hits in 20 free throws

示例：40 次抛硬币的正面次数，20 次罚球的命中次数

• PMF: $b(x;n,p) = {n\choose x}p^x(1-p)^{n-x}$, $x=0,1,\ldots,n$

• Expectation and Variance 期望值和方差值

  $$\mu = E(X) = np, \\ \sigma^2 = var(X) = np(1-p).$$

  • Let $Y_i, i = 1, \ldots, n$ denote the Bernoulli random variable.
    $n$ 表示伯努利随机变量。

  • This is becasue $Y_i$ are independent and identical distributed (iid).
    这是因为 $Y_i$ 是独立同分布的（iid）。

# Another Simple Example: We toss a fair coin 扔一枚公平的硬币

• Each time of the tossing coin is a Bernoulli Experiment. Consider head is success (1).
  每次抛硬币都是伯努利实验。考虑头像面是成功（1）。

• Suppose we toss this coin 10 times, how many heads can we get?
  假设我们抛这枚硬币 10 次，我们能得到几个正面？

  	n <- 10 # number of random experiments
  	x <- c(0,1) # sample space for tossing a coin, 0--Tail, 1--Head
  	coinToss <- sample(x, size=n, replace=TRUE)
  	# generate n outcomes of the experiments
  	sum(coinToss)

  [1] 6
  

• 1000000 times?

  	n <- 10000000 # number of random experiments
  	x <- c(0,1) # sample space for tossing a coin, 0--Tail, 1--Head
  	coinToss <- sample(x, size=n, replace=TRUE)
  	# generate n outcomes of the experiments
  	sum(coinToss)

  [1] 5001636
  

• Let's explore how this value varies in repeated experiments.
  这个值在重复实验中是如何变化的

  	n <- 10000
  	x <- c(0,1)
  	sumsInRepeatedSampling10Reps<-replicate(10, sum(sample(x, size=n, replace=TRUE)))
  	sumsInRepeatedSampling100Reps<-replicate(100, sum(sample(x, size=n, replace=TRUE)))
  	sumsInRepeatedSampling1000Reps<-replicate(1000, sum(sample(x, size=n, replace=TRUE)))
  	par(mfrow=c(1,3))
  	hist(sumsInRepeatedSampling10Reps,main = paste("Number of heads- 10 reps"))
  	hist(sumsInRepeatedSampling100Reps,main = paste("Number of heads- 100 reps"))
  	hist(sumsInRepeatedSampling1000Reps,main = paste("Number of heads- 1000 reps"))

# Conclusions 结论

The number of heads observed--we know it is $X$--when tossing a coin $n$ times is a statistics computed from the sample of $n$ coin tosses.
头像面数 $X$，抛硬币$n$ 次是从抛$n$ 次硬币样本计算的统计数据。

$X$ should follow the Binomial distribution.
$X$ 应该服从二项分布

The experiments above suggest that the sampling distribution of $X$ appears to have a mean around the expected number of heads when a fair coin is tossed, which is about $n/2$ or $np$.
上述实验表明，抽样分布$X$ 当抛一枚公平的硬币时，似乎在预期的正面次数附近有一个平均值，大约是$n/2$ 或者$np$。

The more times we repeat the experiment of $n$ coin tosses, the closer $X$ gets to its expected value -- this can be measured by looking at both the mean and the variance of $X$
我们重复掷硬币的实验次数$n$ 越多，$X$ 离其预期值越近 —— 这可以通过查看 $X$ 的均值和方差来衡量

meansOfX<-c(mean(sumsInRepeatedSampling10Reps),

mean(sumsInRepeatedSampling100Reps),

mean(sumsInRepeatedSampling1000Reps))

varsOfX<-c(var(sumsInRepeatedSampling10Reps),var(sumsInRepeatedSampling100Reps),var(sumsInRepeatedSampling1000Reps))

meansOfX

[1] 4967.200 4993.110 5001.822

varsOfX

[1] 3494.400 2393.452 2355.378

Question: is it possible that something similar to this always happens?
问题：是否有可能总是发生类似的事情？

As we will see, the sampling distribution of $X$ is approximately normal with mean equal to the expected value of $X$. In other words, the example above illustrates a known result--the Central Limit Theorem, one of the cornerstone results used in inference!
正如我们将看到的，抽样分布 $X$ 近似正态，均值等于期望值$X$。 换句话说，上面的例子说明了一个已知的结果 —— 中心极限定理，这是推理中使用的基石结果之一！

# In-class Exercise: Roll A Die

1. Suppose we have a fair die. Modify the code above to simulate rolling a die 10 times, 1000 times, and 100000 times.

   假设我们有一个公平的骰子。修改上面的代码来模拟掷骰子 10 次、1000 次和 100000 次。

   	n <- 10
   	x <- 1:6
   	dieRoll <- sample(x, size=n, replace=TRUE)
   	barplot(table(dieRoll))
   	mean(dieRoll)
   	n <- 1000
   	dieRoll <- sample(x, size=n, replace=TRUE)
   	barplot(table(dieRoll))
   	mean(dieRoll)
   	n <- 100000
   	dieRoll <- sample(x, size=n, replace=TRUE)
   	barplot(table(dieRoll))
   	mean(dieRoll)

2. Do a barplot of the results in 1

   对 1 中的结果进行条形图

   
   
   

3. Calculate the mean of the results in 1.

   计算 1 中结果的平均值。

   [1] 4.2
   [1] 3.435
   [1] 3.49603
   

4. Set $n = 10000$ to calculate the mean value. Then repeat your similuation 10, 100, and 1000 times. Save the results and do a histogram.

   设定 $n = 10000$ 计算平均值。然后重复模拟 10、100 和 1000 次。保存结果并绘制直方图。

   	n <- 10000
   	x = 1:6;
   	meanInRepeatedSampling10Reps<-replicate(10, mean(sample(x, size=n, replace=TRUE)))
   	meanInRepeatedSampling100Reps<-replicate(100, mean(sample(x, size=n, replace=TRUE)))
   	meanInRepeatedSampling1000Reps<-replicate(1000, mean(sample(x, size=n, replace=TRUE)))
   	par(mfrow=c(1,3))
   	hist(meanInRepeatedSampling10Reps,main = paste("Mean of DieRolling- 10 reps"))
   	hist(meanInRepeatedSampling100Reps,main = paste("Mean of DieRolling reps"))
   	hist(meanInRepeatedSampling1000Reps,main = paste("Mean of DieRolling- 1000 reps"))

5. One chanllenge question: Suppose this die is unfair. We can get number 1,3, and 5 with probability $1/9$ and 2, 4, and 6 with probability $2/9$. How can we do the similar experiments as the in-class exercise?

   假设这个骰子是不公平的。我们可以以$1/9$ 概率得到数字 1、3 和 5， $2/9$ 的概率得到 2、4 和 6。 我们如何做类似的实验？