Klyatskin V.I. Lectures on Dynamics of Stochastic Systems
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116 Lectures on Dynamics of Stochastic Systems
transforms the characteristic functional into the joint characteristic function of random
quantities z
k
= z(t
k
)
8
n
(v
1
, . . . , v
n
) =
*
exp
(
i
n
X
k=1
v
k
z(t
k
)
)+
,
whose Fourier transform is the joint probability density of process z(t) at discrete
instants
P
n
(z
1
, t
1
;. . . ;z
n
, t
n
) =
h
δ(z(t
1
) − z
1
) . . . δ(z(t
n
) − z
n
)
i
. (4.81)
Assume that the above instants are ordered according to the line of inequalities
t
1
≥ t
2
≥ ··· ≥ t
n
.
Then, by definition of the conditional probability, we have
P
n
(z
1
, t
1
;. . . ;z
n
, t
n
) = p
n
(z
1
, t
1
|z
2
, t
2
;. . . ;z
n
, t
n
)P
n−1
(z
2
, t
2
;. . . ;z
n
, t
n
),
(4.82)
where p
n
is the conditional probability density of the value of process z(t) at instant
t
1
under the condition that function z(t) was equal to z
k
at instants t
k
for k = 2, . . . , n
(z(t
k
) =z
k
, k=2, . . . , n). If process z(t) is such that the conditional probability density
for all t
1
> t
2
is unambiguously determined by the value z
2
of the process at instant t
2
and is independent of the previous history, i.e., if
p
n
(z
1
, t
1
|z
2
, t
2
;. . . ;z
n
, t
n
) = p(z
1
, t
1
|z
2
, t
2
), (4.83)
then this process is called the Markovian process, or the memoryless process. In this
case, function
p(z, t|z
0
, t
0
) =
h
δ(z(t) − z)|z(t
0
) = z
0
i
(t > t
0
), (4.84)
is called the transition probability density. Setting t = t
0
in Eq. (4.84), we obtain the
equality
p(z, t
0
|z
0
, t
0
) = δ(z − z
0
).
Substituting expression (4.82) in Eq. (4.81), we obtain the recurrence formula for
the n-time probability density of process z(t). Iterating this formula, we find the
Random Quantities, Processes, and Fields 117
expression of probability density P
n
in terms of the one-time probability density
(t
1
≥ t
2
≥ ··· ≥ t
n
)
P
n
(z
1
, t
1
;. . . ;z
n
, t
n
) = p(z
1
, t
1
|z
2
, t
2
) . . . p(z
n−1
, t
n−1
|z
n
, t
n
)P(t
n
, z
n
). (4.85)
Substituting expression (4.82) in Eq. (4.81), we obtain the recurrence formula for
the n-time probability density of process z(t). Iterating this formula, we find the
relationship of probability density P
n
with the one-time probability density (t
1
≥
t
2
≥ ··· ≥ t
n
)
P
n
(z
1
, t
1
;. . . ;z
n
, t
n
) = p(z
1
, t
1
|z
2
, t
2
) . . . p(z
n−1
, t
n−1
|z
n
, t
n
)P(t
n
, z
n
). (4.86)
Nevertheless, statistical analysis of stochastic equations often requires the knowl-
edge of the characteristic functional of random process z(t).
4.3.2 Characteristic Functional of the Markovian Process
For the Markovian process z(t), no closed equation can be derived in the general case
for the characteristic functional 8[t;v(τ )] =
h
ϕ[t;v(τ )]
i
, where
ϕ[t;v(τ )] = exp
i
t
Z
0
dτ z(τ )v(τ )
.
