.edu Exponential Distribution and Poisson Processmetin/Probability/Folios/expoPois.pdfΒ Β· which is...

15
utdallas .edu /~metin Page 1 Exponential Distribution and Poisson Process Outline Continuous-time Markov Process Poisson Process Thinning Conditioning on the Number of Events Generalizations

Transcript of .edu Exponential Distribution and Poisson Processmetin/Probability/Folios/expoPois.pdfΒ Β· which is...

Page 1: .edu Exponential Distribution and Poisson Processmetin/Probability/Folios/expoPois.pdfΒ Β· which is the Markov property. Continuity property – Right-continuous evolutionP lim 𝑑𝑑↓0

utda

llas.edu /~m

etinPage 1Exponential Distribution and Poisson Process

Outline Continuous-time Markov Process Poisson Process Thinning Conditioning on the Number of Events Generalizations

Page 2: .edu Exponential Distribution and Poisson Processmetin/Probability/Folios/expoPois.pdfΒ Β· which is the Markov property. Continuity property – Right-continuous evolutionP lim 𝑑𝑑↓0

utda

llas.edu /~m

etinPage 2Probability and Stochastic Processes

Overview of Random Variables and Stochastic Processes– Discrete-time: {𝑋𝑋𝑛𝑛:𝑛𝑛 β‰₯ 0,𝑛𝑛 ∈ 𝑍𝑍}; Continuous-time: {𝑋𝑋(𝑑𝑑):𝑑𝑑 β‰₯ 0, 𝑑𝑑 ∈ β„œ}– Discrete state-space if each 𝑋𝑋𝑛𝑛 or 𝑋𝑋(𝑑𝑑) 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 Models𝑋𝑋𝑛𝑛+1 = 𝑋𝑋𝑛𝑛 + π‘žπ‘žπ‘›π‘› βˆ’ 𝐷𝐷𝑛𝑛

π‘žπ‘žπ‘›π‘› ,𝐷𝐷𝑛𝑛 ∈ β„œ

DiscreteRandomVariables

ContinuousRandomVariables

Ran

dom

Ve

ctor

sUncountableRange

CountableRange

Probability Stochastic Processes

– Difference between an infinite-length random vector & a discrete-time stochastic process?Β» Similarity, they are both denoted as {𝑋𝑋𝑛𝑛:𝑛𝑛 β‰₯ 0,𝑛𝑛 ∈ 𝑍𝑍}.Β» Difference is in the treatment of dependence:

Random vectors – Only short lengths (2-3 rvs) can be studied if they are dependent; – Or assume structured dependence through copulas. – Or work with independent and identically distributed sequences.

In discrete-time stochastic processes, assume Markov Property to limit the dependence

Page 3: .edu Exponential Distribution and Poisson Processmetin/Probability/Folios/expoPois.pdfΒ Β· which is the Markov property. Continuity property – Right-continuous evolutionP lim 𝑑𝑑↓0

utda

llas.edu /~m

etinPage 3Stochastic Processes in Continuous Time

Discrete-time Continuous-time

DiscreteState Space

ContinuousState Space

MarkovChains

PoissonProcess

BrownianProcess

Inventory Models𝑋𝑋𝑛𝑛+1 = 𝑋𝑋𝑛𝑛 + π‘žπ‘žπ‘›π‘› βˆ’ 𝐷𝐷𝑛𝑛

π‘žπ‘žπ‘›π‘› ,𝐷𝐷𝑛𝑛 ∈ β„œ

DiscreteRandomVariables

ContinuousRandomVariables

Ran

dom

Ve

ctor

sUncountableRange

CountableRange

Probability Stochastic Processes

Difference between a discrete-time stochastic process & continuous-time stochastic process?– Similarity, limited dependence is still sought.– 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 𝑋𝑋(𝑑𝑑) & 𝑋𝑋(𝑑𝑑+ πœ–πœ–) differ.

Β» For random variables, almost sure continuity: P limπœ–πœ–β†’ 0

πœ”πœ” ∈ Ξ©: 𝑋𝑋 𝑑𝑑 = 𝑋𝑋 𝑑𝑑 βˆ’ πœ–πœ– = 1.

Ex: The exchange rate between Euro and US dollar is 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐷𝐷. The stochastic process 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐷𝐷(𝑑𝑑) can be generally continuous except when FED or ECB act to move the exchange rate in a certain direction.

