Suppose that x has a geometric distribution where p(x) = p x−1 (1 − p) for x = 1, 2, 3, 4, 5, . . (a) (bonus) show that the expected value e(x − 1)(x − 2) = 2p 2 (1 − p) 2 by using the facts: x[infinity] x=1 d dpp(x) = 0 and x[infinity] x=1 d 2 dp2 p(x) = 0. hints: (i) e(x − 1) = p[infinity] x=1(x − 1)p(x) = p[infinity] x=1(x − 1)p x−1 (1 − p) (ii) using (i) and the first fact show e(x − 1) = p 1−p (iii) using (ii) and the second fact show e(x − 1)(x − 2) = 2p 2 (1−p) 2 (b) use the answer from part (a) to calculate the variance of x. (c) calculate e(a x) for some number a that is |a| < 1. (d) use the answer from part (c) to calculate e(e tx) for some number t. for which values of t does this expectation not exist?