edu exponential distribution and poisson processmetin/probability/folios/expopois.pdfΒ Β· which is...
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etinPage 1Exponential Distribution and Poisson Process
Outline Continuous-time Markov Process Poisson Process Thinning Conditioning on the Number of Events Generalizations
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etinPage 2Probability and Stochastic Processes
Overview of Random Variables and Stochastic Processesβ Discrete-time: {ππππ:ππ β₯ 0,ππ β ππ}; Continuous-time: {ππ(π‘π‘):π‘π‘ β₯ 0, π‘π‘ β β}β Discrete state-space if each ππππ or ππ(π‘π‘) has a countable range; Otherwise, Continuous state-space.
Discrete-time Continuous-time
DiscreteState Space
ContinuousState Space
MarkovChains
PoissonProcess
BrownianProcess
A lot of OM ModelsE.g., Inventory Modelsππππ+1 = ππππ + ππππ β π·π·ππ
ππππ ,π·π·ππ β β
DiscreteRandomVariables
ContinuousRandomVariables
Ran
dom
Ve
ctor
sUncountableRange
CountableRange
Probability Stochastic Processes
β Difference between an infinite-length random vector & a discrete-time stochastic process?Β» Similarity, they are both denoted as {ππππ:ππ β₯ 0,ππ β ππ}.Β» Difference is in the treatment of dependence:
Random vectors β Only short lengths (2-3 rvs) can be studied if they are dependent; β Or assume structured dependence through copulas. β Or work with independent and identically distributed sequences.
In discrete-time stochastic processes, assume Markov Property to limit the dependence
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etinPage 3Stochastic Processes in Continuous Time
Discrete-time Continuous-time
DiscreteState Space
ContinuousState Space
MarkovChains
PoissonProcess
BrownianProcess
Inventory Modelsππππ+1 = ππππ + ππππ β π·π·ππ
ππππ ,π·π·ππ β β
DiscreteRandomVariables
ContinuousRandomVariables
Ran
dom
Ve
ctor
sUncountableRange
CountableRange
Probability Stochastic Processes
Difference between a discrete-time stochastic process & continuous-time stochastic process?β Similarity, limited dependence is still sought.β Difference is in the continuity of the process in time:
Β» Continuity is not an issue for processes with a discrete state spaceΒ» When the state space is continuous, we wonder how ππ(π‘π‘) & ππ(π‘π‘+ ππ) differ.
Β» For random variables, almost sure continuity: P limππβ 0
ππ β Ξ©: ππ π‘π‘ = ππ π‘π‘ β ππ = 1.
Ex: The exchange rate between Euro and US dollar is πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπ·π·. The stochastic process πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπ·π·(π‘π‘) can be generally continuous except when FED or ECB act to move the exchange rate in a certain direction.
β Immediately after some bank board meetings, πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπ·π·(π‘π‘) can jump. Central bank interventions can cause jumps.
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etinPage 4Seeking Convenience
Convention: A jump that happens at time π‘π‘ is included in ππ(π‘π‘) but not in ππ π‘π‘ β ππ Two common properties
β Independent increments ππ π‘π‘1 β ππ π π 1 β₯ ππ(π‘π‘2)βππ(π π 2) for disjoint intervals (π π 1,π‘π‘1] and (π π 2,π‘π‘2].β Stationary increments P(ππ(π π + ππ) βππ(π π ) = π₯π₯) = P(ππ(π‘π‘+ ππ)βππ(π‘π‘) = π₯π₯).
Ex: Independent increments imply P(ππ(π‘π‘) = π₯π₯π‘π‘|ππ(ππ) = π₯π₯ππ for ππ β (0,π π ]) = P(ππ(π‘π‘) = π₯π₯π‘π‘|ππ(π π ) = π₯π₯π π ), which is the Markov property.
Continuity property
β Right-continuous evolution P limπ‘π‘β0
ππ π‘π‘ + ππ = πππ‘π‘ = 1, i.e., πππ‘π‘ is right-continuous almost surely.