– Immediately after some bank board meetings, 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐷𝐷(𝑑𝑑) can jump. Central bank interventions can cause jumps.

Page 4: .edu Exponential Distribution and Poisson Processmetin/Probability/Folios/expoPois.pdfΒ Β· which is the Markov property. Continuity property – Right-continuous evolutionP lim 𝑑𝑑↓0

utda

llas.edu /~m

etinPage 4Seeking Convenience

Convention: A jump that happens at time 𝑑𝑑 is included in 𝑋𝑋(𝑑𝑑) but not in 𝑋𝑋 𝑑𝑑 βˆ’ πœ–πœ– Two common properties

– Independent increments 𝑋𝑋 𝑑𝑑1 βˆ’ 𝑋𝑋 𝑠𝑠1 βŠ₯ 𝑋𝑋(𝑑𝑑2)βˆ’π‘‹π‘‹(𝑠𝑠2) for disjoint intervals (𝑠𝑠1,𝑑𝑑1] and (𝑠𝑠2,𝑑𝑑2].– Stationary increments P(𝑋𝑋(𝑠𝑠+ π‘Žπ‘Ž) βˆ’π‘‹π‘‹(𝑠𝑠) = π‘₯π‘₯) = P(𝑋𝑋(𝑑𝑑+ π‘Žπ‘Ž)βˆ’π‘‹π‘‹(𝑑𝑑) = π‘₯π‘₯).

Ex: Independent increments imply P(𝑋𝑋(𝑑𝑑) = π‘₯π‘₯𝑑𝑑|𝑋𝑋(π‘Ÿπ‘Ÿ) = π‘₯π‘₯π‘Ÿπ‘Ÿ for π‘Ÿπ‘Ÿ ∈ (0,𝑠𝑠]) = P(𝑋𝑋(𝑑𝑑) = π‘₯π‘₯𝑑𝑑|𝑋𝑋(𝑠𝑠) = π‘₯π‘₯𝑠𝑠), which is the Markov property.

Continuity property

– Right-continuous evolution P lim𝑑𝑑↓0

𝑋𝑋 𝑑𝑑 + πœ–πœ– = 𝑋𝑋𝑑𝑑 = 1, i.e., 𝑋𝑋𝑑𝑑 is right-continuous almost surely.

Ex: Let 𝑋𝑋 𝑑𝑑,πœ”πœ” = 𝐸𝐸𝑑𝑑(πœ”πœ”) and 𝑋𝑋 𝑑𝑑 + πœ–πœ–,πœ”πœ” = 𝐸𝐸𝑑𝑑+πœ–πœ–(πœ”πœ”) for 𝐸𝐸𝑑𝑑 ,𝐸𝐸𝑑𝑑+πœ–πœ– iid from 𝐸𝐸(0,1).– Check for convergence in probability. P 𝑋𝑋 𝑑𝑑 + πœ–πœ– βˆ’ 𝑋𝑋𝑑𝑑 β‰₯ 0.5 = Area of two triangles in unit square = 1

4– 𝑋𝑋 𝑑𝑑 + πœ–πœ– does not converge to 𝑋𝑋𝑑𝑑 in probability. So it does not converge almost surely.– 𝑋𝑋(𝑑𝑑) is not right continuous. It is not left-continuous either.– Can 𝑋𝑋(𝑑𝑑) 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 𝑋𝑋(𝑑𝑑) 𝑋𝑋(𝑠𝑠) = π‘₯π‘₯𝑠𝑠 = π‘₯π‘₯𝑠𝑠 for 𝑠𝑠 ≀ 𝑑𝑑.

Page 5: .edu Exponential Distribution and Poisson Processmetin/Probability/Folios/expoPois.pdfΒ Β· which is the Markov property. Continuity property – Right-continuous evolutionP lim 𝑑𝑑↓0

utda

llas.edu /~m

etinPage 5Counting Process

Assume independent and stationary increments:– The number of events happening over disjoint intervals are independent: 𝑁𝑁 𝑑𝑑1 βˆ’π‘π‘ 𝑠𝑠1 βŠ₯ 𝑁𝑁(𝑑𝑑2)βˆ’π‘π‘(𝑠𝑠2) or

disjoint intervals (𝑠𝑠1,𝑑𝑑1] and (𝑠𝑠2,𝑑𝑑2].– The number of events over two intervals of the same length have the same distribution: P(𝑁𝑁(𝑠𝑠+ π‘Žπ‘Ž)βˆ’π‘π‘(𝑠𝑠) =

π‘˜π‘˜) = P(𝑁𝑁(𝑑𝑑+ π‘Žπ‘Ž) βˆ’π‘π‘(𝑑𝑑) = π‘˜π‘˜).

