Learn discrete random variables

If the Random variable X assumes only finite or countably infinte set of values it is known as discrete random variable.

Probability density function of Discrete Random variable:-

Suppose X is a Random variable which can take at the most a countable number of values X1, X2, X3, ..................... Xn with each value of  " X ". We associate a number

pi = P ( X = Xi ) ; i = 1,2,..............n

which is known as the probability of Xi and satisfies the following conditions:

pi = P ( X = Xi ) `>=` 0  ( i = 1,2,..............n )    i.e., pi 's are all non- negative and
`sum` pi = p1 + p2 +................... + pn = 1
i.e., the total probability is one.

The function pi = P ( X = Xi ) ; i = 1,2,..............n is called the probability function or more precisely probability mass function of the random variable X


Cumulative distribution function of F(x) of discrete random variable


Cumulative distribution function F( x ) of a discrete random variable X is denoted as F( X = xi ) and defined as               F ( X = xi ) = P ( X = xi )

F ( X = xi ) = P ( X = x1 ) + P ( X = x2 ) + ............................ + P ( X = xi )

F ( X = xi ) =    `sum_(n=1)^i` P ( X = xn )


Example of learn discrete random variables


Let x denote the minimum of two numbers that appear when a two dice is thrown once. find the discrete probability distribution?

Solution:-  The sample space S = { ( 1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) }

n ( S ) = 62 = 36

Given that X = min ( a, b )

P( 1 ) = P ( X = 1 ) = { (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (3,1) (4,1) (5,1) (6,1) }

= 11 / 36

P( 2 ) = P ( X = 2 ) = { (2,2) (2,3) (2,4) (2,5) (2,6) (3,2) (4,2) (5,2) (6,2) }

= 9 / 36

P( 3 ) = P ( X = 3 ) = { (3,3) (3,4) (3,5) (3,6) (4,3) (5,3) (6,3) }

= 7 / 36

P( 4 ) = P ( X = 4 ) = { (4,4) (4,5) (4,6) (5,4) ( 6,4) }

= 5 / 36

P( 5 ) = P ( X = 5 ) = { (5,5) (5,6) (6,5) }

= 3 / 36

P( 6 ) = P ( X = 6 ) = { (6,6) }

= 1 / 36

X    1    2    3    4    5    6
P ( X = x)    11/36    9/36    7/36    5/36    3/36    1/36