Ex: Let ππ π‘π‘,ππ = πΈπΈπ‘π‘(ππ) and ππ π‘π‘ + ππ,ππ = πΈπΈπ‘π‘+ππ(ππ) for πΈπΈπ‘π‘ ,πΈπΈπ‘π‘+ππ iid from πΈπΈ(0,1).β Check for convergence in probability. P ππ π‘π‘ + ππ β πππ‘π‘ β₯ 0.5 = Area of two triangles in unit square = 1
4β ππ π‘π‘ + ππ does not converge to πππ‘π‘ in probability. So it does not converge almost surely.β ππ(π‘π‘) is not right continuous. It is not left-continuous either.β Can ππ(π‘π‘) be used to model a stock price?
A process with Independent & Stationary increments, Right-continuous evolution is a LΓ©vy process.
LΓ©vy-ItΓ΄ Decomposition: LΓ©vyProcess = Poisson
Process + BrownianProcess + Martingale
Process + Deterministic driftLinear in Time
A Martingale satisfies E ππ(π‘π‘) ππ(π π ) = π₯π₯π π = π₯π₯π π for π π β€ π‘π‘.
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etinPage 5Counting Process
Assume independent and stationary increments:β The number of events happening over disjoint intervals are independent: ππ π‘π‘1 βππ π π 1 β₯ ππ(π‘π‘2)βππ(π π 2) or
disjoint intervals (π π 1,π‘π‘1] and (π π 2,π‘π‘2].β The number of events over two intervals of the same length have the same distribution: P(ππ(π π + ππ)βππ(π π ) =
ππ) = P(ππ(π‘π‘+ ππ) βππ(π‘π‘) = ππ).
Ex: For a car rental agency, let ππππ be the number of scratches on returned car ππ. Then the associated counting process, which yields the total number of scratches in the first ππ returned cars, is
ππ ππ = βππ=1ππ ππππ .
β ππ(ππ) is indexed by integer ππ but we can still ask if it has independent and stationary increments.β Independence: Scratches generally occur when the cars are rented out to different drivers and driven on different
roads at different timesβ Stationary: Different cars have rental history of being driven under similar conditions.β Assuming independent and stationary increments for ππ(ππ) is equivalent to assuming iid for {ππ1,ππ2, β¦ }.β This highlights the similarity between stochastic processes and random vectors.
A counting process {ππ(π‘π‘): π‘π‘ β₯ 0} is a continuous-time stochastic process with natural numbers as its state-space and ππ(π‘π‘) is the number of events that occur over interval (0, π‘π‘].
β Start with N(0)=0 β ππ(π‘π‘) is non-decreasing and right-continuous.
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etinPage 6Poisson Process: A special counting process A counting process ππ(π‘π‘) is a Poisson process with rate ππ if
i) ππ(0) = 0, ii) ππ(π‘π‘) has independent and stationary increments,
iii) P ππ π‘π‘ = ππ = exp βππ π‘π‘ ππ π‘π‘ ππ
ππ!.
An alternative definition of the process is useful
A function ππ is said to be order of number β and written as ππ βΌ ππ(β) if limββ0
ππ ββ
= 0. β An ππ(β) function drops faster than the linear function β as we get closer to 0.
Alternative definition of Poisson process, maintain properties i) and ii) but replace iii) with iiiβa) P(ππ(β) = 1) = ππβ + ππ(β) and iiiβb) P(ππ(β) = 0) = 1β ππβ + ππ(β).β At most 1 event in a short interval with occurrence probability proportional to the length of interval
Another alternative definition of Poisson process, maintain properties i) and ii) but replace iii) with iiiββ) The interevent times are iid with πΈπΈπ₯π₯πΈπΈππ(ππ).