Ex: For a car rental agency, let 𝑛𝑛𝑖𝑖 be the number of scratches on returned car 𝑖𝑖. Then the associated counting process, which yields the total number of scratches in the first π‘šπ‘š returned cars, is

𝑁𝑁 π‘šπ‘š = βˆ‘π‘–π‘–=1π‘šπ‘š 𝑛𝑛𝑖𝑖 .

– 𝑁𝑁(π‘šπ‘š) is indexed by integer π‘šπ‘š but we can still ask if it has independent and stationary increments.– Independence: Scratches generally occur when the cars are rented out to different drivers and driven on different

roads at different times– Stationary: Different cars have rental history of being driven under similar conditions.– Assuming independent and stationary increments for 𝑁𝑁(π‘šπ‘š) is equivalent to assuming iid for {𝑛𝑛1,𝑛𝑛2, … }.– This highlights the similarity between stochastic processes and random vectors.

A counting process {𝑁𝑁(𝑑𝑑): 𝑑𝑑 β‰₯ 0} is a continuous-time stochastic process with natural numbers as its state-space and 𝑁𝑁(𝑑𝑑) is the number of events that occur over interval (0, 𝑑𝑑].

– Start with N(0)=0 – 𝑁𝑁(𝑑𝑑) is non-decreasing and right-continuous.

Page 6: .edu Exponential Distribution and Poisson Processmetin/Probability/Folios/expoPois.pdfΒ Β· which is the Markov property. Continuity property – Right-continuous evolutionP lim 𝑑𝑑↓0

utda

llas.edu /~m

etinPage 6Poisson Process: A special counting process A counting process 𝑁𝑁(𝑑𝑑) is a Poisson process with rate πœ†πœ† if

i) 𝑁𝑁(0) = 0, ii) 𝑁𝑁(𝑑𝑑) has independent and stationary increments,

iii) P 𝑁𝑁 𝑑𝑑 = 𝑛𝑛 = exp βˆ’πœ†πœ† 𝑑𝑑 πœ†πœ† 𝑑𝑑 𝑛𝑛

𝑛𝑛!.

An alternative definition of the process is useful

A function 𝑔𝑔 is said to be order of number β„Ž and written as 𝑔𝑔 ∼ π‘œπ‘œ(β„Ž) if limβ„Žβ†’0

𝑔𝑔 β„Žβ„Ž

= 0. – An π‘œπ‘œ(β„Ž) function drops faster than the linear function β„Ž as we get closer to 0.

Alternative definition of Poisson process, maintain properties i) and ii) but replace iii) with iii’a) P(𝑁𝑁(β„Ž) = 1) = πœ†πœ†β„Ž + π‘œπ‘œ(β„Ž) and iii’b) P(𝑁𝑁(β„Ž) = 0) = 1βˆ’ πœ†πœ†β„Ž + π‘œπ‘œ(β„Ž).– 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 iii’’) The interevent times are iid with 𝐸𝐸π‘₯π‘₯πΈπΈπ‘œπ‘œ(πœ†πœ†).

0𝐸𝐸1 𝐸𝐸2 𝐸𝐸3

𝑋𝑋1 𝑋𝑋2 𝑋𝑋3

πΈπΈπ‘–π‘–βˆ’1 𝐸𝐸𝑖𝑖

𝑋𝑋𝑖𝑖

Exponential interevent times 𝑋𝑋𝑖𝑖 = 𝐸𝐸𝑖𝑖 βˆ’ πΈπΈπ‘–π‘–βˆ’1 ∼ 𝐸𝐸π‘₯π‘₯πΈπΈπ‘œπ‘œ(πœ†πœ†)

Gamma event times 𝐸𝐸𝑖𝑖 = βˆ‘π‘—π‘—=1𝑖𝑖 𝑋𝑋𝑗𝑗 ∼ πΊπΊπ‘Žπ‘Žπ‘šπ‘šπ‘šπ‘šπ‘Žπ‘Ž(𝑖𝑖, πœ†πœ†)

