Before I get started here I have one really quick basic question:
Lets say I want the probability that an survives two hours, and that the probability an engine will fail in any given hour is .02. Then I can get 1 - .02 - .98(.02) = .9604. This is found by a geometric distribution, 1 - the sum of q^1-1 p - q^2-1 p.
This got me thinking though, if I have (q)^2 = (1-p)^2 this gives me (.98)^2. Is this not equal to 1 - 2p + p^2? I feel like I'm forgetting something basic about order of operations but if I haven't plugged in p yet I don't see why this applies. Why can't I expand then solve rather than solve then expand (though strictly in this case I have nothing more to expand).
P(Y) = q (= 1-p)^Y-1 (p)
My other question is more interesting I suppose: Find E[Y(Y-1)] for a geometric random variable Y by finding the second derivative of the infinite sum (from 1 to infinity) of q^y. Use this result to find the variance. I have no idea where to begin on this question to be honest.