0πΈπΈ1 πΈπΈ2 πΈπΈ3
ππ1 ππ2 ππ3
πΈπΈππβ1 πΈπΈππ
ππππ
Exponential interevent times ππππ = πΈπΈππ β πΈπΈππβ1 βΌ πΈπΈπ₯π₯πΈπΈππ(ππ)
Gamma event times πΈπΈππ = βππ=1ππ ππππ βΌ πΊπΊππππππππ(ππ, ππ)
πΈπΈππ
ππ π‘π‘ β₯ ππ
πΈπΈππ β€ π‘π‘
πΈπΈ0
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etinPage 7
iii) P ππ π‘π‘ = ππ = exp βππ π‘π‘ ππ π‘π‘ ππ/n! β iiiβ) P(ππ(β) = 1) = ππβ + ππ(β) & P(ππ(β) = 0) = 1 β ππβ + ππ(β)
iii) β iiiβ)
P ππ β = 0 = exp βππβ = 1 β ππβ+ ππβ 2
2!β ππβ 3
3!β¦ = 1β ππβ + ππ(β)
P ππ β = 1 = exp βππβ ππβ = ππβ+ β ππβ 2 + ππβ 3
2!β ππβ 4
3!β¦ = ππβ+ ππ(β)
iiiβ) β iii) Let πΈπΈππ π‘π‘ = P(ππ π‘π‘ = ππ) πΈπΈ0(π‘π‘+ β) = P(ππ(π‘π‘) = 0,ππ(π‘π‘ + β)βππ(π‘π‘) = 0) = P(ππ(π‘π‘) = 0)P(ππ(β) = 0) = πΈπΈ0 π‘π‘ πΈπΈ0(β) = πΈπΈ0(π‘π‘)(1β ππβ+ ππ β ),
which leads to πππππ‘π‘πΈπΈ0 π‘π‘ = lim
ββ0
πΈπΈ0(π‘π‘+ β)β πΈπΈ0(π‘π‘)β
= βπππΈπΈ0 π‘π‘ + limββ0
ππ ββ
= βπππΈπΈ0(π‘π‘)
β This differential equation has the solution of the form πΈπΈ0(π‘π‘) = ππ exp(βπππ‘π‘). β Using the initial condition πΈπΈ0(0) = 1, we obtain ππ = 1. Hence, πΈπΈ0(π‘π‘) = exp(βπππ‘π‘).
πΈπΈππ π‘π‘+ β = P ππ π‘π‘ = ππ,ππ π‘π‘ + β β ππ π‘π‘ = 0 + P ππ π‘π‘ = ππ β 1,ππ π‘π‘ + β β ππ π‘π‘ = 1+ P ππ π‘π‘ = ππ β 2,ππ π‘π‘+ β βππ π‘π‘ = 2 + P ππ π‘π‘ = ππ β 3,ππ π‘π‘+ β βππ π‘π‘ = 3 +β―
= πΈπΈππ π‘π‘ 1β ππβ + ππ β + πΈπΈππβ1 π‘π‘ ππβ+ ππ β + ππ β , which leads to, for ππ β₯ 1,
πππππ‘π‘πΈπΈππ π‘π‘ + πππΈπΈππ π‘π‘ = lim
ββ0
πΈπΈππ(π‘π‘+ β) β πΈπΈππ(π‘π‘)β
+ πππΈπΈππ π‘π‘ = πππΈπΈππβ1 π‘π‘ + limββ0
ππ ββ
= πππΈπΈππβ1(π‘π‘)
What remains to show: this equation has the solution πΈπΈππ(π‘π‘) = exp βπππ‘π‘ πππ‘π‘ ππ/ππ!. See the lecture notes.
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iii) P ππ π‘π‘ = ππ = exp βππ π‘π‘ ππ π‘π‘ ππ/n! β iiiββ) The interevent times are iid with πΈπΈπ₯π₯πΈπΈππ(ππ) An algebraic equality used below: Differences of Gammas is Poisson
P πΊπΊππππππππ(ππ, ππ) β€ π‘π‘ = οΏ½0
π‘π‘ππβππππ
πππ₯π₯ ππβ1
ππ β 1 ! πππππ₯π₯ = ππβπππ‘π‘πππ‘π‘ ππ
ππ! + οΏ½0
π‘π‘ππβππππ
πππ₯π₯ ππ
ππ! πππππ₯π₯
= P ππ(π‘π‘) = ππ + P πΊπΊππππππππ(ππ + 1, ππ) β€ π‘π‘It can be established via integration by parts; see lecture notes.