𝐸𝐸𝑛𝑛

𝑁𝑁 𝑑𝑑 β‰₯ 𝑛𝑛

𝐸𝐸𝑛𝑛 ≀ 𝑑𝑑

𝐸𝐸0

Page 7: .edu Exponential Distribution and Poisson Processmetin/Probability/Folios/expoPois.pdfΒ Β· which is the Markov property. Continuity property – Right-continuous evolutionP lim 𝑑𝑑↓0

utda

llas.edu /~m

etinPage 7

iii) P 𝑁𝑁 𝑑𝑑 = 𝑛𝑛 = exp βˆ’πœ†πœ† 𝑑𝑑 πœ†πœ† 𝑑𝑑 𝑛𝑛/n! ↔ iii’) P(𝑁𝑁(β„Ž) = 1) = πœ†πœ†β„Ž + π‘œπ‘œ(β„Ž) & P(𝑁𝑁(β„Ž) = 0) = 1 βˆ’ πœ†πœ†β„Ž + π‘œπ‘œ(β„Ž)

iii) β‡’ iii’)

P 𝑁𝑁 β„Ž = 0 = exp βˆ’πœ†πœ†β„Ž = 1 βˆ’ πœ†πœ†β„Ž+ πœ†πœ†β„Ž 2

2!βˆ’ πœ†πœ†β„Ž 3

3!… = 1βˆ’ πœ†πœ†β„Ž + π‘œπ‘œ(β„Ž)

P 𝑁𝑁 β„Ž = 1 = exp βˆ’πœ†πœ†β„Ž πœ†πœ†β„Ž = πœ†πœ†β„Ž+ βˆ’ πœ†πœ†β„Ž 2 + πœ†πœ†β„Ž 3

2!βˆ’ πœ†πœ†β„Ž 4

3!… = πœ†πœ†β„Ž+ π‘œπ‘œ(β„Ž)

iii’) β‡’ iii) Let 𝐸𝐸𝑖𝑖 𝑑𝑑 = P(𝑁𝑁 𝑑𝑑 = 𝑖𝑖) 𝐸𝐸0(𝑑𝑑+ β„Ž) = P(𝑁𝑁(𝑑𝑑) = 0,𝑁𝑁(𝑑𝑑 + β„Ž)βˆ’π‘π‘(𝑑𝑑) = 0) = P(𝑁𝑁(𝑑𝑑) = 0)P(𝑁𝑁(β„Ž) = 0) = 𝐸𝐸0 𝑑𝑑 𝐸𝐸0(β„Ž) = 𝐸𝐸0(𝑑𝑑)(1βˆ’ πœ†πœ†β„Ž+ π‘œπ‘œ β„Ž ),

which leads to 𝑑𝑑𝑑𝑑𝑑𝑑𝐸𝐸0 𝑑𝑑 = lim

β„Žβ†’0

𝐸𝐸0(𝑑𝑑+ β„Ž)βˆ’ 𝐸𝐸0(𝑑𝑑)β„Ž

= βˆ’πœ†πœ†πΈπΈ0 𝑑𝑑 + limβ„Žβ†’0

π‘œπ‘œ β„Žβ„Ž

= βˆ’πœ†πœ†πΈπΈ0(𝑑𝑑)

– This differential equation has the solution of the form 𝐸𝐸0(𝑑𝑑) = 𝑐𝑐 exp(βˆ’πœ†πœ†π‘‘π‘‘). – Using the initial condition 𝐸𝐸0(0) = 1, we obtain 𝑐𝑐 = 1. Hence, 𝐸𝐸0(𝑑𝑑) = exp(βˆ’πœ†πœ†π‘‘π‘‘).

𝐸𝐸𝑖𝑖 𝑑𝑑+ β„Ž = P 𝑁𝑁 𝑑𝑑 = 𝑖𝑖,𝑁𝑁 𝑑𝑑 + β„Ž βˆ’ 𝑁𝑁 𝑑𝑑 = 0 + P 𝑁𝑁 𝑑𝑑 = 𝑖𝑖 βˆ’ 1,𝑁𝑁 𝑑𝑑 + β„Ž βˆ’ 𝑁𝑁 𝑑𝑑 = 1+ P 𝑁𝑁 𝑑𝑑 = 𝑖𝑖 βˆ’ 2,𝑁𝑁 𝑑𝑑+ β„Ž βˆ’π‘π‘ 𝑑𝑑 = 2 + P 𝑁𝑁 𝑑𝑑 = 𝑖𝑖 βˆ’ 3,𝑁𝑁 𝑑𝑑+ β„Ž βˆ’π‘π‘ 𝑑𝑑 = 3 +β‹―

