Geometric random variable with probability mass function

[playboy]

If you have a geometric random variable with probability mass function:

P(X=n) = p(1-p)^n n = 0,1,2,3...

Find the Mean and the Variance.

----------------------------

Okay, I've looked everywhere and tried everything, however, i just cannot get it.

i think that your supposed to find the sum of this sequence to find the mean. I tried everything. I just tried the integral test. Anybody got any ideas?

 

[HallsofIvy]

You might have tried writing the correct formula for geometric distribution.

It is p^n-1(1-p)ⁿ, not p(1-p)ⁿ.

And you might try looking here:
http://www.math.mcmaster.ca/canty/teaching/stat2d03/lectures4.pdf

 

[playboy]

No.. i actually mean

" P(X=n) = p(1-p)^n n = 0,1,2,3... " That is the geometric one I am looking for.

AND NOT

" p^n-1(1-p)^n " by the way that may not be right, you must mean "p(1-p)^n-1 n = 1,2,3..." take a look at the beginning of this:

http://en.wikipedia.org/wiki/Geometric_distribution

And its actually ironic how that website you gave me is the EXACT course I am doing right now.. that's my prof! :)

But back to the work... i still cannot figure out how to get the sum of that sequence.

 

[happyg1]

We did this by finding the moment generating function and then differentiating at t=0.

 

[happyg1]

If you have a geometric random variable with probability mass function:
P(X=n) = p(1-p)^n n = 0,1,2,3...
Find the Mean and the Variance.
----------------------------
Okay, I've looked everywhere and tried everything, however, i just cannot get it.
i think that your supposed to find the sum of this sequence to find the mean. I tried everything. I just tried the integral test. Anybody got any ideas?

I'm pretty sure the sum of the infinite series there is 1. You factor out the p and then 1-p=q, so you have p*SUMq^n (0<=q<=1) which is (from calculus) p*1/(1-q) and 1-q =p...which is p/p=1. That's the proof that the sum of all of the possible outcomes is 1.