See more on https://en.wikipedia.org/wiki/Level_of_measurement.
The concept of infinity is very broad. Currently, you just need to keep the distinction of countable and uncountable infinities in mind.
How many real numbers are there between 0 and 1?
Here are brief descriptions of mean (expectation), median, mode, variance, standard deviation, quantile.
Mean: \(\bar{X} = \sum_i^N X_i\)
Median: Let’s say \(X_k\) are ordered from smallest to largest and there are \(n\) values in the sample. Median(\(X\))\(=X_{(n+1)/2}\) if n is odd and (usually) Median(\(X\))\(=\dfrac{X_{(n/2)} + X_{(n/2+1)}}{2}\).
Quantile: On an ordered list of values for quantile (\(\alpha\)) provides the \((\alpha*n)^{th}\) smallest value of the list. For instance, if \(\alpha = 70\% = 0.7\) quantile value is the 7th smallest value in a list of 10 values. \(\alpha = 1\) means the maximum. Quantile is an important parameter in especially statistics.
Mode: \(X_k\) with the highest frequency in the sample. In a sample of (\(1,2,2,3,4,5\)), \(2\) is the mode.
Variance: \(V(X) = \dfrac{\sum_i^N (X_i - \bar{X})^2}{n-1}\)
Standard Deviation: \(\sigma(X) = \sqrt{\dfrac{\sum_i^N (X_i - \bar{X})^2}{n-1}}\)
set.seed(231)
#Let's pick 10 values from the numbers between 1 and 50.
numbers <- sample(1:50,10,replace=TRUE)
#The sorted version of the numbers
sort(numbers)
## [1] 1 9 15 16 16 18 26 31 32 35
#The mean values of the numbers
sum(numbers)/10
## [1] 19.9
#or in R
mean(numbers)
## [1] 19.9
#Median of the numbers
median(numbers)
## [1] 17
#Quantile 7/9 of the numbers
quantile(numbers,7/9)
## 77.77778%
## 31
#Quantile 0 of the numbers (also the min)
quantile(numbers,0)
## 0%
## 1
#Quantile 1 of the numbers (also the max)
quantile(numbers,1)
## 100%
## 35
#No simple solution for mode in R
freq_table<-table(numbers)
freq_table
## numbers
## 1 9 15 16 18 26 31 32 35
## 1 1 1 2 1 1 1 1 1
names(freq_table[which.max(freq_table)])
## [1] "16"
#Sample variance of numbers
sum((numbers - mean(numbers))^2)/(10-1)
## [1] 118.7667
#For large values you can take n ~ n-1
#in R
var(numbers)
## [1] 118.7667
#Sample standard deviation of values
sqrt(sum((numbers - mean(numbers))^2)/(10-1))
## [1] 10.89801
#in R
sd(numbers)
## [1] 10.89801
Random variables are the abstractions of uncertain events so that we can generalize events in formal functions instead of explicitly enumerating the outcomes. For instance, assume \(X\) is the number of tails in 2 coin tosses.
\[\begin{align} P(X = 0) &= P(\{H,H\}) = 1/4 \\ P(X = 1) &= P(\{H,T\},\{T,H\}) = 2/4 \\ P(X = 2) &= P(\{T,T\}) = 1/4 \\ \end{align}\]\(X\) can take values 0, 1 and 2. \(X\) is a discrete random variable.
There are also the continuous random variables. Continuous random variables are usually defined in intervals instead of individual values. For instance, define \(Y\) as any real number between 0 and 1 and and all values within the interval are equally probable (i.e. uniform distribution).
\[\begin{align} P(Y \le 0.25) &= 1/4 \\ P(X \le 0.5) &= 2/4 \\ P(X \le 0.75) &= 3/4 \\ \end{align}\]There are several fundamental concepts to keep in mind.
Probability Mass Function (pmf): pmf is the point probability for discrete distributions (i.e. \(f(x) = P(X = x)\)). For instance \(P(X = H) = 1/2\), \(P(X = T) = 1/2\).
\($\sum_i^n f(x_i) = 1\).$
\[\int_{-\infty}^{\infty} f(x) dx = 1\]
Main relationship between cdf and pdf is (\(F(X \le a) = \int_{-\infty}^a f(x) dx\)).
\[\mu = E[X] = \sum_i^n x_i f(x_i)\] for discrete distributions.
\[\mu = E[X] = \int_{-\infty}^{\infty} xf(x)dx\] for continuous distributions.
Example: Calculate the expected value of number of tails in two coin tosses.
\[\begin{align} E[X] &= \sum_i^n x_i f(x_i) = 0 * P(X=0) + 1 * P(X = 1) + 2 * P(X = 2) \\ &= 0 * 1/4 + 1 * 1/2 + 2 * 1/4 \\ &= 1 \end{align}\]\[V(X) = E[(X - \mu)^2] = \sum_i^n (x_i - \mu)^2 f(x_i)\] for discrete distributions.
\[V(X) = E[(X - \mu)^2] = \int_{-\infty}^{\infty} (x - \mu)^2 f(x) dx\] for discrete distributions.