= 𝐸𝐸𝑖𝑖 𝑑𝑑 1βˆ’ πœ†πœ†β„Ž + π‘œπ‘œ β„Ž + πΈπΈπ‘–π‘–βˆ’1 𝑑𝑑 πœ†πœ†β„Ž+ π‘œπ‘œ β„Ž + π‘œπ‘œ β„Ž , which leads to, for 𝑖𝑖 β‰₯ 1,

𝑑𝑑𝑑𝑑𝑑𝑑𝐸𝐸𝑖𝑖 𝑑𝑑 + πœ†πœ†πΈπΈπ‘–π‘– 𝑑𝑑 = lim

β„Žβ†’0

𝐸𝐸𝑖𝑖(𝑑𝑑+ β„Ž) βˆ’ 𝐸𝐸𝑖𝑖(𝑑𝑑)β„Ž

+ πœ†πœ†πΈπΈπ‘–π‘– 𝑑𝑑 = πœ†πœ†πΈπΈπ‘–π‘–βˆ’1 𝑑𝑑 + limβ„Žβ†’0

π‘œπ‘œ β„Žβ„Ž

= πœ†πœ†πΈπΈπ‘–π‘–βˆ’1(𝑑𝑑)

What remains to show: this equation has the solution 𝐸𝐸𝑖𝑖(𝑑𝑑) = exp βˆ’πœ†πœ†π‘‘π‘‘ πœ†πœ†π‘‘π‘‘ 𝑖𝑖/𝑖𝑖!. See the lecture notes.

Page 8: .edu Exponential Distribution and Poisson Processmetin/Probability/Folios/expoPois.pdfΒ Β· which is the Markov property. Continuity property – Right-continuous evolutionP lim 𝑑𝑑↓0

utda

llas.edu /~m

etinPage 8

iii) P 𝑁𝑁 𝑑𝑑 = 𝑛𝑛 = exp βˆ’πœ†πœ† 𝑑𝑑 πœ†πœ† 𝑑𝑑 𝑛𝑛/n! ↔ iii’’) The interevent times are iid with 𝐸𝐸π‘₯π‘₯πΈπΈπ‘œπ‘œ(πœ†πœ†) An algebraic equality used below: Differences of Gammas is Poisson

P πΊπΊπ‘Žπ‘Žπ‘šπ‘šπ‘šπ‘šπ‘Žπ‘Ž(𝑛𝑛, πœ†πœ†) ≀ 𝑑𝑑 = οΏ½0

π‘‘π‘‘π‘’π‘’βˆ’πœ†πœ†πœ†πœ†

πœ†πœ†π‘₯π‘₯ π‘›π‘›βˆ’1

𝑛𝑛 βˆ’ 1 ! πœ†πœ†π‘‘π‘‘π‘₯π‘₯ = π‘’π‘’βˆ’πœ†πœ†π‘‘π‘‘πœ†πœ†π‘‘π‘‘ 𝑛𝑛

𝑛𝑛! + οΏ½0

π‘‘π‘‘π‘’π‘’βˆ’πœ†πœ†πœ†πœ†

πœ†πœ†π‘₯π‘₯ 𝑛𝑛

𝑛𝑛! πœ†πœ†π‘‘π‘‘π‘₯π‘₯

= P 𝑁𝑁(𝑑𝑑) = 𝑛𝑛 + P πΊπΊπ‘Žπ‘Žπ‘šπ‘šπ‘šπ‘šπ‘Žπ‘Ž(𝑛𝑛 + 1, πœ†πœ†) ≀ 𝑑𝑑It can be established via integration by parts; see lecture notes.

iii) β‡’ iii’’) P 𝐸𝐸1 > 𝑑𝑑 = P 𝑁𝑁 𝑑𝑑 = 0 = exp βˆ’πœ†πœ†π‘‘π‘‘ so 𝐸𝐸1 ∼ πΊπΊπ‘Žπ‘Žπ‘šπ‘šπ‘šπ‘šπ‘Žπ‘Ž(1,πœ†πœ†) and 𝑋𝑋1 ∼ 𝐸𝐸π‘₯π‘₯πΈπΈπ‘œπ‘œ(πœ†πœ†) P 𝐸𝐸𝑛𝑛 > 𝑑𝑑 = P 𝑁𝑁 𝑑𝑑 ∈ {0,1, … , 𝑛𝑛 βˆ’ 1} = βˆ‘π‘–π‘–=0

