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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 y k p( y ) t k E (Y k ) (t y )k y p( y ) k! k! y k! 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 0 y k! (t y)k May rearrange, p( y ) since finite for y k 0 k ! |t| < b. et y p( y ) E[ety ] y Moment Generating Function • That is, m(t ) E[ety ] ety p ( y ) y 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 k th 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)!t1E (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 Geometric MGF • For the geometric distribution, m(t ) E[e ] e ( q ty ty y 1 p) y et p e 2t qp e3t q 2 p e 4t q 3 p e p 1 e q e q e q t t 2t 2 3t 3 1 pet e p t t 1 e q 1 qet Common MGFs • The MGFs for some of the discrete distributions we’ve seen include: binomial: m(t ) ( pe q)t n t pe geometric: m(t ) 1 qet r pe t negative binomial: m(t ) t 1 qe ( et 1) Poisson: m(t ) e Recognize the distribution • Identify the distribution having the moment generating function 20 e 3 t m(t ) 4 • Give the mean and variance for this distribution. • Could use the derivatives, but is that necessary? Geometric MGF 1 t t e e • Consider the MGF m(t ) 3 1 3 e 3 2e 2 t t • Use derivatives to determine the first and second moments. t 3e m(t ) 3 2e t 2 And so, 3e0 3 E (Y ) m(0) 3 3 2e 0 2 1 Geometric MGF 3et • Since m(t ) 3 2e t 2 V (Y ) E (Y 2 ) [ E (Y )]2 • We have 3e (3 2e ) t t 15 (3) 6 2 m(t ) 3 2e t 3 And so, 3e (3 2e ) 0 0 E (Y ) m(t ) 2 15 3 2e 0 3 Geometric MGF 1 t t e e • Since m(t ) 3 1 3 e 3 2e 2 t t 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 1 3 3 1 2 13 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[etY ] mW (t ) E[et ( aY b ) ] Construct the MGF for the random variable E[e atY ebt ] W= 2Y + 3, where Y is a geometric random e E[e ] bt atY variable with p = 4/5. 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 e amY (at ) bmY (at ) bt And so, letting t = 0, E (W ) mW (0) e0 amY (0) bmY (0) aE (Y ) b as expected. 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 Remember, it’s just P(| Y | k ) 1 2 a lower bound. k 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

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posted: | 10/5/2012 |

language: | English |

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