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
Y
t
Xn
kn
n
nk
2
p
nk=^2 q
nk=^2 ejkt:
4 : 76
pY
k
n
nk
2
p
nk=^2 q
nk=^2 ; kn;
n 2 ;...;n:
4 : 77
pq1/2.
n!
2 ^1 =^2 ennn
(^12)
4 : 78
pY
k
2
n
1 = 2
ek
(^2) = 2 n
;kn;...;n:
4 : 79
rt
fY
y
1
2 a^2 rt^1 =^2
exp
y^2
2 a^2 rt
; 1<y< 1 ;
4 : 80
yka
D
a^2 r
2
;
fY
y
1
4 Dt^1 =^2
exp
y^2
4 Dt
; 1<y< 1 :
4 : 81