Show that the following function is a probability mass funct

Show that the following function is a probability mass function (pmf) for the random variable X and constant b.f(x)=(1/1+b)(b/1-b)^x,x=0,1,2,3......

Solution

Here, f(x) must be normalized.

Thus,

f(0) + f(1) + f(2) + f(3) ... = 1

This is the condition that needs be satisfied.

Thus,

f(0) + f(1) + f(2) + f(3) ...

= (1/(1+b))(b/(1+b))^0 + (1/(1+b))(b/(1+b))^1 + (1/(1+b))(b/(1+b))^2 + (1/(1+b))(b/(1+b))^3...

= (1/(1+b)) + (1/(1+b))(b/(1+b))^1 + (1/(1+b))(b/(1+b))^2 + (1/(1+b))(b/(1+b))^3...

As we can see, this is a geometric series with

a1 = first term = 1/(1+b)
r = common ratio = b/(a+b)

Thus, as the sum of a geometric series is a1/(1-r),

= (1/(1+b)) + (1/(1+b))(b/(1+b))^1 + (1/(1+b))(b/(1+b))^2 + (1/(1+b))(b/(1+b))^3...

= [1/(1+b)]/[1 - b/(1+b)]

Simplifying the denominator,

= [1/(1+b)]/[(1 + b - b)/(1+b)]

= [1/(1+b)]/[1/(1+b)]

= 1 [SATISFIED!]

Thus, f(x) is a valid pmf. [CONCLUSION]