iii) β iiiββ) P πΈπΈ1 > π‘π‘ = P ππ π‘π‘ = 0 = exp βπππ‘π‘ so πΈπΈ1 βΌ πΊπΊππππππππ(1,ππ) and ππ1 βΌ πΈπΈπ₯π₯πΈπΈππ(ππ) P πΈπΈππ > π‘π‘ = P ππ π‘π‘ β {0,1, β¦ , ππ β 1} = βππ=0
ππβ1P(ππ(π‘π‘) = ππ)= P ππ π‘π‘ = 0 + βππ=1
ππβ1P(ππ(π‘π‘) = ππ)
= ππβπππ‘π‘ +βππ=1ππβ1 β«0π‘π‘ ππβππππ ππππ ππβ1
ππβ1 !πππππ₯π₯ β β«0
π‘π‘ ππβππππ ππππ ππ
ππ!πππππ₯π₯
= ππβπππ‘π‘ + β«0π‘π‘ ππβππππ πππππ₯π₯ β β«0
π‘π‘ ππβππππ ππππ ππβ1
ππβ1 !πππππ₯π₯
= 1 β β«0π‘π‘ ππβππππ ππππ ππβ1
ππβ1 !πππππ₯π₯ πΈπΈππ βΌ πΊπΊππππππππ(ππ, ππ) and ππππ βΌ πΈπΈπ₯π₯πΈπΈππ(ππ)
iiiββ) β iii) P(ππ(π‘π‘) = 0) = P(ππ1 > π‘π‘) = ππβπππ‘π‘ so ππ π‘π‘ = 0 has the Poisson probability P ππ π‘π‘ = ππ = P ππ π‘π‘ β₯ ππ β P ππ π‘π‘ β₯ ππ + 1
= P(πΈπΈππ β€ π‘π‘) β P(πΈπΈππ+1 β€ π‘π‘)
= ππβπππ‘π‘ πππ‘π‘ ππ
ππ! so ππ π‘π‘ = ππ has Poisson probability
P ππ π‘π‘ β₯ ππ = P ππ π‘π‘ = ππ +P(ππ π‘π‘ β₯ ππ + 1)
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etinPage 9Thinning of a Poisson Process
ππ1(π‘π‘) is a the number of type 1 events; ππ2(π‘π‘) is a the number of type 2 events. ππ1 π‘π‘ +ππ2 π‘π‘ = ππ(π‘π‘) and so [ππ1(π‘π‘) = ππ1 |ππ(π‘π‘) = ππ] and [ππ2(π‘π‘) = ππ2|ππ(π‘π‘) = ππ] are dependent
β P ππππ π‘π‘ = ππππ ππ π‘π‘ = ππ = P(π΅π΅ππππ ππ,πΈπΈππ = ππππ) for ππ β {1,2}β Illustration of dependence
Β» P ππ1 π‘π‘ = ππ1 ππ π‘π‘ = ππ P ππ2 π‘π‘ = ππ2 ππ π‘π‘ = ππ = P(π΅π΅ππππ ππ,πΈπΈ1 = ππ1)P(π΅π΅ππππ ππ,πΈπΈ2 = ππ2)Β» P ππ1 π‘π‘ = ππ1,ππ2 π‘π‘ = ππ2 ππ π‘π‘ = ππ = P π΅π΅ππππ ππ,πΈπΈ1 = ππ1 = P(π΅π΅ππππ ππ,πΈπΈ2 = ππ2)
Conditional random variables are dependent; unconditional ones can still be independent, in 2 steps:
1. Unconditional random variable ππ1 π‘π‘ βΌ ππππ(πππΈπΈ1π‘π‘). Similarly ππ2 π‘π‘ βΌ ππππ(πππΈπΈ2π‘π‘) .β P ππ1 π‘π‘ = ππ1 = βππ=ππ1
β P ππ1 π‘π‘ = ππ1 ππ π‘π‘ = ππ P ππ π‘π‘ = ππ = βππ=ππ1β P(π΅π΅ππππ ππ,πΈπΈ1 = ππ1)P ππ π‘π‘ = ππ
= βππ=ππ1β πΆπΆππ1
ππ πΈπΈ1ππ1 πΈπΈ2
ππβππ1ππβπππ‘π‘ πππ‘π‘ ππ
ππ!= ππβππππ1π‘π‘ ππππ1π‘π‘ ππ1
ππ1 !ππβππππ2π‘π‘βππ=ππ1
β 1(ππβππ1)!
πΈπΈ2πππ‘π‘ ππβππ1 = ππβππππ1π‘π‘ ππππ1π‘π‘ ππ1
ππ1 !