Instead, we can derive the closed equation for the functional
9[z, t;v(τ )] =
h
δ(z(t) − z)ϕ[t;v(τ )]
i
, (4.87)
describing correlations of process z(t) with its prehistory. The characteristic functional
8[t;v(τ )] can be obtained from functional 9[z, t;v(τ )] by the formula
8[t;v(τ )] =
∞
Z
−∞
dz9[z, t;v(τ )]. (4.88)
To derive the equation in functional 9[z, t;v(τ )], we note that the following
equality
ϕ[t;v(τ )] = 1 + i
t
Z
0
dt
1
z(t
1
)v(t
1
)ϕ[t
1
;v(τ )] (4.89)
holds. Substituting Eq. (4.89) in Eq. (4.87), we obtain the expression
9[z, t;v(τ )] = P(z, t) + i
t
Z
0
dt
1
v(t
1
)
h
δ(z(t) − z)z(t
1
)ϕ[t
1
;v(τ )]
i
, (4.90)
118 Lectures on Dynamics of Stochastic Systems
where P(z, t) =
h
δ(z(t) − z)
i
is the one-time probability density of random quantity
z(t).
We rewrite Eq. (4.90) in the form
9[z, t;v(τ )] = P(z, t)
+ i
t
Z
0
dt
1
v(t
1
)
∞
Z
−∞
dz
1
z
1
h
δ(z(t) − z)δ(z(t
1
) − z
1
)ϕ[t
1
;v(τ )]
i
.
(4.91)
Taking into account the fact that process z(t) is the Markovian process, we can perform
averaging in Eq. (4.91) to obtain the closed integral equation
9[z, t;v(τ )] = P(z, t) + i
t
Z
0
dt
1
v(t
1
)
∞
Z
−∞
dz
1
z
1
p(z, t|z
1
, t
1
)9[z
1
, t
1
;v(τ )],
(4.92)
where p(z, t;z
0
, t
0
) is the transition probability density.
Integrating Eq. (4.92) with respect to z, we obtain an additional relationship bet-
ween the characteristic functional 8[t;v(τ )] and functional 9[z, t;v(τ )]. This rela-
tionship has the form
1
iv(t)
d
dt
8[t;v(τ )] =
∞
Z
−∞
dz
1
z
1
9[z
1
, t;v(τ )] = 9[t;v(τ )]. (4.93)
Multiplying Eq. (4.92) by z and integrating the result over z, we obtain the relation-
ship between functionals 9[t;v(τ )] and 9[z, t;v(τ)]
9[t;v(τ )] =
h
z(t)
i
+ i
t
Z
0
dt
1
v(t
1
)
∞
Z
−∞
dz
1
h
z(t)|z
1
, t
1
i
9[z
1
, t
1
;v(τ )]. (4.94)
Equation (4.92) is generally a complicated integral equation whose explicit form
depends on functions P(z, t) and p(z, t;z
0
, t
0
), i.e., on parameters of the Markovian
process.
In some specific cases, the situation is simplified. For example, for telegrapher’s
process, we have from Eq. (4.74)
h
z(t)|z
1
, t
1
i
= z
1
e
−2ν(t−t
1
)
,
h
z(t)
i
= 0,
and we obtain Eq. (4.24), page 95.
Random Quantities, Processes, and Fields 119
Problems
Problem 4.1
Derive expression for the characteristic functional of Poisson’s random process
(4.67).
Solution The characteristic functional of Poisson’s random process has the form
8[t;v(τ )] =
*
exp
i
t
Z
0
dτ v(τ )z(τ )
+
z
. (4.95)
We will average Eq. (4.95) in two stages. At first, we perform averaging over
random quantity ξ and positions of random points t
k
:
8[t;v(τ )] =
**
exp
i
t
Z
0
dτ v(τ )
n
X
k=1
ξ
k
g(τ − t
k
)
+
ξ
i
,t
k
+
n
=
*
1
T
T
Z
0
dt
0
∞
Z
−∞
dξ exp
iξ
t
Z
0
dτ v(τ )g(τ − t
0
)
n
+
n
=
*
1
T
T
Z
0
dt
0
W
t
Z
t
0
dτ v(τ )g(τ − t
0
)
n
+
n
, (4.96)
where W(v) =
R
∞
−∞
dξ p(ξ)e
iξv
is the characteristic function of random quantity
ξ. Let t < T. Then Eq. (4.96) can be rewritten in the form
8[t;v(τ )] =
*
1
T
t
Z
0
dt
0
W
t
Z
t
0
dτ v(τ )g(τ − t
0
)
+
T − t
T
n
+
n
. (4.97)
Average now Eq. (4.97) over Poisson’s distribution p
n
=
e
−n
n!
