B+D Process: Examples
Stat 330 (Spring 2015): slide set 19
Figure 1: State diagram of B+D process
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♠ This process sometimes is referred as Random walk, used more in material
science, physics, chemistry and biology.
♥ Conditional on X(t) = k, we either move to state k + 1 or to k − 1,
depending on whether a birth or a death occurs first.
♥ One can visualize the set-up for a B+D process in a state diagram as
movements between consecutive states.
Visualization:
Last update: February 16, 2015
Stat 330 (Spring 2015)
Slide set 19
Stat 330 (Spring 2015): slide set 19
1
11
4.46
job i
arrival time
12
4.66
2
0.40
13
4.68
3
0.78
1
0.22
11
5.31
job i
finishing time
job i
finishing time
12
5.54
2
0.63
13
5.59
3
1.61
2. The printer finishes jobs at:
1
0.10
job i
arrival time
14
5.62
4
1.71
14
4.89
4
1.06
15
5.84
5
1.76
15
5.01
5
1.36
16
6.04
6
1.90
16
5.56
6
1.84
17
6.83
7
2.32
17
5.56
7
1.87
1. Jobs arrive at the following points in time (in h):
18
7.10
8
2.68
18
5.85
8
2.04
19
7.23
9
3.42
19
6.32
9
3.10
3
20
7.39
10
4.67
20
6.99
10
4.42
Stat Printer: The ”heavy-duty” printer in the Stats department gets 3 jobs
per hour. On average, it takes 15 min to complete printing. The printer
queue is monitored for a day (8h total time).
B+D Process: Examples (Cont’d)
Stat 330 (Spring 2015): slide set 19
2. X(t) is still called the state at time t. X(t) is in {0, 1, 2, . . .} for all t.
1. X(t) = k, implies that at time t there are k people/objects in the
system.
Remarks:
Definition:A B+D process X(t) is a stochastic process that monitors the
number of people in a system.
More motivation: The concept of memoryless property is further formalized
by the Markov property.
Motivation: Birth and Death process (B + D) is a generalization of Poisson
process, and it provides for modeling of queues, i.e. we assume that arrivals
stay some time in the system and leave after that.
Review: What is a stochastic process? What is a Poisson Process?
Birth and Death Processes
Stat 330 (Spring 2015): slide set 19
Modeling B & D Processes
state k − 1
state k + 1
is
λk
μ k + λk
μk
is
.
μ k + λk
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♣ This implies, that, given the process is in state k, the probability of
moving to
B < D the move is to state k + 1 at time t + B
2. if
B > D the move is to state k − 1 at time t + D
3. B and D are independent for each state k.
Modeling (Cont’d)
this is the
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♣ Y itself is, again, an exponential variable, its rate is the the sum of the
rates of B and D.
♦ We talked about this discussing the Poisson process:
exponential races!
= 1 − e(λk +μk )y = Expλk +μk (y),
= 1 − e−λk y + 1 − e−μk y − (1 − e−λk y )(1 − e−μk y ) =
= P (B ≤ y) + P (D ≤ y) − P (B ≤ y) · P (D ≤ y)
= P (B ≤ y) + P (D ≤ y) − P (B ≤ y ∩ D ≤ y)
P (Y ≤ y) = P (min(B, D) ≤ y) = P (B ≤ y ∪ D ≤ y)
♦ What can we say about the distribution of Y := min(B, D)?
♣ Notice that Y = min(B, D) is the time the system will be in state k
until the move.
Stat 330 (Spring 2015): slide set 19
Stat 330 (Spring 2015): slide set 19
= 5) = (5.31 − 5.01) + (5.59 − 5.56) = 0.33 = 0.04125.
P (X(t)
8
8
The empirical probability for 5 jobs in the printer is the time, X(t) is in
state 5 divided by the total time:
5
1. The model for a birth or a death is given, conditional on X(t) = k, as:
B = time till a potential birth ∼ Exp(λk )
N.B. (P (B = D) = 0!)
D = time till a potential death ∼ Exp(μk )
Model:
Stat 330 (Spring 2015): slide set 19
2. What is the (empirical) probability that there are 5 jobs in the printer
and its queue at some time t?
Stat Printer (Cont’d)
4
1. Can you draw a graph of X(t) for the value monitored?
♣ Let X(t) be the number of jobs in the printer and its queue at time t.
X(t) is a Birth & Death process.
Stat Printer (Cont’d)
Stat 330 (Spring 2015): slide set 19
k pk = 1.
♥ How to compute pk ?
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♣ The pk probabilities are called the steady state probabilities of the B+D
process, and they form a probability mass function for X.
where the pk are numbers between 0 and 1, with
t→∞
k
total time t
state k by time t
→
→
use (‡)
pk (λk + μk )
pk
pk
· (λk + μk ) = λk pk
Similarly, μk pk is the long run rate of transitions from k → k − 1.
is the long run rate of transitions from k → k + 1.
λk
λk +μk pk
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The long rate of visits to state k is pk (λk + μk ); but we know that a fraction
k
visits to state k results in moves to state k + 1. So
of λ λ+μ
k
in state k
total time t
# of visits to state k by time t
time in state k until time t
→
total time t
# of visits to mean stay
Stat 330 (Spring 2015): slide set 19
Stat 330 (Spring 2015): slide set 19
Sketched arguments:
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♦ Mathematically, the notion of a steady state translates to
lim P (X(t) = k) = pk for all k,
Stat 330 (Spring 2015): slide set 19
♦ N.E. In the picture, three different simulations of Birth & Death processes
are shown. Only in the first case, the process is stable (birth rate < death
rate). The other two processes are unstable (birth rate = death rate (2nd
process) and birth rate > death rate (3rd process)).
p
Examples:
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Equilibrium: Only if the B+D process is stable, it will reach an equilibrium
after some time - this is called the steady state of the B+D process.
More on stability:
• A lot depends on the ratio of births and deaths:
• The analysis of this model for small t is mathematically difficult because
of “start-up” effects - but in some cases, we can compute the “large t”
behaviour.
• N.E. A Poisson process with rate λ is a special case of a Birth & Death
process, where the birth rates and death rates are constant, λk = λ and
μk = 0 for all k.
• ‡ The mean staying time therefore is 1/(λk + μk ).
• The mean staying time in state k is the expected value of an exponential
distribution with rate λk + μk .
• Knowing the distribution of Y , the staying time in state k, gives us, e.g.
the possibility to compute the mean staying time in state k.
Remarks: