Unformatted text preview: EE420/500 Class Notes 07/08/09 John Stensby Chapter 6  Random Processes
Recall that a random variable X is a mapping between the sample space S and the
extended real line R+. That is, X : S → R+.
A random process (a.k.a stochastic process) is a mapping from the sample space into an
ensemble of time functions (known as sample functions). To every ρ ∈ S, there corresponds a
function of time (a sample function) X(t;ρ). This is illustrated by Figure 61. Often, from the
notation, we drop the ρ variable, and write just X(t). However, the sample space ρ variable is
always there, even if it is not shown explicitly.
For a fixed t = t0, the quantity X(t0;ρ) is a random variable mapping sample space S into
the real line. For fixed ρ0 ∈ S, the quantity X(t;ρ0) is a welldefined, nonrandom, function of
time. Finally, for fixed t0 and ρ0, the quantity X(t0;ρ0) is a real number.
Example 61: X maps Heads and Tails
Consider the coin tossing experiment where S = {H, T}. Define the random function X(t;Heads) = sin(t)
X(t;Tails) = cos(t) X(t;ρ1)
X(t;ρ2)
time X(t;ρ3)
X(t;ρ4)
Figure 61: Sample functions of a random process. Updates at http://www.ece.uah.edu/courses/ee420500/ 61 EE420/500 Class Notes 07/08/09 John Stensby Continuous and Discrete Random Processes
For a continuous random process, probabilistic variable ρ takes on a continuum of
values. For every fixed value t = t0 of time, X(t0;ρ) is a continuous random variable.
Example 62: Let random variable A be uniform in [0, 1]. Define the continuous random
process X(t;ρ) = A(ρ)s(t), where s(t) is a unitamplitude, Tperiodic square wave. Notice that
sample functions contain periodicallyspaced (in time) jump discontinuities. However, the process is continuous.
For a discrete random process, probabilistic variable ρ takes on only discrete values. For
every fixed value t = t0 of time, X(t0;ρ) is a discrete random variable.
Example 63: Consider the coin tossing experiment with S = {H, T}. Then X(t;H) = sin(t),
X(t;T) = cos(t) defines a discrete random process. Notice that the sample functions are continuous functions of time. However, the process is discrete.
Distribution and Density Functions
The firstorder distribution function is defined as F(x,t) = P[X(t) ≤ x]. (61) The firstorder density function is defined as f ( x; t ) ≡ dF(x, t)
.
dx (62) These definitions generalize to the nthorder case. For any given positive integer n, let x1,
x2, ... , xn denote n “realization” variables, and let t1, t2, ... , tn denote n time variables. Then,
define the nthorder distribution function as F(x1, x2, ... , xn; t1, t2, ... , tn ) = P[X(t1) ≤ x1, X(t2) ≤ x2, ... , X(tn) ≤ xn]. Updates at http://www.ece.uah.edu/courses/ee420500/ (63) 62 EE420/500 Class Notes 07/08/09 John Stensby Similarly, define the nthorder density function as f (x1, x2 , ... , x n ; t1, t 2 , ... , t n ) = ∂ n F(x1, x 2 , ... , x n ; t1, t 2 , ... , t n )
∂x1∂x 2 ... ∂x n (64) In general, a complete statistical description of a random process requires knowledge of all
order distribution functions.
Stationary Random Process A process X(t) is said to be stationary if its statistical properties do not change with
time. More precisely, process X(t) is stationary if F(x1, x2, ... , xn; t1, t2, ... , tn) = F(x1, x2, ... , xn; t1+c, t2+c, ... , tn+c) (65) for all orders n and all time shifts c.
Stationarity influences the form of the first and secondorder distribution/density
functions. Let X(t) be stationary, so that F(x; t) = F(x; t+c) (66) for all c. This implies that the firstorder distribution function is independent of time. A similar
statement can be made concerning the firstorder density function. Now, consider the secondorder distribution of stationary X(t); for all t1, t2 and c, this function has the property
F(x1, x 2 ; t1, t 2 ) = F(x1, x 2 ; t1 + c, t 2 + c) = F(x1, x 2 ; t1 + c, {t 2 + c}{t1 + c}+{t1 + c})
τ = t 2 − t1 (67) = F(x1, x 2 ; t1 + c, {t1 + c}+ τ). Updates at http://www.ece.uah.edu/courses/ee420500/ 63 EE420/500 Class Notes 07/08/09 John Stensby This must be true for all t1, t2 and c. Hence, F(x1,x2;t1,t2) depends on the time difference τ ≡ t2 –
t1; the secondorder distribution does not depend on absolute t1 and t2. In F(x1,x2;t1,t2), you will
only see t1 and t2 appear together as t2 – t1, which we define as τ. Often, for stationary processes,
we change the notation and define
F(x1, x 2 ; τ) ≡ F(x1, x 2 ; t1, t1 +τ) .
