If Y has a geometric distribution with success probability p, show that

P(Y= an odd integer)= p/(1-q^2)

Thanks.

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- February 13th 2008, 04:34 PMfulltwist8Probability- Geometric Prob. Distribution
If Y has a geometric distribution with success probability p, show that

P(Y= an odd integer)= p/(1-q^2)

Thanks. - February 13th 2008, 05:42 PMmr fantastic
- February 13th 2008, 07:09 PMfulltwist8
That IS exactly as it's stated in the book. q=1-p. it doesn't say what constitutes a success.

- February 13th 2008, 07:49 PMmr fantastic
Without knowing what event constitutes a success, the question makes no sense - to me, anyway. It's very .... odd (ha ha) - to me, anyway - that the question would not define it. And it doesn't even define the random variable Y .....

By the way, if you make what would seem the obvious assumption that success is getting an odd number, then the probability of getting an odd number eventually is going to be 1, I'd have thought (unless p = 0) ......

Are you sure there's not some lead in, some introduction, some broader context ...... to the question. Because as is, the question is (Puke) - April 4th 2008, 02:04 AMpuzzellled
- April 4th 2008, 03:23 AMOpalg