derive mgf of geometric distribution

Answer 1

The moment generating function (mgf) of a geometric distribution is given by:

M(t) = E[e^(tX)] = Σ e^(tx) * p(x)

where X is a random variable following a geometric distribution with probability of success p(success) = p(X = 1) = p.

For the geometric distribution, the possible values of X are x = 1, 2, 3, ... and their probabilities are p(x) = (1-p)^(x-1) * p.

Substitute p(x) into the formula for the mgf, we get:

M(t) = Σ e^(tx) * (1-p)^(x-1) * p

To simplify this expression, we recognize that it is a geometric series and use the formula for the sum of a geometric series to get:

M(t) = p * Σ e^(tx) * (1-p)^(x-1)
= p * Σ e^(t) * ((1-p)e^t)^(x-1)
= p * e^t / (1 - (1-p)e^t)
= p * e^t / (1-p + p * e^t)

Therefore, the moment generating function (mgf) of the geometric distribution with parameter p is given by:

M(t) = p * e^t / (1-p + p * e^t)