" new " notation (68) " old " notation Similar statements can be made concerning the secondorder density function.
These conditions on firstorder F(x) and secondorder F(x1, x2; τ) are Be careful! necessary conditions; they are not sufficient to imply stationarity. For a given random process,
suppose that the first order distribution/density is independent of time and the secondorder
distribution/density depends only on the time difference. Based on this knowledge alone, we
cannot conclude that X(t) is stationary.
First and SecondOrder Probabilistic Averages First and secondorder statistical averages are useful. The expected value of general
random process X(t) is defined as η(t) = E[X(t)] = z ∞ ∞ x f(x; t) dx . (69) In general, this is a timevarying quantity. The expected value is often called a firstorder
statistic since it depends on a firstorder density function. The autocorrelation function of X(t) is defined as R ( t1, t 2 ) = E[X( t1 )X( t 2 )] = zz
∞ ∞ x x f (x1, x 2 ; t1, t 2 ) dx1 dx2 .
−∞ −∞ 1 2 (610) In general, R depends on two time variables, t1 and t2. Also, R is an example of a secondorder Updates at http://www.ece.uah.edu/courses/ee420500/ 64 EE420/500 Class Notes 07/08/09 John Stensby statistic since it depends on a secondorder density function. Suppose X(t) is stationary. Then the mean z η = E[X(t)] = ∞ ∞ (611) x f(x) dx is constant, and the autocorrelation function R ( τ ) = E[X( t )X( t + τ )] = zz
∞ ∞ x x f (x1, x 2 ; τ ) dx1 dx 2
−∞ −∞ 1 2 (612) depends only on the time difference τ = t2 – t1 (it does not depend on absolute time). However,
the converse is not true: the conditions η a constant and R(τ) independent of absolute time do not
imply that X(t) is stationary.
Wide Sense Stationarity (WSS) Process X(t) is said to be widesense stationary (WSS) if
1) Mean η = E[X(t)] is constant, and
2) Autocorrelation R(τ) = E[X(t)X(t+τ)] depends only on the time difference.
Note that stationarity implies widesense stationarity. However, the converse is not true: WSS
does not imply stationarity.
Ergodic Processes A process is said to be Ergodic if all orders of statistical and time averages are
interchangeable. The mean, autocorrelation and other statistics can be computed by using any
sample function of the process. That is η = E[X( t )] = z ∞ −∞ R ( τ ) = E[X( t )X( t + τ )] = z 1T
X( t ) dt
T→∞ 2T − T x f ( x )dx = limit zz
∞ ∞ −∞ −∞ 1T
X( t )X( t + τ ) dt .
T→∞ 2T − T x1x2 f ( x1, x2 ; τ )dx1dx2 = limit Updates at http://www.ece.uah.edu/courses/ee420500/ z (613) 65 EE420/500 Class Notes 07/08/09 All
Processes John Stensby Wide Sense Stationary
Stationary
Ergodic Figure 62: Hierarchy of random processes. This idea extends to higherorder averages as well. Since we are averaging over absolute time,
the ensemble averages (all orders) cannot depend on absolute time. This requires that the
original process must be stationary. That is, ergodicity implies stationarity. However, the
converse is not true: there are stationary processes that are not ergodic. The hierarchy of random
processes is abstractly illustrated by Figure 62.
Example 64: Let X(t) = A, where A is uniformly distributed in the interval [0, 1]. Sample functions of X are straight lines, as shown by Figure 63. Clearly, X(t) is not ergodic since the
time average of each sample function is different.
Example 65: Random Walk The random walk is the quintessential example of a Markov process, a type of process
that has many applications in engineering and the physical sciences. Many versions of the 1 X(t;ρ1)
X(t;ρ2)
X(t;ρ3)
time Three sample functions of X(t) Figure 63: Sample functions of a nonergodic random process.
