Question

Let X Geom(p). The MGF was derived during the class. d g(t)

Using the MGF, show that E(X) = 1/p.

Answer 1

To show that E(X) = 1/p using the moment generating function (MGF) of a geometric distribution, we can find the MGF of X and differentiate it to find E(X). Simplifying the expression, we get E(X) = 1/p.

Step-by-step explanation:

To show that E(X) = 1/p using the moment generating function (MGF) of a geometric distribution, we can start by finding the MGF of X. The MGF of X is given by g(t) = E(etX). For a geometric distribution with parameter p, the probability mass function (PMF) is P(X = k) = (1-p)k-1p. Using this PMF, we can find the MGF:

g(t) = E(etX) = Σ etkP(X = k) = Σ etk(1-p)k-1p = pΣ (et(1-p))k-1

To find E(X), we need to differentiate g(t) and evaluate it at t = 0:

E(X) = g'(0) = pΣ (k-1)(et(1-p))k-2 * (1-p)(1-p)etk at t = 0
= pΣ (k-1)(1-p)k-2 e0
= pΣ (k-1)(1-p)k-2

Now, we can simplify the expression:

E(X) = pΣ (k-1)(1-p)k-2
= p((1-p)0 + 2(1-p)1 + 3(1-p)2 + ...)
= p(1 + 2(1-p) + 3(1-p)2 + ...)
= 1/p