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...

Hello!

Prove this for b>0: then 0<b/(1+b)<1.

We have to prove that 1>f(x)>0 for all x and the sum of all f(x) is 1.

f(x)>0 is obvious and if we prove that sum of all f(x) is 1 it will mean also that f(x)<=1 for all x.

The sum of [b/(1+b)]^x is the sum of infinite geometrical progression and is equal to

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

So the sum of f(x) is (1+b)/(1+b) = 1.

For b=0 we can assume f(0)=1 and all other f(x)=0. Then iw is also a pmf. For b<0 it isn't a pmf.

Secure Information

Content will be erased after question is completed.

Enter the email address associated with your account, and we will email you a link to reset your password.

Forgot your password?

Sign Up