Updates at http://www.ece.uah.edu/courses/ee420500/ 66 EE420/500 Class Notes 07/08/09 John Stensby Xd(N) N Figure 64: Two sample functions of a random walk process. random walk have been studied over the years (i.e., the gambler's ruin, drunken sailor, etc.).
At first, a discrete random walk is introduced. Then, it is shown that a limiting form of the
random walk is the wellknown continuous Wiener process. Finally, simple equations are
developed that provide a complete statistical description of the discrete and limiting form of the
random walk.
Suppose a man takes a random walk by starting at a designated origin on a straight line
path. With probability p (alternatively, q ≡ 1  p), he takes a step to the right (alternatively, left).
Suppose that each step is of length meters, and each step is completed in τ s seconds. After N steps (completed in Nτ s seconds), the man is located Xd(N) steps from the origin; note that − N ≤
Xd(N) ≤ N since the man starts at the origin. If Xd(N) is positive (alternatively, negative), the
man is located to the right (alternatively, left) of the origin. The quantity P[Xd(N) = n], − N ≤ n
≤ N, denotes the probability that the man's location is n steps from the origin after he has taken N
steps. Figure 64 depicts two sample functions of Xd(N), a discrete random process.
The calculation of P[Xd(N) = n] is simplified greatly by the assumption, implied in the
previous paragraph, that the man takes independent steps. That is, the direction taken at the Nth Updates at http://www.ece.uah.edu/courses/ee420500/ 67 EE420/500 Class Notes 07/08/09 John Stensby step is independent of Xd(k), 0 ≤ k ≤ N − 1, and the directions taken at all previous steps. Also
simplifying the development is the assumption that p does not depend on step index N. Under
these conditions, it is possible to write
P[X d (N + 1) = X d (N) + 1] = p (614) P[X d (N + 1) = X d (N) − 1] = 1 − p = q . Let Rn0 and Ln0 denote the number of steps to the right and left, respectively, that will
place the man n, −N ≤ n ≤ N, steps from the origin after he has completed a total of N steps.
Integers Rn0 and Ln0 depend on integers N and n; the relationship is given by
R n 0 − Ln 0 = n (615) R n 0 + Ln 0 = N since − N ≤ n ≤ N. Integer values for Rn0 and Ln0 exist only if −N ≤ n ≤ N and the pair of
integers (N + n), (N  n) are even. When an integer solution of (615) exists it is given by Rn 0 =
Ln 0 = N+n
2 , (616) N−n
2 for N ≤ n ≤ N, and (N+n), (Nn) even integers. After taking a total of N steps, it is not possible
to reach n steps (to the right of the origin) if integer values for Rn0 and Ln0 do not exist. Also, if
it is possible to reach n steps (to the right of the origin after taking a total of N steps), then it is
not possible to reach n ± 1 steps (to the right of the origin).
Of course, there are multiple sequences of N steps, Rn0 to the right and Ln0 to the left, that
the man can take to insure that he is n steps to the right of the origin. In fact, the number of such Updates at http://www.ece.uah.edu/courses/ee420500/ 68 EE420/500 Class Notes 07/08/09 John Stensby sequences is given by F N I N!
GG JJ = R ! L ! .
H R n 0K n 0 n 0 (617) This quantity represents the number of subsets of size Rn0 that can be formed from N distinct
objects. These sequences are mutually exclusive events. Furthermore, they are equally probable, and the probability of each of them is P [R n0 (Ln0 ) Steps To Right (Left) in a Specific Sequence] = p R n0 q Ln0 . (618) The desired probability P[Xd(N) = n] can be computed easily with the use of (616),
(617) and (618). From the theory of independent Bernoulli trials, the result
N!
⎧
R n0 L n0
⎪ R ! L ! p q , if integers R n0 and L n0 exit
⎪
P[X d (N) = n] = ⎨ n 0 n 0
⎪
if integers R n0 and L n0 do not exit
⎪0,
⎩ (619) follows easily. If there are no integer solutions to (615) for given values of n and N (i.e.,
integers Rn0 and Ln0 do not exist), then it is not possible to arrive at n steps from the origin after
taking N steps and P[Xd(N) = n] = 0. Note that (619) is just the probability that the man takes
Rn0 steps to the right given that he takes N independent steps.
The analysis leading to (619) can be generalized to include a nonzero starting location.
