Download Poisson: telegraph wave, spatial, non-stationary, compound

ECE 6960: Adv. Random Processes & Applications
Lecture Notes, Fall 2010
Lecture 14
Today: Poisson Processes II
• HW Set 6 “due” Thu Oct 21
• Readings posted. Problem section from Leon-Garcia Ch 9
posted.
• Overview of remaining R.P. topics: Poisson processes, Gaussian processes, Markov chains and hidden Markov chains, Bayesian
networks; Applications with queueing problems, finance, games,
networking, tracking, data mining.
• Project presentations due Dec. 7 and 9 (seven weeks away).
0.1
Random Telegraph Signal
Consider a signal which switches between +1 and −1 at each arrival in a Poisson process. This switching part we might write as
(−1)N (t) , where N (t) is a Poisson process. This was originally used
to model the signal sent over telegraph lines. Today it is still useful
in digital communications, and digital control systems.
Figure 1: The telegraph wave process is generated by switching
between +1 and -1 at every arrival of a Poisson process.
Next, let the initial state, X(0), be plus or minus 1, and random.
So,
X(t) = X(0)(−1)N (t)
Where X(0) is -1 with prob. 1/2, and 1 with prob. 1/2, and N (t)
is a Poisson counting process with rate λ, (the number of arrivals in
a Poisson process at time t). X(0) is independent of N (t) for any
time t. See Figure 1.
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ECE 6960-002 Fall 2010
1. What is EX(t) [X(t)]?
i
h
i
h
µX (t) = EX(t) X(0)(−1)N (t) = EX [X(0)] EN (−1)N (t)
i
h
(1)
= 0 · EN (−1)N (t) = 0
2. What is RX (t, δ)? (Assume τ ≥ 0.)
i
h
RX (t, τ ) = EX X(0)(−1)N (t) X(0)(−1)N (t+τ )
i
h
= EX (X(0))2 (−1)N (t)+N (t+τ )
i
h
= EN (−1)N (t)+N (t+τ )
i
h
= EN (−1)N (t)+N (t)+(N (t+τ )−N (t))
i
h
= EN (−1)2N (t) (−1)(N (t+τ )−N (t))
i
h
i
h
= EN (−1)2N (t) EN (−1)(N (t+τ )−N (t))
i
h
= EN (−1)(N (t+τ )−N (t))
Remember the trick you see inbetween lines 3 and 4? N (t)
and N (t + τ ) represent the number of arrivals in overlapping intervals. Thus (−1)N (t) and (−1)N (t+τ ) are NOT independent. But N (t) and N (t + τ ) − N (t) DO represent the
number of arrivals in non-overlapping intervals, so we can proceed to simplify the expected value of the product (in line 6)
to the product of the expected values (in line 7). This difference is just the number of arrivals in a period τ , call it
K = N (t + τ ) − N (t), and it must have a Poisson pmf
with
parameter λ and time τ . Thus the expression is EK (−1)K
is given by
RX (t, τ ) =
∞
X
(−1)k
k=0
−λτ
= e
(λτ )k −λτ
e
k!
∞
X
(−λτ )k
k=0
k!
= e−λτ e−λτ = e−2λτ
(2)
If τ < 0, we would have had e2λτ . Thus
RX (t, τ ) = e−2λ|τ | = RX (τ )
It is WSS. Note RX (0) ≥ 0, that it is also symmetric and
decreasing as it goes away from 0. What is the power in this
R.P.? (Answer: Avg power = RX (0) = 1.) Does that make
sense?
You could also have considered RX (t, τ ) to be a question of,
whether or not X(t) and X(t + τ ) have the same sign – if so,
their product will be one, if not, their product will be zero.
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ECE 6960-002 Fall 2010
0.2
Spatial Poisson Processes
(From Ross, Exercise 94) A two-dimensional Poisson process is a
process of randomly occurring events in the plane such that
1. For any region A with area |A|, the number of events in that
region has a Poisson distribution with mean λ|A|,
2. The number of events in non-overlapping regions are independent.
1
0.5
0
0
0.5
1
Figure 2: A realization of a spatial Poisson process.
Consider an arbitrary point on the plane. What is the distribution of the distance to the nearest event? This is analogous to the
inter-arrival time in a (one-dimensional) Poisson process.
What is P [X > x]? Denoting the circle with radius x as set A,
P [X > x] = P [N (A) = 0] = e−λ|A|
(λ|A|)0
2
= e−λπx .
0!
What is the pdf of X?
fX (x) =
∂
∂x (1
2
2
− e−λπx ) = 2λπxe−λπx .
1
This is the Rayleigh pdf with parameter α = √2λπ
. Knowing the
mean and variance of a Rayleigh r.v., we find that
p
1
EX [X] = α π/2 = √
2 λ
π 2 4 − π
α =
Var [X] =
2−
2
4πλ
We can extend other results from Poisson processes to the 2-D
Poisson process. For example, given that there is one event in a
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ECE 6960-002 Fall 2010
particular region A, what is the distribution of the event within the
region?
For any region B ⊂ A,
P [N (B) = 1|N (A) = 1] =
=
P [N (B) = 1 ∩ N (A) = 1]
P [N (A) = 1]
P [N (B) = 1] P N (A ∩ B C ) = 0
P [N (A) = 1]
1
=
−λ|A∩B
e−λ|B| (λ|B|)
1! e
e−λ|A| (λ|A|)
C|
=
|B|
|A|
As we expect, the probability is uniform in the region A.
Note we can easily extend the same spatial Poisson process to
3-D or higher dimensions.
