Stat 330 (Spring 2015): slide set 18
2
5. Exponential race: P (min(S, T ) > t) = P (S > t, T > t) = e−(λ+μ)t if
T, S independent. What about P (min(T1, · · · , Tn) > t)?
(this is key for Poisson process later)
P (T > t + s|T > t) = P (T > s)
4. Lack of memory property for Exponential:
3. pdf of Exponential distribution: fT (t) = λe−λt for t ≥ 0, and λ is the
rate, 1/time. What is E(T ), and Var(T )? What is E(X) and Var(X))
2. Poisson: P (X = k) = e−λλx/x! where X is the number of observations
of rare event during certain time period (or space).
1. Exponential: P (T ≤ t) = 1 − e−λt for all t ≥ 0 where T is waiting time
for rare event to happen (once).
Review: What is Exponential distribution? and Poisson distribution?
Poisson Process
Last update: February 16, 2015
Stat 330 (Spring 2015)
Slide set 18
Stat 330 (Spring 2015): slide set 18
Poisson process
3
Jargon: X(t) is a “counting process” with independent Poisson increments.
X(t2) − X(t1) is independent from X(t4) − X(t3)
3. non-overlapping intervals are independent for any 0 ≤ t1 < t2 ≤ t3 < t4
X(t2) − X(t1) ∼ P oλ(t2−t1)
2. distribution depends only on length of interval for any 0 ≤ t1 < t2:
1. for t > 0, X(t) takes values in {0, 1, 2, 3, . . .}.
Definition: A stochastic process X(t) is called homogenous Poisson process
with rate λ, if
Stat 330 (Spring 2015): slide set 18
1
The example about ’hits on a webpage’ is a typical example of stochastic
process, and it has a special name: Poisson Process.
3. Values of X(t) are called states, the set of all possible values for X(t) is
called the state space.
2. Modeling usually requires somehow specifying the joint distribution
(X(t1), · · · , X(tk )) or P (X1 ∈ A1, · · · , Xk ∈ Ak )
1. Stochastic process is a mathematical model of reality.
Some remarks:
Definition: A stochastic process is a set of random variables indexed by
some indices, particularly time t, and is usually denoted by X(t).
Review: What is a Random variable?
Stochastic Processes
Stat 330 (Spring 2015): slide set 18
6
The time until the k th hit Ok is therefore given as the sum of inter-arrival
times Ok = I1 + . . . + Ik .
• What is a nonhomogeneous process?
7
• We mention homogeneous Poisson process; What is meant by
homogeneous?
• Why Poisson so important?!
Note: This theorem is very important! - it links the Poisson, Exponential,
and Erlang distributions tightly together! Some thoughts:
The time until the kth hit Ok is an Erlangk,λ distributed variable, ⇐⇒
X(t) is a Poisson process with rate λ.
and the inter-arrival time between successive hits:
Ij = Oj − Oj−1 for j = 1, 2, . . .
Corollary:
O0 = 0, Oj = time of the j thoccurrence = the first t for which X(t) ≥ j
X(t) is a Poisson process with rate λ iff the inter-arrival times I1, I2, . . .
are i.i.d. Expλ.
For a given Poisson process X(t) we define occurrences
Stat 330 (Spring 2015): slide set 18
Stat 330 (Spring 2015): slide set 18
Equivalence theorem:
5
Equivalence theorem
X(t) = X(t) − X(0) ∼ P oλ(t−0)
4. The distribution of X(t) is Poisson with rate λt, since:
3. With the same argument, X(0) = 0 - ALWAYS!
2. Similarly, X(t2) − X(t1) is the number of occurrences in the interval
(t1, t2].
Based on the last example:
Example (Cont’d)
Stat 330 (Spring 2015): slide set 18
1. X(t) can be thought of as the number of occurrences until time t.
Remarks
Example (cont’d)
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♦ For example, we find that X(t) = 3 for t ∈ [5, 8] minutes; i.e., only 3
hits upto any time within 5 to 8 minutes.
♥ Here arrival times are generated from Exp(2). X(t) counts numbers of
hits until time t min.
