Re: Geometric distribution

Quote:

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.

Re: Geometric distribution

Quote:

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*.

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.

Re: Geometric distribution

Quote:

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 :)