(
n
)
n
of random
quantity n:
8[t;v(τ)] =
∞
X
0
e
−n
n!
(
n
)
n
*
1
T
t
Z
0
dt
0
W
t
Z
t
0
dτ v(τ)g(τ − t
0
)
+
T − t
T
n
+
n
= exp
−n +
n
T
t
Z
0
dt
0
W
t
Z
t
0
dτ v(τ)g(τ − t
0
)
+ T − t
.
120 Lectures on Dynamics of Stochastic Systems
As a result, we obtain the expression (4.67) for the characteristic functional of
Poisson’s random process.
Problem 4.2
Show that random process
ξ
N
(t) = z
1
(t) + ··· + z
N
(t),
where all z
k
(t) are statistically independent telegrapher’s processes with zero-
valued means and correlation functions
h
z(t)z(t + τ)
i
=
σ
2
N
e
−α|τ |
in the limit N → ∞ is the Gaussian Markovian random process
ξ(t) = lim
N→∞
ξ
N
(t)
with the exponential correlation function
h
ξ(t)ξ(t + τ )
i
= σ
2
e
−α|τ |
.
Remark Consider the differential equation for the characteristic functional of
random process ξ
N
(t)
8
N
[t;v(τ )] =
*
exp
i
t
Z
0
dτ ξ
N
(τ )v(τ )
+
.
Solution The characteristic functional of process z
k
(t) satisfies Eq. (4.79)
d
dt
8[t;v(τ )] = −
σ
2
N
v(t)
t
Z
0
dt
1
e
−α(t−t
1
)
v(t
1
)8[t
1
;v(τ )],
from which follows that 8[t;v(τ )] → 1 for N → ∞.
For the characteristic functional of random process ξ
N
(t), we have the
expression
8
N
[t;v(τ )] =
*
exp
i
t
Z
0
dτ ξ
N
(τ )v(τ )
+
=
{
8[t;v(τ )]
}
N
.
Random Quantities, Processes, and Fields 121
Consequently, it satisfies the equation
d
dt
ln 8
N
[t;v(τ )] = −σ
2
v(t)
t
Z
0
dt
1
e
−α(t−t
1
)
v(t
1
)
8[t
1
;v(τ )]
8[t;v(τ )]
.
In the limit N → ∞, we obtain the equation
d
dt
ln 8
∞
[t;v(τ )] = −σ
2
v(t)
t
Z
0
dt
1
e
−α(t−t
1
)
v(t
1
),
which means that process ξ(t) is the Gaussian random process.
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Lecture 5
Correlation Splitting
5.1 General Remarks
For simplicity, we content ourselves here with the one-dimensional random processes
(extensions to multidimensional cases are obvious). We need the ability of calculating
correlation
h
z(t)]R[z(τ )]
i
, where R[z(τ )] is the functional that can depend on process
z(t) both explicitly and implicitly.
To calculate this average, we consider auxiliary functional R[z(τ ) + η(τ )], where
η(t) is arbitrary deterministic function, and calculate the correlation
h
z(t)R[z(τ ) + η(τ )]
i
. (5.1)
The correlation of interest will be obtained by setting η(τ ) = 0 in the final result.
We can expand the above auxiliary functional R[z(τ ) + η(τ )] in the functional
Taylor series with respect to z(τ ). The result can be represented in the form
R
[
z(τ ) + η(τ)
]
= exp
∞
Z
−∞
dτ z(τ )
δ
δη(τ )
R
[
η(τ )
]
,
where we introduced the functional shift operator. With this representation, we can
obtain the following expression for correlation (5.1)
h
z(t)R[z(τ ) + η(τ )]
i
=
t;
δ
iδη(τ )
h
R[z(τ ) + η(τ)]
i
, (5.2)
where functional
[t; v(τ )] =
*
z(t) exp
i
∞
Z
−∞
dτ z(τ )v(τ )
+
*
exp
i
∞
Z
−∞
dτ z(τ )v(τ )
+
=
δ
iδv(t)
2[v(τ )]. (5.3)
Lectures on Dynamics of Stochastic Systems. DOI: 10.1016/B978-0-12-384966-3.00005-2
Copyright
c
2011 Elsevier Inc. All rights reserved.
124 Lectures on Dynamics of Stochastic Systems
Here 2[v(τ )] = ln 8[v(τ )] and 8[v(τ )] is the characteristic functional of random
process z(t).
