Moment Generating FunctionsDate: 06/20/98 at 12:03:10 From: Kenneth W. Hodgkinson Subject: Moment Generating Functions I would appreciate any help you can give me on explaining Moment Generating functions, specifically these questions: 1. Prove that M(x+y)(t) = Mx(t) * My(t). What assumptions are needed? Here, Mx means the moment generating function of x, My means the moment generating function of y and M(x+y) means the moment generating function of x+y. 2. Find the moment generating function of the exponential distribution with mean a. What is the restriction on t and why is it needed? 3. Prove Var(x + y) = Var(x) + Var(y). What assumptions are needed, and why? Sincerely, K.W. Hodgkinson Date: 06/20/98 at 16:03:22 From: Doctor Anthony Subject: Re: Moment Generating Functions Problem 1: By definition: Mx(t) = E(e^(tx)) and My(t) = E(e^(ty)) Then: Mx(t) * My(t) = E[e^(tx) e^(ty)] provided x and y are independent. = E[e^(t(x+y))] = M(x+y)(t) The necessary condition is that x and y are independent. Problem 2: The probability distribution function (pdf) of an exponential distribution with parameter a is f(x) = ae^(-ax). So: Mx(t) = INT(0 to inf)[e^(tx) * ae^(-ax) * dx] = a * INT[e^(x(t-a)) dx] This integral converges only if t < a. So the mgf exists only for those values of t. Then we have: Mx(t) = a/(t-a) e^(x(t-a)) for 0 to infinity = a/(a-t) t < a Problem 3: Provided x and y are independent, M(x+y)(t) = Mx(t) My(t). We have: e^(tx) = 1 + tx + (tx)^2/2! + (tx)^3/3! + ... Taking expectations Mx(t) = 1 + tE(x) + t^2 E(x^2)/2! + t^3 E(x^3)/3! + ... My(t) = 1 + tE(y) + t^2 E(y^2)/2! + t^3 E(y^2)/3! + ... To find M(x+y)(t) = Mx(t) My(t), we multiply the two series: M(x+y)(t) = 1 + t[E(x) + E(y)] + (t^2/2!)[E(x^2) + 2.E(x) E(y) + E(y^2)] + ... Then: Var(x) = E(x^2) - [E(x)]^2 = coeff of t^2/2! - [coeff of t]^2 in the series for Mx(t) Similarly: Var(y) = E(y^2) - [E(y)]^2 = coeff of t^2/2! - [coeff of t]^2 in the series for My(t) Similarly, Var(x+y) = coeff of t^2/2! - [coeff of t]^2 in the series M(x+y)(t) = [E(x^2 + 2E(x) E(y) + E(y^2)] - [E(x) + E(y)]^2 = E(x^2) + E(y^2) + 2E(x) E(y) - [E(x)]^2 - 2E(x) E(y) - [E(y)]^2 = E(x^2) - [E(x)]^2 + E(y^2) - [E(y)]^2 = Var(x) + Var(y) Remember that this only applies if x and y are independent. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 06/21/98 at 15:18:11 From: Ken Hodgkinson Subject: Moment Generating Functions I am having problems understanding the Moment Generating Function. 1) If a random variable x has the discrete uniform distribution f(x;k) = 1/k, x = 1,2,3,...k and 0 elsewhere, can it be shown that the MGF of x is Mx(t) = (e^t(1-e^kt)/k(1-e^t) ? 2) If x has a Poisson distribution p(x;m) = e^-m m^x/x! for x = 0, 1, 2 ... can it be shown the MGF is = e^(m(e^t-1)) ? 3) Can it be shown that the mean and variance of the chi-squared distribution with v degrees of freedom are, respectively, v and 2v? Thank you for your time and help. Sincerely, K.W. Hodgkinoson Date: 06/21/98 at 16:20:16 From: Doctor Anthony Subject: Re: Moment Generating Functions Problem 1: We define Mx(t) = E(e^(xt)). Then for the uniform distribution given, Mx(t) = SUM[e^(xt)(1/k)] = (1/k)[e^t + e^(2t) + e^(3t) + ..... + e^(kt)] (a geometric progression) = (1/k)e^t[1-e^(kt)]/(1-e^t) as required. We have used the formula for the sum of a geometric progression: a(1-r^n)/(1-r) Problem 2: Mx(t) = SUM[e^(tx) m^x e^(-m)/x!] x = 0 to infinity = e^(-m) SUM[(me^t)^x (1/x!)] = e^(-m) e^(me^t) = e^(m(e^t - 1)) Problem 3: The mgf of the X^2 (chi-squared) distribution with v degrees of freedom is: Mx(t) = (1 - 2t)^(-v/2) So: Mx(t) = 1 + (-v/2)(-2t) + (-v/2)(-v/2 -1)(-2t)^2/2! + ..... = 1 + t(v) + (t^2/2!)[(v^2/4 + v/2)4] + .... = 1 + t(v) + t^2/2!)[v^2 + 2v] + ..... Then E(x) = coefficient of t = v and E(x^2) = coefficient of t^2/2 = v^2 + 2v So, the mean = v and variance = E(x^2) - mean^2 = v^2 + 2v - v^2 = 2v And, as required, we get mean = v and variance = 2v. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 07/01/98 at 18:24:46 From: Ken Hodgkinson Subject: Probability Dr. Math, I was wondering if you could help me with another question: Explain why the "units" of a density f(x) are "probability per unit length." This is based on what assumption? I appreciate any help you might be able to give me. Sincerely, Ken Hodgkinson Date: 07/02/98 at 05:18:18 From: Doctor Anthony Subject: Re: Probability The area standing on any interval of the density function represents the probability that x lies in that interval. For example the probability that X lies in the interval x to x+dx is f(x) dx. This means that f(x) is the probability per unit length at that value of x. [i.e. if dx = 1 unit, then probability is f(x), so f(x) is the probability per unit length]. - Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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