# Thread: Geometric distribution

1. ## Geometric distribution

Hi, I am studying discrete random variables and have been given the question:

Suppose that X is a geometrically distributed random variable with parameter p €(0,1) i.e for all n=1,2..... P(X=n)=(1-p)^(n-1)p

Show that for k=0,1,......
P(X>k)=(1-p)^k

In the previous part of the question I proved that E(X)=1/p, will the formula
E(X)=sum(P(X>n)) where n € 0,1,2..... help me?

Thanks for your help

2. ## Re: Geometric distribution

E(X)=sum(P(X>n)) where n € 0,1,2..... help me?
I haven't seen that before and its not true in general. if its true for this distribution (and i haven't checked) then it might help but its probably just as easy to note the following:

P(X>k) = P(X=k+1) + P(X=k+2) ....
if you write out the terms you will see a geometric progression. There is a standard formula for summing those.

3. ## Re: Geometric distribution

Originally Posted by SpringFan25
I haven't seen that before and its not true in general. if its true for this distribution (and i haven't checked) then it might help but its probably just as easy to note the following:

P(X>k) = P(X=k+1) + P(X=k+2) ....
if you write out the terms you will see a geometric progression. There is a standard formula for summing those.
The "..." makes it a bit more difficult than if we look at $\displaystyle 1 - P(X \leq k) = P(X > k)$. This way we have a finite sum of probabilites up to k for each k.

4. ## Re: Geometric distribution

i guess its a matter of style/backround whether you prefer to sum a finite or infinite geometric series.

Personally i would consider the infinite case simpler, as there is no risk of counting the wrong number of terms (which i have an irrational fear of doing). In addition I would point out that the formula for the finite geometric sum is just the difference between 2 infinite sums anyway.

5. ## Re: Geometric distribution

Originally Posted by SpringFan25
i guess its a matter of style/backround whether you prefer to sum a finite or infinite geometric series.

Personally i would consider the infinite case simpler, as there is no risk of counting the wrong number of terms (which i have an irrational fear of doing). In addition I would point out that the formula for the finite geometric sum is just the difference between 2 infinite sums anyway.

thanks for your help guys, I used the infinite case