Chapter. 5_Probability Distributions Discrete Random Variables

Random Variable (X): It is a real-valued function, defined on the Sample Space (S), which assigns a Real Value (x) for each basic outcome of (S) Example1: Toss a balanced coin twice S={HH, HT, TH, TT} Lets define Random Variable X such as X= the total number of heads X=1 if basic outcome is either HT or TH X=2 if basic outcome is HH X=0 if basic outcome is TT When basic outcome is not a numerical value, Random Variable (X) helps us to assign one & only numerical value (x) for each basic outcome. 1st coin 2nd coin Basic outcomes Random Variable X Head(H) H HH 2 T HT 1 Tail (T) TH TT

If Random Variable (X) can take countable number of values we call it as DISCRETE RANDOM VARIBALE. In our example, it is seen that X=x, X=0 or X=1 or X=2, such that x is an INTEGER. Thus X may take three different value!  FINITE Number of Values Example2: Toss a coin up to Head appears X=how many times we toss the coin X=1 or X=2 or X=3, ….upto Head appears  X may take INFINITE NUMBER of VALUES but still X is COUNTABLE

A random variable is a continuous random variable if it can take any value in an interval. INCOUNTABLE, INFINITE Example: X is defined in interval [0,1] It means 0<x<1 here X=x can be X=0.01, or X=0.002 ..etc.  incountable! & infinite (there are infinite number of real values between 0 and 1)

Probability Distribution The probability distribution function, P(X=x), of a discrete random variable X expresses the probability that X takes the value x, as a function of x. That is, Continue Example 1: Basic outcomes Probability of each basic outcome Random Variable X Pr(X=x) HH 1/4 2 P(X=2) = P(2) = P(HH) = 1/4 HT 1 TH P(X=1) = P(1) = P(HT,TH) = 1/4 + 1/4 =1/2 TT P(X=0) = P(0) = P(TT) = 1/4 X P(x) 1/4 1 1/2 2 SUM =

Properties of Probability Distribution Functions Let X be a discrete random variable with probability distribution function P(x). Then 1) 0 <= P(x) <= 1 for any value x, and 2) The individual probabilities sum to 1, For our Example 1: X P(x) F(x) 1/4 F(0)=P(X<=0) = P(X=0) = 1/4 1 1/2 F(1)=P(X<=1) =P(X=1)+P(X=0)=1/2+1/4=3/4 2 F(2)=P(X<=2) =P(X=2)+ P(X=1) + P(X=0)=1/4 + 1/2+1/4=1

CUMULATIVE PROBABILITY FUNCTION The cumulative probability function, F(x0) for a random variable X, expresses the probability that X does not exceed the value x0, as a function of x0. That is For our Example 1: X P(x) F(x) 1/4 F(0)=P(X<=0) = P(X=0) = 1/4 1 1/2 F(1)=P(X<=1) =P(X=1)+P(X=0)=1/2+1/4=3/4 2 F(2)=P(X<=2) =P(X=2)+ P(X=1) + P(X=0)=1/4 + 1/2+1/4=1

Practice on Probability Distribution & Cumulative Probability Function X 1 2 3 4 5 6 P(x) 0.1 0.15 0.25 0.2 0.05 EXAMPLE: The number of cars sold per day at a small town is defined by the following probability distribution: a) P(X=3)=? b)F(5)=? F(5)= P(X<=5)= P(X=0)+P(X=1)+P(X=2)+P(X=3)+ P(X=4)+P(X=5) = c) P(2<X<6)= ? =P(X=3)+P(X=4)+P(X=5) d)P(4<=X<5)= ? e)P(2<X<=6)=?

Properties of Discrete Random Variable Expected Value (or mean) of a discrete distribution (Weighted Average) Variance and Standard Deviation Example1: Toss a balanced coin twice S={HH, HT, TH, TT} X P(x) µ = xP(x) (x-µ)2 P(x) (x-µ)2 1/4 (-1)2 0.25 1 1/2 0.5 (0)2 2 (1)2 Mean (µ) Variance Lets define Random Variable X such as X= the total number of heads E(x) = (0 x 1/4) + (1 x 1/2) + (2 x 1/4)= 1.0

Function of Random Variables If P(x) is the probability function of a discrete random variable X , and g(X) is some function of X , then the expected value of function g, E(g(X)), is

Function of Random Variables Continue to Example 1:Toss a balanced coin twice S={HH, HT, TH, TT} Define Random Variable X such as: X= the total number of heads Suppose that you claim to pay (5X + 200)$ for any realization of random variable X thus we define following pay of function g(x) such that g(x)= 5x + 200 g(0)= 5.0+200 = 200 g(1)=5.1 +200 = 205 g(2)= 5.2 +200 = 210 What is the expected amount (mean) that you pay off? Therefore we can consider g(x) as an another RANDOM VARIABLE, lets call as Y, such that Y=G(x) and the underlying probability of Y depends of X=x X P(x) Y= g(x)=5x+200 P(y) 1/4 200 ¼ 1 1/2 205 ½ 2 210

Linear Function of Random Variables Let random variable X have mean µx and variance σ2x & Let a and b be any constants. i.e., if a random variable always takes the value ‘a’, it will have mean ‘a’ and variance ‘0’. Example: x P(x) xP(x) (x-mean)2P(x) 5 1/4 5/4 mean variance

Linear Function of Random Variables Please see below Table as an example. Here X is a r.v. and Y is a linear function of X such that Y = 10*X X P(x) E(X)=X.P(X) Y=10*X E(Y)= YP(x) 1/4 1 1/2 10 10/2 2 2/4 20 20/4

Linear Function of Random Variables Let random variable X have mean µx and variance σ2x & Let a and b be any constants. We define a linear function Y=a+bX such that Then the mean of Y is the and variance and standard deviation of Y are

Solve the problem applying the rules of Linear Function of Random Variables Continue to Example 1:Toss a balanced coin twice S={HH, HT, TH, TT} Define Random Variable X such as: X= the total number of heads Suppose that you claim to pay (5X + 200)$ for any realization of random variable X thus we define following pay of function g(x) such that g(x)= 5x + 200 g(0)= 5.0+200 = 200 g(1)=5.1 +200 = 205 g(2)= 5.2 +200 = 210 What is the expected amount , mean, that you pay off? X P(x) Y= g(x)=5x+200 P(y) 1/4 200 ¼ 1 1/2 205 ½ 2 210