Show that the following function is a probability mass function(pmf) for the random variable x and constant b
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.
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