Instead of starting at the origin, assume that the man starts his random walk at m steps to the
right of the origin. Then, after the man has completed N independent steps, P[Xd(N) = n⎮Xd(0)
= m] denotes the probability that he is n steps to the right of the origin given that he started m
steps to the right of the origin. A formula for P[Xd(N) = n⎮Xd(0) = m] is developed in what Updates at http://www.ece.uah.edu/courses/ee420500/ 69 EE420/500 Class Notes 07/08/09 John Stensby follows.
Let ν ≡ n − m. The quantity ν denotes the man's net increase in the number of steps to
the right after he has completed N steps. Also, Rnm (alternatively, Lnm) denotes the number of
steps to the right (alternatively, left) that are required if the man starts and finishes m and n,
respectively, steps to the right of the origin. Note that Rnm + Lnm = N and Rnm  Lnm = ν so that R nm =
L nm = N +ν
2 (620) N −ν
.
2 Solution (620) is valid only if ⎮ν⎮ ≤ N and integers (N + ν), (N − ν ) are even. Otherwise,
integers Rnm and Lnm do not exist, and it is not possible to start at m (steps to the right of the
origin), take N independent steps, and find yourself at n (steps to the right of the origin).
Finally, suppose that integers Rnm and Lnm exist for some n and m; that is, it is possible to go
from m to n (steps to the right of the origin) in a total of N steps. Then, it is not possible to go
from m to n ± 1 steps in a total of N steps.
The desired result follows by substituting Rnm and Lnm for Rn0 and Ln0 in (619); this
procedure leads to
P[ X d (N) = n⎮X d (0) = m ] = P[ R nm steps to the right out of N steps ] = N!
pR nm q Lnm
R nm!(N − R nm )! (621) ⎛N⎞
⎟ p R nm q Lnm
=⎜
⎜
⎟
⎝ R nm ⎠ if integers Rnm and Lnm exist, and Updates at http://www.ece.uah.edu/courses/ee420500/ 610 EE420/500 Class Notes 07/08/09 John Stensby Y P[X d (N) = n X d ( 0) = m] = 0 (622) if Rnm and Lnm do not exist.
To simplify the developments in the remainder of this chapter, it is assumed that p = q =
1/2. Also, Rnm is assumed to exist in what follows. Otherwise, results (that use Rnm) given
below can be modified easily if n and m are such that Rnm does not exist.
Process Xd(N) has independent increments. That is, consider integers N1, N2, N3 and N4,
where N1 < N2 ≤ N3 < N4. Then Xd(N2)  Xd(N1) is statistically independent of Xd(N4)  Xd(N3).
The Wiener Process As a Limit of the Random Walk
Recall that each step corresponds to a distance of meters, and each step is completed in
τ s seconds. At time t = Nτ s , let X(Nτ s ) denote the man's physical displacement (in meters) from
the origin. Then X(Nτ s ) is a random process given by X(Nτ s ) ≡ Xd(N), since Xd(N) denotes the
number of steps the man is from the origin after he takes N steps. Note that X(Nτ s ) is a discretetime random process that takes on only discrete values.
For large N and small and τ s , the probabilistic nature of X (Nτ s ) is of interest. First, note that P[ X (Nτ s ) = n⎮X(0) = m ] = P[X d (N) = n⎮Xd(0) = m]; this observation and the
Binomial distribution function leads to the result (use p = q = ½)
P[X (Nτs ) ≤ n ⎮ X (0) = m ] = P[X d (N) ≤ n ⎮ X d (0) = m]
= P[from N steps, the number k taken to right is ≤ R nm ] = (623) R nm ⎛ N ⎞ ⎟( 1 ) k ( 1 ) N − k .
⎟2 2
k=0 ⎝ k ⎠ ∑⎜
⎜ For large N, the DeMoivreLaplace theorem (see Chapter 1 of these class notes) leads to the
approximation Updates at http://www.ece.uah.edu/courses/ee420500/ 611 EE420/500 Class Notes
P[ X (Nτs ) ≤ n Y X ( 0) = 07/08/09
m] ≈ G = John Stensby FG R nm − N/ 2IJ = GFG ν IJ ,
H N/ 4 K H N K z ν/ N 1
2π −∞ (624) exp − 1 u 2 du
2 where G is the distribution function for a zeromean, unitvariance Gaussian random variable.
