Fundamentals of Probability and Statistics for Engineers

(John Hannent) #1

Letting k 2 i n, we get


Comparing Equation (4.76) with the definition in Equation (4.46) yields the
mass function


Note 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 In order to consider this sp ecial case, we need to use Stirling’s
formula, which states that, for large n,


Substituting this approximation into Equation (4.77) gives


A further simplification results when the length of each step is small. Assuming
that r steps occur in a unit time (i.e. n ) and letting a be the length of each
step, then, as n becomes large, random variable Y approaches a continuous
random variable, and we can show that Equation (4.79) becomes


where. On letting


we have


The probability density function gi ven above belongs to a Gaussian or normal
random variable. This result is an illustration of the central limit theorem, to be
discussed in Section 7.2.


Expectations and Moments 107


ˆ 

Y…t†ˆ

Xn

kˆn

n
n‡k
2



p…n‡k†=^2 q…nk†=^2 ejkt:… 4 : 76 †

pY…k†ˆ

n
n‡k
2



p…n‡k†=^2 q…nk†=^2 ; kˆn;…n 2 †;...;n: … 4 : 77 †



pˆqˆ1/2.

n!… 2 †^1 =^2 ennn‡

(^12)
… 4 : 78 †
pY…k†


2

n

 1 = 2

ek

(^2) = 2 n
;kˆn;...;n: … 4 : 79 †
ˆrt
fY…y†ˆ


1

… 2 a^2 rt†^1 =^2

exp 

y^2
2 a^2 rt



; 1<y< 1 ; … 4 : 80 †

yˆka


a^2 r
2

;

fY…y†ˆ

1

… 4 Dt†^1 =^2

exp 

y^2
4 Dt



; 1<y< 1 : … 4 : 81 †
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