Random Variable: A random variable is a real-valued function defined on a sample space. Discrete Random Variable: When the possi- ble values of a random variable are countable, then we say the random variable is a discrete random variable. It is relatively easier to understand discrete random variables. Note that countable means that either the set has finite many elements or we can establish a one-one correspondence be- tween the elements and the integers 1, 2, 3,.... So we can list all the elements in the countable set one by one. For example, an infinite-countable-many set usually has the form a ,a ,a ,...,a ,..., . { 1 2 3 k } 2 Notation 1. We use ω to denote the outcome in the sample space S. 2. Uppercase letters are usually used to denote random variables. For example, X is a random variable. We use lowercase letter x to denote the value of X. 3. We use to denote the collection of the X possible values of the random variable X. So = X(ω), ω S . X { ∈ } 4. If there are more than two random variables, like X and Y , then we add a subscript like XX to distinguish them. XY 5. Often we use the abbreviation r.v. instead of random variable. 3 Discrete Random Variable Defined on a Finite Sample Space Let us consider a finite sample space. S = ω ,ω ,...,ωn . { 1 2 } Let X be a random variable on S, which means that X is a real-valued function whose argu- ments are ωs in S. Let xi = X(ωi) for i = 1, 2,...,n. Then = x ,x ,...,xn . X { 1 2 } Note that it is ok to have some xis that are the same. Any well-defined rule can be used to define a random variable X on S. We see some exam- ples in the next few pages. 4 Examples of Discrete R.V. Ex1: Toss a coin. S = H,T . { } Let X(T ) = 1 and X(T ) = 0 then we define a r.v. X. = 1, 0 . X { } Ex2: Roll a die. S = 1, 2, 3, 4, 5, 6 . { } Let X(i) = i for i = 1, 2, 3, 4, 5, , 6. = XX 1, 2, 3, 4, 5, 6 . { } Or we can define another r.v. Y by Y (1) = Y (3) = Y (5) = 1 and Y (2) = Y (4) = Y (6) = 0. = 1, 0 . In this example we see different XY { } arguments could give the same value to Y . Ex3: Toss 2 coins. S = (H,H), (H,T ), (T,H), (T,T ) . { } Let X be the number of heads. Then X((H,H)) = 2, X((H,T )) = X((T,H)) = 1, X((T,T )) = 0. = 2, 1, 0 . X { } 5 Probability Function Since the argument of the random variable X is random, the value of X is random too. It is natural to ask what is the probability that the value of X is equal to x?(x is some real number). Probability Function: Probability function (p.f.) of the discrete random variable X is defined as f(x) = Pr(X = x)= p(ω). ω:XX(ω)=x p(ω) is the probability distribution on the finite sample space. We use probability function f(x) to describe the distribution of X. If A is a subset of then X Pr(X A)= f(xi). ∈ x A Xi∈ 6 In Ex1, f(1) = Pr(X = 1) = p(H) = 0.5 and f(0) = Pr(X =0)= p(T ) = 0.5. 1 In Ex2, Pr(X = i)= p(i)= 6. i = 1, 2, 3, 4, 5, 6. 1 Pr(Y =1)= p(1) + p(3) + p(5) = , 2 1 Pr(Y =0)= p(2) + p(4) + p(6) = . 2 1 In Ex3, f(2) = Pr(X = 2) = p((H,H)) = 4 and f(1) = Pr(X =1)= p((H,T ))+p((T,H)) = 1 1 4 + 4 = 1/2, f(0) = Pr(X = 0) = p((T,T )) = 1 4. Ex4: Roll 2 dice. Let X be the sum of the two numbers. Find Pr(X = 5)? Solution: f(5) = Pr(X =5)= p((4, 1)) + p((1, 4)) + p((3, 2)) + 4 1 p((2, 3)) = 36 = 9. 