2. Independence: ππ1 π‘π‘ β₯ ππ2 π‘π‘β P(ππ1(π‘π‘) = ππ1, ππ2(π‘π‘) = ππ2) = P(ππ1(π‘π‘) = ππ1, ππ2(π‘π‘) = ππ2|ππ(π‘π‘) = ππ1 + ππ2) P(ππ(π‘π‘) = ππ1 + ππ2)
= P ππ1 π‘π‘ = ππ1 ππ π‘π‘ = ππ1 + ππ2 P ππ π‘π‘ = ππ1 + ππ2 = πΆπΆππ1ππ1+ππ2 πΈπΈ1
ππ1 πΈπΈ2ππ2ππβπππ‘π‘
πππ‘π‘ ππ1+ππ2
(ππ1+ ππ2)!
= ππβππππ1π‘π‘πππΈπΈ1π‘π‘ ππ1
ππ1!ππβππππ2π‘π‘
πππΈπΈ2π‘π‘ ππ2
ππ2!= P ππ1 π‘π‘ = ππ1 P(ππ2(π‘π‘) = ππ2)
πΈπΈ1 πΈπΈ2 πΈπΈ3 πΈπΈππβ1 πΈπΈππ πΈπΈππ
Some events are blue, some are green; with probabilities πΈπΈ1 , πΈπΈ2 and πΈπΈ1 + πΈπΈ2 = 1. All events occur according to Poisson process. Occurrence of green or of blue events according to thinning of Poisson process.
With probability πΈπΈ1
With probability πΈπΈ2Poisson process
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etinPage 10Number of Emails Until a Coupon
Ex: A graduate student is interested in Texas style horseback riding but does not have much money to spend on this. o The student enrolls in livingsocial.com's Dallas email list to receive a discount offer for horseback riding. o Livingsocial sends offers for other events such as piloting, karate lessons, gun carriage license training, etc. o Livingsocial includes 1 discount offer in every email and that offer can be for any one of the events listed above.
What is the expected number of livingsocial emails that the student has to receive to find an offer for horseback riding? β Probability πΈπΈ1 is for each email to include a discount offer for horseback riding. Let ππ1 be the index of the first
email including a horseback riding offer. Then P(ππ1 = ππ1) = πΈπΈ1 1β πΈπΈ1 ππ1β1,ππ1 is a geometric random variable.
Ex: Livingsocial sends offers for 5 types of events with probabilities πΈπΈ1 + πΈπΈ2 + πΈπΈ3 + πΈπΈ4 + πΈπΈ5 = 1. What is the expected number of emails that the student has to receive to obtain a discount offer for each event type?
β Let ππππ be the index of the first email including a discount offer for event type ππ. β A particular realization can have [ππ1 = 3,ππ2 = 7,ππ3 = 1,ππ4 = 2,ππ5 = 5],
Β» the 1st email has an offer for event 3, Β» the 2nd has an offer for event 4, Β» the 3rd has an offer for event 1, Β» the 4th and 6th must include offers for events 1, 3 and 4. Β» the 5th has an offer for event 5Β» the 7th has an offer for event 2
β In this realization, the number of emails to endure to obtain an offer for each event is max{3, 7, 1, 2, 5} = 7. β In general,
ππ = max{ππ1,ππ2,ππ3,ππ4,ππ5} for dependentππππsis the number emails to obtain an offer for each event.
β We want E(ππ), but ππππs are dependent even if they have marginal pmfβs that are geometric distributions.
β How to break the dependence?
3 4 1
25
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Amount of Time until a Coupon for EachContinuous-time Analog
Suppose emails are sent at the rate of 1 per day. Number of emails ππ π‘π‘ βΌ ππππ(1π‘π‘) with interevent time ππππ βΌ πΈπΈπ₯π₯πΈπΈππ(1). ππππ π‘π‘ is the number of coupons of type ππ obtained by time π‘π‘. ππππ π‘π‘ βΌ ππππ(1πΈπΈπππ‘π‘) ππππ π‘π‘ is obtained from ππ(π‘π‘) via thinning so π΅π΅ππ β₯ π΅π΅ππ β₯ β― β₯ π΅π΅ππ ππππ is the interevent time for ππππ π‘π‘ process. ππππ βΌ πΈπΈπ₯π₯πΈπΈππ(1πΈπΈππ)
β ππππ is the time between two emails both containing coupon ππ Amount of time until a coupon for each event
ππ = max ππ1,ππ2,ππ3,ππ4,ππ5 for πΏπΏππ β₯ πΏπΏππ β₯ β― β₯ πΏπΏππ. Relating the number of emails to the time of the emails.