π‘›π‘›βˆ’1P(𝑁𝑁(𝑑𝑑) = 𝑖𝑖)= P 𝑁𝑁 𝑑𝑑 = 0 + βˆ‘π‘–π‘–=1

π‘›π‘›βˆ’1P(𝑁𝑁(𝑑𝑑) = 𝑖𝑖)

= π‘’π‘’βˆ’πœ†πœ†π‘‘π‘‘ +βˆ‘π‘–π‘–=1π‘›π‘›βˆ’1 ∫0𝑑𝑑 π‘’π‘’βˆ’πœ†πœ†πœ†πœ† πœ†πœ†πœ†πœ† π‘–π‘–βˆ’1

π‘–π‘–βˆ’1 !πœ†πœ†π‘‘π‘‘π‘₯π‘₯ βˆ’ ∫0

𝑑𝑑 π‘’π‘’βˆ’πœ†πœ†πœ†πœ† πœ†πœ†πœ†πœ† 𝑖𝑖

𝑖𝑖!πœ†πœ†π‘‘π‘‘π‘₯π‘₯

= π‘’π‘’βˆ’πœ†πœ†π‘‘π‘‘ + ∫0𝑑𝑑 π‘’π‘’βˆ’πœ†πœ†πœ†πœ† πœ†πœ†π‘‘π‘‘π‘₯π‘₯ βˆ’ ∫0

𝑑𝑑 π‘’π‘’βˆ’πœ†πœ†πœ†πœ† πœ†πœ†πœ†πœ† π‘›π‘›βˆ’1

π‘›π‘›βˆ’1 !πœ†πœ†π‘‘π‘‘π‘₯π‘₯

= 1 βˆ’ ∫0𝑑𝑑 π‘’π‘’βˆ’πœ†πœ†πœ†πœ† πœ†πœ†πœ†πœ† π‘›π‘›βˆ’1

π‘›π‘›βˆ’1 !πœ†πœ†π‘‘π‘‘π‘₯π‘₯ 𝐸𝐸𝑛𝑛 ∼ πΊπΊπ‘Žπ‘Žπ‘šπ‘šπ‘šπ‘šπ‘Žπ‘Ž(𝑛𝑛, πœ†πœ†) and 𝑋𝑋𝑛𝑛 ∼ 𝐸𝐸π‘₯π‘₯πΈπΈπ‘œπ‘œ(πœ†πœ†)

iii’’) β‡’ iii) P(𝑁𝑁(𝑑𝑑) = 0) = P(𝑋𝑋1 > 𝑑𝑑) = π‘’π‘’βˆ’πœ†πœ†π‘‘π‘‘ so 𝑁𝑁 𝑑𝑑 = 0 has the Poisson probability P 𝑁𝑁 𝑑𝑑 = 𝑛𝑛 = P 𝑁𝑁 𝑑𝑑 β‰₯ 𝑛𝑛 βˆ’ P 𝑁𝑁 𝑑𝑑 β‰₯ 𝑛𝑛 + 1

= P(𝐸𝐸𝑛𝑛 ≀ 𝑑𝑑) βˆ’ P(𝐸𝐸𝑛𝑛+1 ≀ 𝑑𝑑)

= π‘’π‘’βˆ’πœ†πœ†π‘‘π‘‘ πœ†πœ†π‘‘π‘‘ 𝑛𝑛

𝑛𝑛! so 𝑁𝑁 𝑑𝑑 = 𝑛𝑛 has Poisson probability

P 𝑁𝑁 𝑑𝑑 β‰₯ 𝑛𝑛 = P 𝑁𝑁 𝑑𝑑 = 𝑛𝑛 +P(𝑁𝑁 𝑑𝑑 β‰₯ 𝑛𝑛 + 1)

