Report

Generating Functions The Moments of Y • We have referred to E(Y) and E(Y2) as the first and second moments of Y, respectively. In general, E(Yk) is the kth moment of Y. • Consider the polynomial where the moments of Y are incorporated into the coefficients k 2 2 3 3 t t E (Y ) t E (Y ) k E (Y ) 1 t E (Y ) 2! 3! k 0 k ! Moment Generating Function • If the sum converges for all t in some interval |t| < b, the polynomial is called the moment-generating function, m(t), for the random variable Y. t 2 E (Y 2 ) t 3 E (Y 3 ) m(t ) 1 t E (Y ) 2! 3! • And we may note that for each k, t k E (Y k ) k! t k y k p( y ) y k! (t y)k p( y ) k! y Moment Generating Function • Hence, the moment-generating function is given by t 2 E (Y 2 ) t 3 E (Y 3 ) m(t ) 1 t E (Y ) 2! 3! (t y )k p( y ) k! k 0 y (t y)k y k 0 k ! p( y ) May rearrange, since finite for |t| < b. et y p( y ) E[ety ] y Moment Generating Function • That is, m(t ) E[ety ] t 2 E (Y 2 ) t 3 E (Y 3 ) 1 t E (Y ) 2! 3! is the polynomial whose coefficients involve the moments of Y. The th k moment • To retrieve the kth moment from the MGF, evaluate the kth derivative at t = 0. d k [m(t )] k !t 0 E (Y k ) (k 1)!t 1E (Y k 1 ) t 2 E (Y k 2 ) dt k! (k 1)! 2! • And so, letting t = 0: k d [m(t )] E (Y k ) dt t 0 Common MGFs • The MGFs for some of the discrete distributions we’ve seen include: binomial: m(t ) ( pet q)n pet geometric: m(t ) t 1 qe Poisson: m(t ) e ( et 1) Geometric MGF 1 3 e t t e • Consider the MGF m(t ) t t 2 1 3 e 3 2e • Use derivatives to determine the first and second moments. t m(t ) 3e 3 2e t 2 And so, E (Y ) m(0) 3e0 3 2e 0 2 3 3 1 Geometric MGF • Since m(t ) 3et 3 2e t 2 V (Y ) E (Y 2 ) [ E (Y )]2 • We have m(t ) t 3 2e t 3 And so, E (Y ) m(t ) 2 15 (3)2 6 3e (3 2e ) t 3e (3 2e ) 0 0 3 2e 0 3 15 Geometric MGF 1 3 e t t e • Since m(t ) t t 2 1 3 e 3 2e is for a geometric random variable with p = 1/3, our prior results tell us E(Y) = 1/p and V(Y) = (1 – p)/p2. 1 1 1 3 2 9 E (Y ) 3 and V (Y ) 6 2 13 1 3 3 1 which do agree with our current results. All the moments • Although the mean and variance help to describe a distribution, they alone do not uniquely describe a distribution. • All the moments are necessary to uniquely describe a probability distribution. • That is, if two random variables have equal MGFs, (i.e., mY(t) = mZ(t) for |t| < b ), then they have the same probability distribution. m(aY+b)? • For the random variable Y with MGF m(t), consider W = aY + b. m(t ) mY (t ) E[e ] tY mW (t ) E[et ( aY b ) ] E[e atY ebt ] e E[ e ] bt atY e mY (at ) bt E(aY+b) • Now, based on the MGF, we could again consider E(W) = E(aY + b). d bt mW (t ) e mY (at ) ebt mY (at )(a) mY (at )bebt dt bt e amY (at ) bmY (at ) And so, letting t = 0, E (W ) mW (0) e0 amY (0) bmY (0) aE(Y ) b as expected. V(aY+b) • Now, based on the MGF, can you again consider V(W) = V(aY + b). 2 mW (0) E (W ) ? • …and so V(W) = V(aY + b) = a2V(Y). Tchebysheff’s Theorem • For “bell-shaped” distributions, the empirical rule gave us a 68-95-99.7% rule for probability a value falls within 1, 2, or 3 standard deviations from the mean, respectively. • When the distribution is not so bell-shaped, Tchebysheff tells use the probability of being within k standard deviations of the mean is at least 1 – 1/k2, for k > 0. 1 P(| Y | k ) 1 2 k Remember, it’s just a lower bound. A Skewed Distribution • Consider a binomial experiment with n = 10 and p = 0.1. P(| Y 1| 2(0.95)) 1 1 2 0.75 2 A Skewed Distribution • Verify Tchebysheff’s lower bound for k = 2: P(| Y 1| 2(0.95)) P(0.9 Y 2.9) 1 1 2 0.75 2 P(0.9 Y 2.9) 0.34868 0.38742 0.19371 0.93