Mathematical Methods for Physics and Engineering - Matematica.NET

30.5 PROPERTIES OF DISTRIBUTIONSWe note that the notation µ k and ν k for the moments and central momentsrespectively is not universal. Indeed, in some books their meanings are reversed.We can write the kth central moment of a distribution in terms of its kth andlower-order moments by expanding (X − µ) k in powers of X. We have alreadynoted that ν 2 = µ 2 − µ 2 1 , and similar expressions may be obtained for higher-ordercentral moments. For example,In general, it is straightforward to show thatν 3 = E [ (X − µ 1 ) 3]= E [ X 3 − 3µ 1 X 2 +3µ 2 1X − µ 3 ]1= µ 3 − 3µ 1 µ 2 +3µ 2 1µ 1 − µ 3 1= µ 3 − 3µ 1 µ 2 +2µ 3 1. (30.53)ν k = µ k − k C 1 µ k−1 µ 1 + ···+(−1) r k C r µ k−r µ r 1 + ···+(−1) k−1 ( k C k−1 − 1)µ k 1.(30.54)Once again, direct evaluation of the sum or integral in (30.52) can be rathertedious for higher moments, and it is usually quicker to use the moment generatingfunction (see subsection 30.7.2), from which the central moments can be easilyevaluated as well.◮The PDF for a Gaussian distribution (see subsection 30.9.1) with mean µ and varianceσ 2 is given byf(x) = 1 [ ]σ √ 2π exp (x − µ)2− .2σ 2Obtain an expression for the kth central moment of this distribution.As an illustration, we will perform this calculation by evaluating the integral in (30.52)directly. Thus, the kth central moment of f(x) isgivenbyν k =∫ ∞−∞= 1σ √ 2π= 1σ √ 2π(x − µ) k f(x) dx∫ ∞−∞∫ ∞−∞](x − µ) k (x − µ)2exp[−2σ 2dx)y k exp(− y2dy, (30.55)2σ 2where in the last line we have made the substitution y = x − µ. It is clear that if k isodd then the integrand is an odd function of y and hence the integral equals zero. Thus,ν k =0ifk is odd. When k is even, we could calculate ν k by integrating by parts to obtaina reduction formula, but it is more elegant to consider instead the standard integral (seesubsection 6.4.2)I =∫ ∞−∞exp(−αy 2 ) dy = π 1/2 α −1/2 ,1149