Trying to derive the expectation for the geometric distribution. Can someone help me derive the identity of the series n(p)^(n-1) = 1/(1-p)^2?

Printable View

- Aug 10th 2007, 12:47 AMCrazyAsianEvaluate series np(1-p)^k
Trying to derive the expectation for the geometric distribution. Can someone help me derive the identity of the series n(p)^(n-1) = 1/(1-p)^2?

- Aug 10th 2007, 05:24 AMTKHunny
I'm not convinced you have written what you want. Anyway, it's called "Geometric" for a reason. Everything you ever leared about Geometric Series applies, here.

Remember this?

$\displaystyle \frac{a}{1-r} = a + ar + ar^{2} + ar^{3} + ...$

You ALMOST have that. You're only additional concern is that pesky linearly increasing coefficient. It is easily removed by an old trick.

If $\displaystyle p + 2pq + 3pq^{2} + 4pq^{3} + ... = S$

Then $\displaystyle pq + 2pq^{2} + 3pq^{3} + 4pq^{4} + ... = Sq$

Subtracting the second from the first gives:

$\displaystyle p + pq + pq^{2} + pq^{3} + ... = S - Sq = S(1-q)$

Are we getting anywhere? Let's work with the left hand side - WAIT! That's just our friend the Geometric Series. We know its sum.

$\displaystyle \frac{p}{1-q} = S(1-q)$

Solve for S, substitute p = 1-q, and I think you're done.

Note: Before solving for S and before finding the sum of the left hand side, one MAY notice that these are simply the probabiltiies of the possible outcomes of the Geometric Distribution. The sum of those had better be one (1).