Page 9: .edu Exponential Distribution and Poisson Processmetin/Probability/Folios/expoPois.pdfΒ Β· which is the Markov property. Continuity property – Right-continuous evolutionP lim 𝑑𝑑↓0

utda

llas.edu /~m

etinPage 9Thinning of a Poisson Process

𝑁𝑁1(𝑑𝑑) is a the number of type 1 events; 𝑁𝑁2(𝑑𝑑) is a the number of type 2 events. 𝑁𝑁1 𝑑𝑑 +𝑁𝑁2 𝑑𝑑 = 𝑁𝑁(𝑑𝑑) and so [𝑁𝑁1(𝑑𝑑) = 𝑛𝑛1 |𝑁𝑁(𝑑𝑑) = 𝑛𝑛] and [𝑁𝑁2(𝑑𝑑) = 𝑛𝑛2|𝑁𝑁(𝑑𝑑) = 𝑛𝑛] are dependent

– P 𝑁𝑁𝑖𝑖 𝑑𝑑 = 𝑛𝑛𝑖𝑖 𝑁𝑁 𝑑𝑑 = 𝑛𝑛 = P(𝐡𝐡𝑖𝑖𝑛𝑛 𝑛𝑛,𝐸𝐸𝑖𝑖 = 𝑛𝑛𝑖𝑖) for 𝑖𝑖 ∈ {1,2}– Illustration of dependence

Β» P 𝑁𝑁1 𝑑𝑑 = 𝑛𝑛1 𝑁𝑁 𝑑𝑑 = 𝑛𝑛 P 𝑁𝑁2 𝑑𝑑 = 𝑛𝑛2 𝑁𝑁 𝑑𝑑 = 𝑛𝑛 = P(𝐡𝐡𝑖𝑖𝑛𝑛 𝑛𝑛,𝐸𝐸1 = 𝑛𝑛1)P(𝐡𝐡𝑖𝑖𝑛𝑛 𝑛𝑛,𝐸𝐸2 = 𝑛𝑛2)Β» P 𝑁𝑁1 𝑑𝑑 = 𝑛𝑛1,𝑁𝑁2 𝑑𝑑 = 𝑛𝑛2 𝑁𝑁 𝑑𝑑 = 𝑛𝑛 = P 𝐡𝐡𝑖𝑖𝑛𝑛 𝑛𝑛,𝐸𝐸1 = 𝑛𝑛1 = P(𝐡𝐡𝑖𝑖𝑛𝑛 𝑛𝑛,𝐸𝐸2 = 𝑛𝑛2)

Conditional random variables are dependent; unconditional ones can still be independent, in 2 steps:

1. Unconditional random variable 𝑁𝑁1 𝑑𝑑 ∼ π‘ƒπ‘ƒπ‘œπ‘œ(πœ†πœ†πΈπΈ1𝑑𝑑). Similarly 𝑁𝑁2 𝑑𝑑 ∼ π‘ƒπ‘ƒπ‘œπ‘œ(πœ†πœ†πΈπΈ2𝑑𝑑) .– P 𝑁𝑁1 𝑑𝑑 = 𝑛𝑛1 = βˆ‘π‘›π‘›=𝑛𝑛1

∞ P 𝑁𝑁1 𝑑𝑑 = 𝑛𝑛1 𝑁𝑁 𝑑𝑑 = 𝑛𝑛 P 𝑁𝑁 𝑑𝑑 = 𝑛𝑛 = βˆ‘π‘›π‘›=𝑛𝑛1∞ P(𝐡𝐡𝑖𝑖𝑛𝑛 𝑛𝑛,𝐸𝐸1 = 𝑛𝑛1)P 𝑁𝑁 𝑑𝑑 = 𝑛𝑛

= βˆ‘π‘›π‘›=𝑛𝑛1∞ 𝐢𝐢𝑛𝑛1

𝑛𝑛 𝐸𝐸1𝑛𝑛1 𝐸𝐸2

π‘›π‘›βˆ’π‘›π‘›1π‘’π‘’βˆ’πœ†πœ†π‘‘π‘‘ πœ†πœ†π‘‘π‘‘ 𝑛𝑛

𝑛𝑛!= π‘’π‘’βˆ’πœ†πœ†π‘π‘1𝑑𝑑 πœ†πœ†π‘π‘1𝑑𝑑 𝑛𝑛1

𝑛𝑛1 !π‘’π‘’βˆ’πœ†πœ†π‘π‘2π‘‘π‘‘βˆ‘π‘›π‘›=𝑛𝑛1

∞ 1(π‘›π‘›βˆ’π‘›π‘›1)!