Applications:
• Ecology & Weather : plant distribution, fires, lightning and
rainfall,
• Health / Epidemiology: disease and environmental risk factors,
• Sensor Networks: locations of deployed sensors, events to be
detected, RF obstructions.
0.3
Non-homogeneous Poisson processes
One key assumption is that arrivals have a constant average rate
λ for all time. This we can imagine is not really true for many
processes. For example, arrivals in the queue at a bank will peak
at the lunch hour. Or, the packet arrival rate at a router is low at
4 am compared to 4 pm. If we consider the case that the average
arrival rate is also a function of time, we call it a non-homogeneous
Poisson process with average arrival rate λ(t).
The sum of two independent non-stationary Poisson processes
remains a non-stationarygeneous Poisson process. Consider:
• {N1 (t), t ≥ 0}, a non-stationary Poisson process with rate
λ1 (t).
• {N2 (t), t ≥ 0}, a non-stationary Poisson process with rate
λ2 (t).
• N (t) = N1 (t) + N2 (t).
You can prove that
• N (t) is a non-stationary Poisson process with rate λ(t) =
λ1 (t) + λ2 (t).
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ECE 6960-002 Fall 2010
• Given that an event occurs in the N (t) process, it resulted
from process 1 with probability λ1 (t)/(λ1 (t) + λ2 (t)).
Consider a non-homogeneous Poisson process N1 (t) with rate
λ(t) that is bounded (that is, there is a maximum rate λ such that
λ(t) < λ for all time t. We can consider it a special case of a
Poisson process with rate λ which we, upon each arrival, decide
whether it came from non-homogeneous R.P. N1 (t) or N2 (t) (with
rate λ−λ(t)). We “count” an arrival as part of N1 (t) with probability λ(t)/λ. Thus we can seeRthat N1 (t) is still a non-homogeneous
t
Poisson process, with mean 0 λ(y)dy.
Def ’n: Mean Value Function
The mean value function m(t) = RE [N ([0, t])] of the non-stationary
t
Poisson process N (t) is given by 0 λ(τ )dτ .
The mean number of arrivals in any interval (s, t], s < t, is
Z t
λ(τ )dτ.
E [N (t) − N (s)] = m(t) − m(s) =
s
0.4
Compound Poisson Processes
In a typical Poisson process N (t), each arrival counts equally. We
assign it a value ‘1’, but as long as each event has a known constant
value, then N (t) indicates the total value of the arrivals.
Consider the process which events or arrivals in the Poisson
process are not all equivalent, and have some variable ‘value’.
Def ’n: Compound Poisson Process
A random process {X(t), t ≥ 0} is a compound Poisson process if
it can be represented as
N (t)
X(t) =
X
Yi
i=1
where {N (t), t ≥ 0} is a Poisson process and {Yi , i = 1, 2, . . .} is a
family of i.i.d. r.v.s that is also independent of {N (t), t ≥ 0}.
Let E [Y ] and E Y 2 denote the mean and second moment of
the distribution of the {Yi }. Then


N (t)
X
E [X(t)] = EN (t) EX(t) [X(t)|N (t)] = EN (t) 
EYi [Yi ]
i=1
E [X(t)] = EN (t) [N (t)E [Y ]] = E [Y ] EN (t) [N (t)] = E [Y ] λt
To find the variance, we need the conditional variance formula,
which states that for two r.v.s X and N ,
Var [X] = EN [VarX [X|N ]] + VarN [EX [X|N ]]
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ECE 6960-002 Fall 2010
One may prove the conditional variance formula using the conditional mean formula E [X] = EN [EX [X|N ]] (good exam problem?). Here, it implies that
Var [X(t)] = EN (t) VarX(t) [X(t)|N (t) = n] + VarN (t) EX(t) [X(t)|N (t) = n]
h
i
= EN (t) n[E Y 2 − E [Y ]2 ] + VarN (t) [nE [Y ]]
i
h = λt E Y 2 − E [Y ]2 + λtE [Y ]2 = λtE Y 2
Example: Ross book, Exercise 68
Suppose that electric shocks having random amplitudes occur at
times distributed according to a Poisson process {N (t), t ≥ 0}, with
rate λ. Suppose that the initial amplitudes Ai , i = 1, 2, . . ., of
successive shocks are i.i.d and with CDF FA (a) with mean µ, and
that {Ai } are independent of the arrival times. Suppose that the
amplitude of a shock decreases over time with an exponential rate α,
meaning that a shock of initial amplitude A will have value Ae−αx
after an additional time x has passed. If we let A(t) be the sum of
all amplitudes at time t, then
N (t)
A(t) =
X
Ai e−α(t−Si )
i=1
where Si is the arrival time of shock i. Find E [A(t)] by conditioning
on N (t).
Solution: Using conditional expectation,
(3)
E [A(t)] = EN (t) EA(t),Si [A(t)|N (t) = n]
Note that the inner expectation takes an expectation w.r.t. both
{Ai } and {Si }. But given N (t)n arrivals, the {Si } are uniform on
[0, t]. Thus
#
" n
X
−α(t−Si )
Ai e
EA(t) [A(t)|N (t) = n] = EA(t)
i=1
=
n
X
i=1
eαSi
EAi [Ai ] e−αt ESi eαSi
but E
is the MGF of Si which is uniform on [0, t]. So E eαSi =
eαt −1
αt , and
EA(t) [A(t)|N (t) = n] =
n
X
i=1
µe−αt
eαt − 1
1 − e−αt
= nµ
αt
αt
Finally, plugging back into (3) with n = N (t),
E [A(t)] = EN (t) [N (t)] µ
1 − e−αt
λµ
=
[1 − e−αt ]
αt
α
(4)