β How many emails are received in ππ amount of time? How much time does it take to receive ππ emails?ππ(π‘π‘ = ππ) or ππ = βππ=1ππ ππππ
β Proceeding in either way: E ππ ππ = 1E ππ = E(ππ) or E ππ = E ππππ E ππ = 1E(ππ), so E ππ = E ππ . To finish, we need E ππ . First equality is from independence,
P ππ β€ π‘π‘ = οΏ½ππ=1
5
P ππππ β€ π‘π‘ =οΏ½ππ=1
5
1 β ππβπππππ‘π‘
E ππ = β«0β
P ππβ₯ π‘π‘ πππ‘π‘ = οΏ½0
β1βοΏ½
ππ=1
5
1β ππβπππππ‘π‘ πππ‘π‘
β For the number of event types approaching infinity, E ππ β β. For a single event type, E(ππ) = 1.
Discrete-time:Number of emailsto obtain coupon ππ
Continuous-time:Amount of timeto obtain coupon ππ
3 4 1
25
ππ3 ππ4ππ1ππ5 ππ2
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Distribution of Event Times Conditioned on Number of Events
Recall from order statistics: {ππ1,ππ2, β¦ ,ππππ} is an iid sequence, where ππππ βΌ ππππ, and ππππ is the ππth smallest element in the sequence. The joint pdf of ππ = [ππ1,ππ2, β¦ ,ππππ] is, for π¦π¦1 β€ π¦π¦2 β€ β― β€ π¦π¦ππ,
ππππ(π¦π¦1,π¦π¦2, β¦ ,π¦π¦ππ) = ππ!οΏ½ππ=1
ππ
ππππ(π¦π¦ππ)
Ex: What is the joint pdf of the order statistics vector ππ if each of underlying random variable ππππ βΌ πΈπΈ(0,π‘π‘)? β The pdf of ππππ is ππππ(π₯π₯) = πΌπΌππβ(0,π‘π‘] /π‘π‘. Then the joint pdf of the order statistics vector ππ is, for 0 < π¦π¦1 < π¦π¦2 < β― < π¦π¦ππ < π‘π‘,
ππππ π¦π¦1,π¦π¦2, β¦ ,π¦π¦ππ = ππ!οΏ½ππ=1
ππ1π‘π‘
=ππ!π‘π‘ππ
Regardless of the rate of Poisson, the conditional joint pdf of ([πΈπΈ1,πΈπΈ2, β¦ , πΈπΈππ]|ππ(π‘π‘) = ππ) is
ππ[ππ1 ,ππ2 ,β¦,ππππ]|ππ(π‘π‘)=ππ π π 1,π π 2, β¦ , π π ππ =ππ!π‘π‘ππ
for 0 < π π 1 < β― < π π ππ < π‘π‘.
β First note [πΈπΈ1 = π π 1,πΈπΈ2 = π π 2, β¦ , πΈπΈππ = π π ππ,ππ(π‘π‘) = ππ] iff [ππ1 = π π 1,ππ2 = π π 2 β π π 1, β¦ ,ππππ = π π ππ β π π ππβ1,ππππ+1 β₯ π‘π‘ β π π ππ].
β Hence,ππππ1 ,ππ2,β¦,ππππ ,ππ(π‘π‘) π π 1,π π 2, β¦ , π π ππ,ππ = ππππ π π 1 ππππ π π 2β π π 1 β¦ ππππ π π ππ β π π ππβ1 1βπΉπΉππ π‘π‘β π π ππ
= ππ ππβπππ π 1ππππβππ π π 2βπ π 1 β¦ππππβππ π π ππβπ π ππβ1 ππβππ π‘π‘βπ π ππ = ππππππβπππ‘π‘.β Then going forward mechanically
ππ[ππ1 ,ππ2,β¦,ππππ]|ππ(π‘π‘)=ππ π π 1,π π 2, β¦ , π π ππ =ππππ1 ,ππ2 ,β¦,ππππ ,ππ π‘π‘ π π 1,π π 2, β¦ , π π ππ,ππ
P ππ π‘π‘ = ππ=
ππππππβπππ‘π‘
(πππ‘π‘)ππππβπππ‘π‘/ππ!=ππ!π‘π‘ππ
.