The discrete random walk process outlined above has a continuous process as a formal
limit. To see this, let → 0, τ s → 0 and N → ∞ in such a manner that
2 2τs →D t = N τs
x=n (625) x0 = m
X (t) = X ( N τs ) , where D is known as the diffusion constant. In terms of D, x, x0 and t, the results of (625) can
be used to write
ν
(x − x 0 ) /
(x − x 0 )
=
=
.
N
t / τs
2 Dt (626) Process X(t) is a continuous random process.
The probabilistic nature of the limiting form of X(t) is seen from (624) and (626). In
the limit, the process X(t) is described by the firstorder conditional distribution function Y z (xx 0 )/ 2 Dt F( x; t x0 ) = 1
2π ∞ exp − 1 u 2 du ,
2 Updates at http://www.ece.uah.edu/courses/ee420500/ (627) 612 EE420/500 Class Notes 07/08/09 John Stensby f(x,t⎮x0) Y f( x, t x 0 ) = LM
NM ( x − x 0 )2
1
exp −
4Dt
4π Dt OP
QP t = .1 second t = 1 second t = 3 second x03 x02 x01 x0 x0+1 x0+2 x0+3 Figure 65: Density function for a diffusion process with D = 1. and the firstorder conditional density function Y f(x, t x0 ) = Y d
F( x; t x0 ) =
dx LM
MN OP
PQ 1
(x − x 0 )2
exp −
,
4 Dt
4π Dt (628) a result depicted by Figure 65. Often, f(x,t⎮x0) is known as the transition density function for
the process since f(x,t⎮x0)Δx is the probability of making the transition from x0 to the interval
(x, x+Δx) by time t. Equation (628) describes the conditional probability density function of a
continuoustime Wiener process. Clearly, process X (t) is Gaussian, it has a mean of x0, and it
has a variance that grows with time (hence, it is nonstationary). Finally, as t → 0+, note that
f(x,t⎮x0) → δ(x – x0), as expected. Updates at http://www.ece.uah.edu/courses/ee420500/ 613 EE420/500 Class Notes 07/08/09 John Stensby X(t)
x0
t Figure 66: A hypothetical sample function of a Wiener process X(t).
Figure 66 attempts to depict a sample function of a Wiener process. While such drawings are “nice” to look at (and they are fun to draw!), they cannot depict accurately all
attributes of a Wiener process sample function. As it turns out, Wiener processes have continuous (in time) sample functions; there won’t be a step or jump in the sample function of a
Wiener process. However, in the traditional Calculus sense, the sample functions are differentiable nowhere. That is, in the classical Calculus sense, the derivative dX/dt does not
exist at any value of time (actually, it’s just a little more complicated than this). A generalized
derivative of the Wiener process does exit, however. In engineering and the physical sciences, it
is known as white Gaussian noise.
Process X (t) has independent increments. Let (t1, t2), (t3, t4) be nonoverlapping intervals (t1 < t2 ≤ t3 < t4). Then increment X (t2)  X (t1) is independent of increment X (t4) X (t3). Finally, increment X(t2)  X(t1) has a mean of zero and a variance of 2D⎮t2  t1⎮. The Diffusion Equation For the Transition Density Function In terms of physical displacement (from the origin) X, the conditional probability
P[ X (Nτ s ) = n⎮X(0) = m ] describes the probabilistic nature of the discrete time random walk problem outlined above. In what follows, this conditional probability is denoted by the short
hand notation P[ n , Nτ s ⎮ m ]. For the case p = q = 1/2, it is easy to see that it satisfies the
difference equation P[ n, ( N + 1)τs Y m] = 1 P[ (n − 1), NτsY m] + 1 P[ (n + 1), NτsY m] .
2
2 Updates at http://www.ece.uah.edu/courses/ee420500/ (629) 614 EE420/500 Class Notes 07/08/09 John Stensby That is, to get to n at time (N+1)τs, you can be at (n1) at time Nτs and take a step to the right
(this occurs with probability equal to 1/2), or you can be at (n+1) at time Nτs and take a step to
the left (this occurs with probability equal to 1/2). Equation (629) can be applied twice to
obtain
P[ n, (N+2)τs ⎮ m] = 1 P[ (n − 1), (N+1)τs ⎮ m] + 1 P[ (n + 1), (N+1)τs ⎮ m]
2 2 = 1 ⎡ 1 P[ (n − 2), Nτs ⎮ m] + 1 P[ n, Nτs ⎮ m]⎤
2 ⎣2
2
⎦ (630) + 1 ⎡ 1 P[ n, Nτs ⎮ m] + 1 P[ (n + 2), Nτs ⎮ m]⎤
2 ⎣2
2
⎦ This last result can be simplified to obtain
P[ n, (N+2)τs ⎮ m] .