♣ A counter of the number of hits on our webpage is an example for a
Poisson process with rate λ = 2/min.
Example
Stat 330 (Spring 2015): slide set 18
+ 10 · e
−10
+ 10 /2e
2
−10
= 0.0028.
= 0.1889.
8
Stat 330 (Spring 2015): slide set 18
)=e
−5/3
P (X > 650)
Then:
Z≤
650 − 600
√
600
N (0, 1).
≈
10
= 1 − 0.9798 = 0.0202. Webpage table, p.789 or Baron p. 386
≈ 1 − Φ(2.05)
approx
X−600
√
∼
600
= 1 − P (X ≤ 650) = 1 − P
Then X ∼ N (600, 600) → Z :=
approx
Recall that a Poisson distribution with large rate λ can be approximated
by a normal distribution with mean μ = λ and variance σ 2 = λ.
5. The number of hits in the first hour is Poisson with mean 600.
You would like to know the probability of more than 650 hits. Exact
calculation isn’t really feasible. So approximate this probability and justify
your approximation.
P (Y ≥ 10) = 1 − P (Y ≤ 10) = 1 − (1 − e
−10·1/6
Let Y be the time until the first hit - then Y has an Exponential
distribution with parameter λ = 10 per minute or λ = 1/6 per second.
2. Evaluate the probability that the time till the first hit exceeds 10 seconds.
(You may also check the Poisson cdf table).
P (X ≤ 2) = P o10(2) = e
−10
Let X be the number of hits in the first minute, then X is a Poisson
variable with λ = 10:
1. Evaluate the probability of 2 or less hits in the first minute.
Hits on a website: Hits on a popular Web page occur according to a Poisson
Process with a rate of 10 hits/min. One begins observation at exactly noon.
Example
Stat 330 (Spring 2015): slide set 18
k
4
=
= 0.4 minutes
λ 10
k
4
=
= 0.04minutes2.
λ2 100
where X ∼ P oi(λ · t)
9
11
♠ This tells us a way to simulate a Poisson process with rate λ on the
interval (0, T ).
♣ In other word, given that there were k arrivals, the set of arrival times is
the same as the locations of k darts thrown at random on the interval [0, t].
Theorem: Let X(t) be a Poisson process. Given that X(T ) = k, the
conditional distribution of the time of the k occurrences O1, . . . , Ok is the
same as the distribution of k ordered independent standard uniform variables
U(1), U(2), . . . , U(k).
Poisson process possesses an interesting property that is consistent with
thinking of it as ”random occurrences” in time t, which leads to the
conditioning theorem
Poisson Process: Conditioning
Stat 330 (Spring 2015): slide set 18
= P o4(3) = 0.433 Website table,p.786 or Baron p.384
= P (X ≤ 3) where X ∼ P oi(10 · 0.4)
P (T > 0.4) = P (X < 4)
4. Evaluate the probability that the time till the 4th hit exceeds 24 seconds.
Need P (T > 24/60) where T ∼ Erlang(4, 10) and T is in minutes; so
we’ll use the Gamma-Poisson formula:
V ar[T ] =
E[T ] =
Let T be the time till the 4th hit. Then T has an Erlang distribution with
stage parameter k = 4 and λ = 10 per minute.
3. Evaluate the mean and the variance of the time till the 4th hit.
Stat 330 (Spring 2015): slide set 18
second, generate w many standard uniform values u1, . . . , uw
define oi = T ·u(i), where u(i) is the ith smallest value among u1, . . . , uw .
•
•
12
♣ There, we have departures as well and, related to that, the time each
surfer stays - which we will call service time (from the perspective of the
web server).
♦ So far, we are looking only at arrivals of events. Besides that, we could,
for example, look at the number of surfers that are on our web site at the
same time.
♥ The above theorem tells us, that, if we pick k values at random from
an interval (0, t) and order them, we can assume that the distance between
two successive values has an exponential distribution with rate λ = k/t.
first, draw a Poisson value w from P oλT . ( This tells us, how many
uniform values Ui we need to simulate )
•
Simulating a Poisson Process