Geometric random variable with probability mass function

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  • #1
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, ive 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?
 
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  • #3
playboy
No.. i actually mean

" P(X=n) = p(1-p)^n n = 0,1,2,3.... " That is the geometric one im 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 im doing right now.. thats my prof! :)

But back to the work... i still cannot figure out how to get the sum of that sequence.
 
  • #4
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We did this by finding the moment generating function and then differentiating at t=0.
 
  • #5
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playboy said:
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, ive 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.
 

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