= 1 P[
4 (n − 2), Nτs ⎮ m] + 1 P[
2 n, Nτs ⎮ m] + 1 P[
4 (631) (n + 2), Nτs ⎮ m] The continuous conditional density f(x,t ⎮x0) given by (628) satisfies a partial
differential equation. To obtain this equation, first note that the difference equation (631) can
be used to write
P[ n, (N+2)τs⎮ m] − P[ n, Nτs⎮ m]
2τs (632)
2⎡
P[ (n + 2), Nτs⎮ m] − 2P[ n, Nτs⎮ m] + P[ (n − 2), Nτs⎮ m] ⎤
=
⎢
⎥
2τs ⎢
(2 )2
⎥
⎣
⎦ (substitute (631) for the first term on the lefthand side of (632) to verify the expression).
Now, in the sense described by (625), the formal limit of (632) can be found. To find this Updates at http://www.ece.uah.edu/courses/ee420500/ 615 EE420/500 Class Notes 07/08/09 John Stensby limiting form, we must consider two results. First, note that (624) and (625) imply
P[ n, Nτs⎮ m] = P[X (Nτs ) ≤ n⎮ X (0) = m] − P[X (Nτs ) ≤ (n − 2)⎮ X (0) = m]
≈ = = 2
⎡⎛
ν⎞
ν−2⎞
⎞⎤
G⎛
− G⎛
≈ 1 exp ⎢ − 1 ⎜ ν ⎟ ⎥ 2
⎜
⎟
⎜
⎟
⎝ N⎠
⎝ N⎠
2π
⎣ 2⎝ N ⎠ ⎦ N
2⎤
⎡
exp ⎢ − 1 ⎛ ν ⎞ ⎥
⎜
⎟
⎛ 2⎞
⎣ 2⎝ N ⎠ ⎦
4 π⎜
Nτs
⎜ 2τ ⎟
⎟
⎝ s⎠ 2 (633) 2 (x − x 0 ) ⎤
⎡
2
= 2 f (x; t⎮x 0 )
exp ⎢ − 1
⎣ 2 2Dt ⎥
⎦
4πD t This last equation shows that, in the limit describe by (625), P[ n, Nτs
2 f(x;t⎮x0). That is, as
P[ n, Nτs . → 0, τs → 0, 2 Y m] approaches /2τs → D as described by (625), we know that Y m] approaches zero according to P[ n, Nτs⎮ m] → 2 f (x; t⎮x 0 ) → 0 . (634) Second, we must review expressions for derivatives; let g(x) be an ordinary function of x, and
recall that the first partial derivative of g can be expressed as
∂
g(x + Δx) − g(x − Δx)
.
g(x) = limit
∂x
2Δx
Δx → 0 (635) Use this formula to express the second derivative of g as Updates at http://www.ece.uah.edu/courses/ee420500/ 616 EE420/500 Class Notes
∂2 g(x) = limit
∂x 2
Δx → 0
= limit
Δx → 0 07/08/09 John Stensby [g(x + 2Δx) − g(x)] − [g(x) − g(x − 2Δx)]
(2Δx) 2 (636) g(x + 2Δx) − 2g(x) + g(x − 2Δx)
(2Δx) 2 From (634), (635) and (636), the formal limit of (632) is ∂
∂2
f(x, t x0 ) = D 2 f(x, t x0 ) ,
∂t
∂x Y Y (637) where f(x,t⎮x0) denotes the conditional probability density function given by (628). Note that
(637) is identical in form to the sourcefree, onedimensional heat equation. Probability diffuses just like heat and electronic charge (and many other physical phenomenon)!
Equation (637) is a onedimensional diffusion equation. It describes how probability
diffuses (or flows) with time. It implies that probability is conserved in much the same way that
the wellknow continuity equation implies the conservation of electric charge. To draw this
analogy, note that f describes the density of probability (or density of probability particles) on
the onedimensional real line. That is, f can be assigned units of particles/meter. Since D has
units of meters2/second, a unit check on both sides of (637) produces
2 1
meter
1
e second je particles j = e second je meter j e particles j .
meter
meter
2 (638) Now, write (637) as
∂
f = −∇ ⋅ ℑ ,
∂t (639) where Updates at http://www.ece.uah.edu/courses/ee420500/ 617 EE420/500 Class Notes
ℑ≡ −D 07/08/09 John Stensby ∂
f,
∂x (640) and ∇ is the divergence operator. The quantity ℑ is a onedimensional probability current, and
it has units of particles/second. Note the similarity between (639) and the wellknown continuity equation for electrical charge.