Variance can also be calculated as \(V(X) = E[X^2] - (E[X])^2\).
It can also be called “single coin toss distribution”. For a single event with probability of success \(p\) and failure \(q = 1 - p\), the distribution is called Bernoulli.
pmf: \(f(x = 0;p) = q\), \(f(x = 1) = p\)
\(E[X] = 0*(1-p) + 1*p = p\)
\(V[X] = pq\)
Example: Coin Toss
\(p = 0.5\), \(q = 1 - p = 0.5\)
pmf: \(f(x = 0) = 0.5\), \(f(x = 1) = 0.5\)
\(E[X] = 0*(1-0.5) + 1*0.5 = 0.5\)
\(V(X) = 0.5*0.5 = 0.25\)
Think of multiple Bernoulli trials (e.g. several coin tosses).
pmf: \(f(x;p,n) = \binom{n}{x} p^xq^{(n-x)}\)
\(E[X] = np\)
\(V(X) = npq\)
cdf: \(F(X \le x) = \sum_{i=0}^n f(i)\)
Example: Multiple Coin Tosses (x5 coins, p = 0.5)
#R way
#(d)ensity(binom)ial
dbinom(x=3,size=5,prob=0.5)
## [1] 0.3125
\(E[X] = 5*0.5 = 2.5\)
\(V(X) = 5*0.5*0.5 = 1.25\)
cdf: \(F(X \le 3;n=5) = \sum_{i=0}^5 f(i) = 0.8125\)
#R way
pbinom(q=3,size=5,prob=0.5)
## [1] 0.8125
Now suppose there is not one probability (\(p\)) but there are many probabilities (\(p_1, p_2, \dots, p_k\)).
pmf: \(f(x_1, \dots , x_k;p_1, \dots, p_k;n) = \binom{n}{x_1, \dots , x_k} p_1^{x_1}*\dots *p_k^{x_k}\)
where \(\binom{n}{x_1, \dots , x_k} = \dfrac{n!}{x_1! \dots x_k!}\), \(\sum_i^k x_i = n\) and \(\sum_i^k p_i = 1\).
Example: Customers of a coffee shop prefer Turkish coffee with probability 0.4, espresso 0.25 and filter coffee 0.35. What is the probability that out of the first 10 customers, 3 will prefer Turkish coffee, 5 will prefer espresso and 2 will prefer filter coffee?
\(f(3,5,2;0.4,0.25,0.35;10) = \binom{10}{3,5,2} * 0.4^3 * 0.25^5 * 0.35^10 = 4.3 * 10^{-6} = 0.0193\)
#Explicit form
factorial(10)/(factorial(3)*factorial(5)*factorial(2))*0.4^3 * 0.25^5 * 0.35^2
## [1] 0.01929375
#Density multinomial
dmultinom(x=c(3,5,2),prob=c(0.4,0.25,0.35))
## [1] 0.01929375
Binomial distribution is a special case of multinomial distribution.
Hypergeometric distribution can be used in case the sample is divided in two such as defective/nondefective, white/black, Ankara/Istanbul. Suppose there are a total of \(N\) items, \(k\) of them are from group 1 and \(N-k\) of them are from group 2. We want to know the probability of getting \(x\) items from group 1 and \(n-k\) items from group 2.
pmf: \(f(x,n;k,N) = \dfrac{\binom{k}{x}\binom{N-k}{n-x}}{\binom{N}{n}}\)
\(E[X] = \dfrac{nk}{N}\)
\(V[X] = \dfrac{N-n}{N-1}*n*\dfrac{k}{N}*(1-\dfrac{k}{N})\)
Example: Suppose we have a group of 20 people, 12 from Istanbul and 8 from Ankara. If we randomly select 5 people from it what is the probability that 1 of them is from Ankara and 4 of them from Istanbul.
\(f(1,5;8,20) = \dfrac{\binom{8}{1}\binom{20-8}{5-1}}{\binom{20}{5}} = 0.256\)
#Explicit form
x=1
n=5
k=8
N=20
(choose(k,x)*choose(N-k,n-x))/choose(N,n)
## [1] 0.255418
#Density hypergeometric, see ?dhyper for explanations
dhyper(x=1,m=8,n=12,k=5)
## [1] 0.255418
Negative Binomial distribution answers the question “What is the probability that k-th success occurs in n trials?”. Differently from the binomial case, we fix the last attempt as success.
Example: Suppose I’m repeatedly tossing coins. What is the probability that 3rd Heads come in the 5th toss?
$f(3;0.5,5) = 0.530.5{(5-3)} = $
#Explicit form
choose(5-1,3-1)*0.5^3*0.5^(5-3)
## [1] 0.1875
#Binomial way
dbinom(3-1,5-1,0.5)*0.5
## [1] 0.1875
#Negative binomial way
dnbinom(x=5-3,size=3,prob=0.5)
## [1] 0.1875
Geometric distribution answers “What is the probability that first success comes in the n-th trial?”
pmf: \(f(x;p,n) = q^{(n-1)}p\)
\(E[X] = 1/p\)
\(V[X] = \dfrac{1-p}{p^2}\)