1. Chapter. 5_Probability Distributions
Discrete Random Variables

2. 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

3. 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

4. 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= etc.  incountable! & infinite (there are infinite number of real values between 0 and 1)

5. 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 =

6. 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

7. 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

8. Practice on Probability Distribution & Cumulative Probability FunctionX 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)=?

9. 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

10. 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

11. 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)= = 200
g(1)= = 205
g(2)= = 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 1
1/2
205 2
210

12. 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

13. 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

14. 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

15. 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)= = 200
g(1)= = 205
g(2)= = 210
What is the expected amount , mean, that you pay off? X
P(x)
Y= g(x)=5x+200
P(y)
1/4
200 1
1/2
205 2
210