Probability current ℑ ( x,t ⎮x0) indicates the rate of particle flow past point x at time t.
Let (x1, x2) denote an interval; integrate (639) over this interval to obtain
∂
∂ x2
P[x1 < X (t) ≤ x 2⎮x 0 ] =
f(x, t⎮x 0 ) dx = − ℑ (x 2 , t⎮x 0 ) + ℑ (x1, t⎮x 0 )] .
∂t
∂ t ∫x1 (641) As illustrated by Figure 67, the lefthand side of this equation represents the time rate of
probability buildup on (x1, x2). That is, between the limits of x1 and x2, the area under f is
changing at a rate equal to the lefthand side of (641). As depicted, the righthand side of (641)
represents the probability currents entering the ends of the interval (x1, x2).
The Wiener process is a simple example of a diffusion process. Diffusion processes are
important in the study of communication and control systems. As it turns out, the state vector
that describes a system (such as a circuit, PLL, springmass, etc.) driven by white Gaussian noise
is a diffusion process. Also, this state vector is described statistically by a density function that f (x, t⎮x0) Y Y ℑ (x1,t x0 ) − ℑ (x2,t x0 )
x1 Y z x2
x2 Y Y Y ∂ P[x < X(t) ≤ x x ] = ∂ f (x, t x )dx = ℑ (x ,t x ) − ℑ (x ,t x )
2 0 ∂t
0
0
0
1
2
∂t 1
x1 Figure 67: Probability buildup on the interval x1, x2 due to
probability current entering the ends of the interval.
Updates at http://www.ece.uah.edu/courses/ee420500/ 618 EE420/500 Class Notes 07/08/09 John Stensby satisfies a partial differential equation known as the FokkerPlanck equation (of which (637) is
a simple example). Finally, it should be pointed out that a diffusion process is a special case of a
Markov process, a more general process (see Papoulis for the definition of a Markov process). Solution of Diffusion Equation by Transform Techniques
The onedimensional diffusion equation (637) can be solved by using transform
techniques. First, initial and boundary conditions must be specified. The desired initial condition is
f ( x, tYx 0 )Y = δ( x − x 0 ) , Yt = 0 (642) which means that random process x starts at x0. What are known as natural boundary conditions
are to be used; that is, we require
f ( x, tYx 0 )Y Yx= ∞ = 0. (643) Consider the transform of f(x,t⎮x0) defined by ϕ(s, t ) = z ∞ −∞ f ( x, tYx 0 )e jxs dx . (644) With respect to t, differentiate (644) to obtain z R
S
T U
V
W z R
S
T U
V
W ∞∂
∞
∂ϕ(s, t )
∂2
jxs
=
f ( x, tYx 0 ) e jxs dx = D
f ( x, tYx 0 ) e dx .
2
−∞ ∂t
−∞
∂t
∂x (645) Now, use Updates at http://www.ece.uah.edu/courses/ee420500/ 619 EE420/500 Class Notes 07/08/09 John Stensby lim it f ( x , tY x 0 ) = 0
x→ ∞ ∂ f ( x , tY x 0 )
lim it
=0
∂x
x→∞ , (646) and integrate by parts twice to obtain
∂ϕ
= − Ds2ϕ .
∂t (647) This equation can be solved easily to obtain
ϕ(s, t ) = exp − Ds2 t ϕ(s,0 ) . (648) However, from the initial condition (642), we have ϕ(s,0) = exp( jx 0s ) . (649) Finally, combine (648) and (649) to obtain
ϕ(s, t ) = exp jx 0s − Ds2 t . (650) But, this is the wellknown characteristic function of the Gaussian density function Y f(x, t x0 ) = LM
NM OP
QP 1
(x − x 0 )2
exp −
.
4 Dt
4π Dt (651) This same technique can be used to solve higherorder, more complicated diffusion equations. Updates at http://www.ece.uah.edu/courses/ee420500/ 620 ...
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 Summer '09
 JohnStensby
 Probability theory, John Stensby

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