Statistics: Proofs and Problems for Random Variables and their Distributions

[StopWatch]

Homework Statement

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.

 

[lippyka]

Any guidance would be greatly appreciated. Thanks.Homework Equations P(Y)= q (= 1-p)^Y-1 (p) The Attempt at a Solution For the first question, yes, you can expand and then solve. The order of operations does not matter here because you are dealing with algebraic equations. You can expand the equation to get: 1 - 2p + p^2 = (1 - p)^2 which is the same as 1 - p = (1 - p)^2 so you can solve for p to get p = 1. For the second question, we can use the formula for the expected value of a geometric random variable: E[Y] = \frac{q}{p} So, the expected value of Y(Y-1) is: E[Y(Y-1)] = E[Y^2] - E[Y] Now, we can compute the variance of Y, which is the expected value of Y^2 minus the square of the expected value of Y: Var[Y] = E[Y^2] - (E[Y])^2 To compute the expected value of Y^2, we can use the formula for the second derivative of an infinite sum: \frac{d^2}{dx^2} \sum_{n=1}^{\infty} a_nx^n = \sum_{n=3}^{\infty} n(n-1)a_nx^{n-2} So, for our case, we can set a_n = q^n and x = q, to get: \frac{d^2}{dq^2} \sum_{n=1}^{\infty} q^{n} = \sum_{n=3}^{\infty} n(n-1)q^{n-2} We can compute this sum by taking the limit as n goes to