Mean. The geometric distribution also has its own mean and variance formulas for Y. So the variance is var(X) = E[X2]− E[X] ... with the mgf, and we will not use the probability generating function. For , , because cannot be smaller than . Proof . MOMENT GENERATING FUNCTION (mgf) •Let X be a rv with cdf F X (x). The mean is defined as the use of a moment generating function. ELEMENTS OF PROBABILITY DISTRIBUTION THEORY For the exponential distribution we have fX(x) = λe−λxI (0,∞)(x), where λ ∈ R+. 1. First though let’s first back up to the concept of center of gravity (cog) from mechanics. As you know multiple different moments of the distribution, you will know more about that distribution. The mgf need not be deﬁned for all t. We saw an example of this with the geometric distribution where it was deﬁned only if et(1 − p) < 1, i.e, t < −ln(1 − p). Proof. For example, you can completely specify the normal distribution by the first two moments which are a mean and variance. Geometric(p) p (1 ¡ p) x ¡ 1; p 2 (0; 1) 1 p 1 ... Notes: If Y is gamma(ﬁ;ﬂ), X is Poisson(x ﬂ), and ﬁ is an integer, then P (X ‚ ﬁ)= P (Y • y). The variance of a geometric random variable is. Moments provide a way to specify a distribution. (1) Find the mgf of X ∼ Exp(λ) and use results of Theorem 1.7 to obtain the mean and variance of X. The shifted geometric distribution. Whenever you compute an MGF, plug in t = 0 and see if you get 1. The moment generating function (mgf) of X, denoted by M X (t), is provided that expectation exist for t in some neighborhood of 0. The distribution function of a geometric random variable is. In fact, it need not be deﬁned for any t other than 0. For , we have. The mean and the variance of a random variable X with a binomial probability distribution can be difficult to calculate directly. The mean (E( Y ) or μ ) is the weighted average of all potential values of Y . 20 CHAPTER 1. By deﬁnition the mgf can be written as MX(t) = E (etX) = Z ∞ −∞ etxf X(x)dx. That is, there is h>0 such that, for all t in h