Replacing differentiation with respect to η(τ ) by differentiation with respect to z(τ )
and setting η(τ ) = 0, we obtain the expression
h
z(t)R[z(τ )
i
=
t;
δ
iδz(τ )
R[z(τ )]
. (5.4)
If we expand functional 2[v(τ)] in the functional Taylor series (4.32), page 98,
and differentiate the result with respect to v(t), we obtain
[t; v(τ )] =
∞
X
n=0
i
n
n!
∞
Z
−∞
dt
1
. . .
∞
Z
−∞
dt
n
K
n+1
(t, t
1
, . . . , t
n
)v(t
1
) . . . v(t
n
)
and expression (5.4) will assume the form
z(t
0
)R[z(τ )]
=
∞
X
n=0
1
n!
∞
Z
−∞
dt
1
. . .
∞
Z
−∞
dt
n
K
n+1
(t
0
, t
1
, . . . , t
n
)
δ
n
R[z(τ )]
δz(t
1
) . . . δz(t
n
)
. (5.5)
In physical problems satisfying the condition of dynamical causality in time, statis-
tical characteristics of the solution at instant t depend on the statistical characteristics
of process z(τ ) for 0 ≤ τ ≤ t, which are completely described by the characteristic
functional
8[t; v(τ )] = exp
{
2[t; v(τ )]
}
=
*
exp
i
t
Z
0
dτ z(τ )v(τ )
+
.
In this case, the obtained formulas hold also for calculating statistical averages
z(t
0
)R[t; z(τ )]
for t
0
< t, τ ≤ t, i.e., we have the equality
z(t
0
)R[t; z(τ )]
=
t
0
, t;
δ
iδz(τ )
R[t; z(τ )]
(0 < t
0
< t), (5.6)
where
[t
0
, t; v(τ )] =
δ
iδv(t
0
)]
2[t; v(τ )]
=
∞
X
n=0
i
n
n!
t
Z
0
dt
1
. . .
t
Z
0
dt
n
K
n+1
(t
0
, t
1
, . . . , t
n
)v(t
1
) . . . v(t
n
). (5.7)
Correlation Splitting 125
For t
0
= t − 0, formula (5.6) holds as before, i.e.,
h
z(t)R[t; z(τ )]
i
=
t, t;
δ
iδz(τ )
R[t; z(τ )]
. (5.8)
However, expansion (5.7) not always gives the correct result in the limit t
0
→ t − 0
(which means that the limiting process and the procedure of expansion in the func-
tional Taylor series can be non-commutable). In this case,
[t, t; v(τ )] =
*
z(t) exp
i
t
Z
0
dτ z(τ )v(τ )
+
*
exp
i
t
Z
0
dτ z(τ )v(τ )
+
=
d
iv(t)dt
2[t; v(τ )], (5.9)
and statistical averages in Eqs. (5.6) and (5.8) can be discontinuous at t
0
= t − 0.
Consider several examples of random processes.
5.2 Gaussian Process
In the case of the Gaussian random process z(t), all formulas obtained in the previ-
ous section become significantly simpler. In this case, the logarithm of characteristic
functional 8[v(τ )] is given by Eq. (4.41), page 103, (we assume that the mean value
of process z(t) is zero), and functional 2[t, v(τ )] assumes the form
2[t, v(τ )] = −
1
2
∞
Z
−∞
∞
Z
−∞
dτ
1
dτ
2
B(τ
1
, τ
2
)v(τ
1
)v(τ
2
).
As a consequence, functional [t; v(τ )] (5.3) is the linear functional
[t; v(τ )] = i
∞
Z
−∞
dτ
1
B(t, τ
1
)v(τ
1
), (5.10)
and Eq. (5.2) assumes the form
h
z(t)R[z(τ ) + η(τ )]
i
=
∞
Z
−∞
dτ
1
B(t, τ
1
)
δ
δη(τ
1
)
h
R[z(τ ) + η(τ)]
i
. (5.11)