7 Three Common Discrete Distributions Let x1,x2,...,xk be distinct real numbers. If we have a function f(x) such that f(x ) 0, i ≥ for all i = 1, 2,...,n, and n f(xi) = 1. iX=1 Then f can be a probability function of X which takes value from = x ,x ,...,x . X { 1 2 k} Ex1: X takes value from 1, 0 . Let Pr(X = { } 1) = f(1) = p and Pr(X = 0) = f(0) = 1 − p = q. Then We say X is a Bernoulli random variable whose probability of success is p. We write X Ber(p). ∼ 8 Ex2: Let = 0, 1, 2,...,n . Take a positive X { } real number p, 0 <p< 1. Let n k n k f(k)= p (1 p) − . k − By Binomial theorem we check n n n k n k n f(k)= p (1 p) − = (p+1 p) = 1. k − − kX=0 kX=0 Thus we can use f(k) to define a distribu- { } tion. We say X is a binomial random vari- able with parameters (n,p) if X takes value from 0, 1, 2,...,n and Pr(X = k) = f(k) = { } n pk(1 p)n k. We write X Bin(n,p). k − − ∼ Ex3: We say X is a uniform random variable on x1,x2,x3,...,xk if Pr(X = xi) = f(xi) = 1 { } k for i = 1, 2,...,k. Remark: Two random variables can have the same distribution. 9 Some Bernoulli examples: 1. In a state gov- ernor election voters vote for two candidates, R and D. Randomly pick a voter, let X =1 if the voter votes for R and X = 0 if the voter votes for D. Then X is a Bernoulli random vari- able. What is the probability of success p? p should be the probability that a randomly se- lected voter votes for R. # of voters for R p = . # of all voters 2. Let A be some event. We define a random variable X as follows 1 if ω A X(ω)= ∈ ( 0 otherwise X is a Bernoulli random variable. Its probability of success is p = Pr(X = 1) = p(ωi) ωi:XX(ωi=1) = p(ωi)= p(A). ω A X∈ 10 Some Binomial examples: 1. Toss a biased coin 100 times p(H) = 0.6 p(T ) = 0.4. Let X be the number of heads in 100 tosses. 100 k 100 k Pr(X = k)= 0.6 0.4 − . k So X Bin(100, 0.6). ∼ 2. Let X be the number of women in a Jury. n = 12. X takes value from 0, 1, 2,..., 12 . { } Let p be the probability that a random chosen person is a woman and 1 p be the probability − that a random chosen person is a man. Then 12 k 12 k Pr(X = k)= p (1 p) − . k − So X Bin(12,p). ∼ What is p? p is the percentage of women in the district. 11 Some Uniform r.v. examples: 1. Roll a die. 1 f(k)= 6, k = 1, 2, 3, 4, 5, 6. 2. The last digit of your SSN. It is uniformly distributed on 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 . { } Normalizing constant: Suppose that X has probability function c f(x)= , x = 1, 2, 3, 4, 5 x2 Find the value of the constant c. Solution: Since f is a p.f. we must have 5 c 1= ( )= f x 2. x x X xX=1 Thus c = ( 5 1 ) 1 = 0.683. x=1 x2 − P Remark: c is often called the normalizing con- stant of the distribution. In some applications we know the probability function upto a con- stant. Then a practically important question is to compute the value of that constant, as we did in the above example. 12 Distribution Functions We start with the discrete random variables. Definition: Let X be a discrete random vari- able. = x ,x ,x ,..., and x < x < x < X { 1 2 3 } 1 2 3 . .. The distribution function of X is given by F (x) = Pr(X x) for <x< ≤ −∞ ∞ Note that in terms of the probability function of X F (x)= f(xi). x :x x i Xi≤ Sometimes we also call F (x) the CDF of X. 13 Example Suppose x1 = 0,x2 = 1,x3 = 2,x4 = 3.f(x1)= 1 2 2 3 8, f(x2) = 8, f(x3) = 8, f(x4) = 8. f is the p.f. of X. Let’s compute the distribution func- tion of X. F ( 1) = 0. − 1 1 F (0) = f(x1)= 8, F (0.5) = f(x1)= 8. 3 F (1) = f(x1)+ f(x2) = 8, F (1.5) = f(x1)+ 3 f(x2)= 8. 5 F (2) = f(x1)+ f(x2)+ f(x3) = 8, F (2.5) = 5 f(x1)+ f(x2)+ f(x3)= 8. F (3) = f(x1) + f(x2) + f(x3) + f(x4) = 1, F (3.25) = f(x1)+ f(x2)+ f(x3)+ f(x4) = 1. 14 o F(x) o o o 0.0 0.2 0.4 0.6 0.8 1.0 −1 0 1 2 3 4 x The above figure shows the sketch F (X). Form the graph we can see that F (x) has three prop- erties. If x <x then F (x ) F (x ). ”=” can 1 2 1 ≤ 2 be taken.