How do I correctly prove the memoryless property of a geometric distribution?

[cookiesyum]

Homework Statement

Let X have a geometric distribution. Show that

P(x>or= k+j|x>or=k) = P(X>or=j)

where k and j are nonnegative integers.

The Attempt at a Solution

P(x>or= k+j|x>or=k) = P(x>or= k+j intersect x>or=k)/P(x>or=k) = P(x>or=k+j)/P(x>or=k) for j>or= 0 = [1-P(x=k+j)]/[1-P(x=k)] = [1 - p(1-p)^(k+j)]/[1-p(1-p)^k] = [1-p(1-p)^k(1-p)^j]/[1-p(1-p)^k] = (1-p)^j which is equal to P(x>or=j) if it had another p so that it was p(1-p)^j...so where am I losing this p in my proof?

 

[Billy Bob]

= [1-P(x=k+j)]/[1-P(x=k)]

That's an error. It should be = [1-P(x<k+j)]/[1-P(x<k)]. However, don't do it this way.

= [1-p(1-p)^k(1-p)^j]/[1-p(1-p)^k] = (1-p)^j

That's an algebra error. Are you saying (a+bcd)/(a+bc) would be equal to d? It isn't.

(1-p)^j which is equal to P(x>or=j)

This is right, and it is the key. Since you know this, you should go back and use this same idea in the expression P(x>or=k+j)/P(x>or=k).