Fundamentals of Probability and Statistics for Engineers

Expectations and Moments 107Letting k ˆ 2i n, we get Y …t† ˆ Xnkˆ nn‡kp …n‡k†=2 q …n k†=2 e jkt : …4:76†2Comparing Equation (4.76) with the definition in Equation (4.46) yields themass functionp Y …k† ˆ n n‡kp …n‡k†=2 q …n k†=2 ; k ˆ n; …n 2†; ...; n: …4:77†2Note that, if n is even, k must also be even, and, if n is odd k must be odd.Considerable importance is attached to the symmetric case in which k n,and p ˆ q ˆ 1/2. In order to consider this special case, we need to use Stirling’sformula, which states that, for large n,Substituting this approximation into Equation (4.77) givesnn! …2† 1=2 e n n n‡1 2 …4:78†p Y …k† 2 1=2e k2 =2n ; k ˆ n; ...; n: …4:79†nA further simplification results when the length of each step is small. Assumingthat r steps occur in a unit time (i.e. n ˆ rt) and letting a be the length of eachstep, then, as n becomes large, random variable Y approaches a continuousrandom variable, and we can show that Equation (4.79) becomes1f Y …y† ˆ…2a 2 rt† exp y 2 1=2 2a 2 ; 1 < y < 1; …4:80†rtwhere y ˆ ka. On lettingwe havef Y …y† ˆD ˆ a2 r2 ;1…4Dt† exp y 2 ; 1 < y < 1: …4:81†1=2 4DtThe probability density function given above belongs to a Gaussian or normalrandom variable. This result is an illustration of the central limit theorem, to bediscussed in Section 7.2.TLFeBOOK