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We now expand the exponential in the definition of the Laplace transform of the waiting-
time density to obtain
∞
1
2
ψ(
e
−
ut
t
2
u
2
u
)
=
dt
ψ(
t
)
=
1
−
t
u
+
+···
,
(5.141)
0
which, on its insertion into (
5.139
), keeping only the lowest-order terms in
u
, and inverse
Laplace transforming, yields the average displacement
≈
μ
1
q
;
t
t
(5.142)
t
and the constant average velocity is given by
d
q
;
t
≈
μ
1
.
(5.143)
dt
t
In the same way the second moment of the process, in one dimension, can be written
k
=
0
2
P
∗
(
}= −
∂
,
)
k
u
q
2
LT
{
;
t
;
u
(5.144)
k
2
∂
and evaluating the second moment using the structure function
k
=
0
=
2
μ
2
=
∂
p
(
k
)
q
2
p
(
q
)
(5.145)
∂
k
2
q
inserted into (
5.144
) and taking the appropriate derivative yields
μ
1
ψ(
2
μ
2
ψ(
u
)
1
u
u
)
q
2
u
1
)
+
LT
{
;
t
;
u
}=
.
(5.146)
−
ψ(
−
ψ(
)
u
1
u
We again make use of the Taylor expansion of the exponential function in the Laplace
transform of the waiting-time distribution in (
5.141
), which we insert into (
5.146
), keep-
ing only the lowest-order term in
u
. The inverse Laplace transform of the resulting
expression yields the variance
q
2
μ
2
−
μ
1
+
t
2
−
2
t
2
2
t
t
2
;
−
q
;
t
≈
t
.
(5.147)
3
t
Note that we are assuming that the second moment of the waiting-time distribution is
finite. There are two contributions to the variance of the displacement of the random
walker; one is due to the variance in the step size and the other is due to the variance
in the waiting-time distribution. If either of these two variances diverges the resulting
variance of the random walk also diverges. We have assumed that the second moment
of the waiting-time distribution is finite and in so doing both the mean and the variance
of the random walk are determined to increase linearly with time. Consequently, the
dimensionless variable, the ratio of the standard deviation to the mean, goes to zero as
the square root of the time,
q
2
t
−
2
;
q
;
t
1
√
t
,
≈
(5.148)
q
;
t