From the last two examples, the conditional joint pdf of ([πΈπΈ1,πΈπΈ2, β¦ , πΈπΈππ]|ππ(π‘π‘) = ππ) and the pdf of order statistics [ππ1,ππ2, β¦ ,ππππ] with an underlying iid ππππ βΌ πΈπΈππ(0,π‘π‘] have the same distribution.
πΈπΈππ ππ π‘π‘ = ππ = ππππ(πΈπΈ1(0, π‘π‘], β¦ ,πΈπΈππ(0, π‘π‘]) in distribution
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etinPage 13Expected Total Waiting Time for a Bus
Ex: A DART bus departs from UTD to grocery markets every ππ minutes. o Passengers arrive according to Poisson process ππ(π‘π‘) with rate ππ and wait for the departure.o E.g.,, the ππth passenger that arrives over (0,π‘π‘] arrives at πΈπΈππ and waits for ππ β πΈπΈππ. o The expected total waiting time is
E οΏ½ππ=1
ππ ππ
ππ β πΈπΈππ = E E οΏ½ππ=1
ππ ππ
ππ β πΈπΈππ |ππ ππ = ππ = οΏ½ππ=0
β
E οΏ½ππ=1
ππ ππ
ππ β πΈπΈππ |ππ ππ = ππ(ππππ)ππππβππππ
ππ!
= οΏ½ππ=0
β
E οΏ½ππ=1
ππ
ππ β ππππ πΈπΈ1 0, ππ , β¦ ,πΈπΈππ 0,ππ P ππ ππ = ππ
= οΏ½ππ=0
β
οΏ½ππ=1
ππ
E ππ β ππππ (πΈπΈ1(0, ππ], β¦ ,πΈπΈππ(0,ππ]) P(ππ ππ = ππ)
= οΏ½ππ=0
β
ππππ β E οΏ½ππ=1
ππ
ππππ πΈπΈ1 0, ππ , β¦ ,πΈπΈππ 0, ππ P(ππ ππ = ππ)
= οΏ½ππ=0
β
ππππ β ππππ2 P ππ ππ = ππ =
12 πππ ππ ππ =
ππππ π»π»π»π»π»π»
ππππ
ππ
Arrival time πΈπΈππ|ππ(π‘π‘) = ππππ(πΈπΈ1(0,ππ], β¦ ,πΈπΈππ(0,ππ])
Waiting time
Area of the triangleis the expected total
waiting time
Sum of order statistics = Sum of sample
E οΏ½ππ=1
ππ
ππππ = E οΏ½ππ=1
ππ
πΈπΈππ
= πππ πΈπΈππ = ππππ2
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etinPage 14
Number of calls arriving to a call centerin 15 minute time intervals
Generalizations
Nonhomogeneous Poisson process.β Non-stationary Stationary Incrementsβ Time dependent event rate ππ(π₯π₯) at time π₯π₯β The expected number of events over (0,π‘π‘] is
οΏ½0
π‘π‘ππ π₯π₯ πππ₯π₯
β The number of events during (0,π‘π‘] has
ππππ οΏ½0
π‘π‘ππ π₯π₯ πππ₯π₯
β Independent increments?β To model varying traffic intensity
Compound Poisson process: A Poisson process ππ(π‘π‘) and an iid sequence {ππππ} can be combined to obtain the compound Poisson process:
ππ(π‘π‘) = οΏ½ππ=1
ππ π‘π‘
ππππ
β Independent increments?β To model batch arrivals of customers to a store (or buying multiple units at once)
Earlymorning
Earlyevening
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etinPage 15Summary
Continuous-time Markov Process Poisson Process Thinning Conditioning on the Number of Events Generalizations