𝐸𝐸2πœ†πœ†π‘‘π‘‘ π‘›π‘›βˆ’π‘›π‘›1 = π‘’π‘’βˆ’πœ†πœ†π‘π‘1𝑑𝑑 πœ†πœ†π‘π‘1𝑑𝑑 𝑛𝑛1

𝑛𝑛1 !

2. Independence: 𝑁𝑁1 𝑑𝑑 βŠ₯ 𝑁𝑁2 𝑑𝑑– P(𝑁𝑁1(𝑑𝑑) = 𝑛𝑛1, 𝑁𝑁2(𝑑𝑑) = 𝑛𝑛2) = P(𝑁𝑁1(𝑑𝑑) = 𝑛𝑛1, 𝑁𝑁2(𝑑𝑑) = 𝑛𝑛2|𝑁𝑁(𝑑𝑑) = 𝑛𝑛1 + 𝑛𝑛2) P(𝑁𝑁(𝑑𝑑) = 𝑛𝑛1 + 𝑛𝑛2)

= P 𝑁𝑁1 𝑑𝑑 = 𝑛𝑛1 𝑁𝑁 𝑑𝑑 = 𝑛𝑛1 + 𝑛𝑛2 P 𝑁𝑁 𝑑𝑑 = 𝑛𝑛1 + 𝑛𝑛2 = 𝐢𝐢𝑛𝑛1𝑛𝑛1+𝑛𝑛2 𝐸𝐸1

𝑛𝑛1 𝐸𝐸2𝑛𝑛2π‘’π‘’βˆ’πœ†πœ†π‘‘π‘‘

πœ†πœ†π‘‘π‘‘ 𝑛𝑛1+𝑛𝑛2

(𝑛𝑛1+ 𝑛𝑛2)!

= π‘’π‘’βˆ’πœ†πœ†π‘π‘1π‘‘π‘‘πœ†πœ†πΈπΈ1𝑑𝑑 𝑛𝑛1

𝑛𝑛1!π‘’π‘’βˆ’πœ†πœ†π‘π‘2𝑑𝑑

πœ†πœ†πΈπΈ2𝑑𝑑 𝑛𝑛2

𝑛𝑛2!= P 𝑁𝑁1 𝑑𝑑 = 𝑛𝑛1 P(𝑁𝑁2(𝑑𝑑) = 𝑛𝑛2)

𝐸𝐸1 𝐸𝐸2 𝐸𝐸3 πΈπΈπ‘–π‘–βˆ’1 𝐸𝐸𝑖𝑖 𝐸𝐸𝑛𝑛

Some events are blue, some are green; with probabilities 𝐸𝐸1 , 𝐸𝐸2 and 𝐸𝐸1 + 𝐸𝐸2 = 1. All events occur according to Poisson process. Occurrence of green or of blue events according to thinning of Poisson process.

With probability 𝐸𝐸1

With probability 𝐸𝐸2Poisson process

Page 10: .edu Exponential Distribution and Poisson Processmetin/Probability/Folios/expoPois.pdfΒ Β· which is the Markov property. Continuity property – Right-continuous evolutionP lim 𝑑𝑑↓0

utda

llas.edu /~m

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? – Probability 𝐸𝐸1 is for each email to include a discount offer for horseback riding. Let 𝑁𝑁1 be the index of the first

email including a horseback riding offer. Then P(𝑁𝑁1 = 𝑛𝑛1) = 𝐸𝐸1 1βˆ’ 𝐸𝐸1 𝑛𝑛1βˆ’1,𝑁𝑁1 is a geometric random variable.

Ex: Livingsocial sends offers for 5 types of events with probabilities 𝐸𝐸1 + 𝐸𝐸2 + 𝐸𝐸3 + 𝐸𝐸4 + 𝐸𝐸5 = 1. What is the expected number of emails that the student has to receive to obtain a discount offer for each event type?

– Let 𝑁𝑁𝑖𝑖 be the index of the first email including a discount offer for event type 𝑖𝑖. – A particular realization can have [𝑁𝑁1 = 3,𝑁𝑁2 = 7,𝑁𝑁3 = 1,𝑁𝑁4 = 2,𝑁𝑁5 = 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

– In this realization, the number of emails to endure to obtain an offer for each event is max{3, 7, 1, 2, 5} = 7. – In general,

𝑁𝑁 = max{𝑁𝑁1,𝑁𝑁2,𝑁𝑁3,𝑁𝑁4,𝑁𝑁5} for dependent𝑁𝑁𝑖𝑖sis the number emails to obtain an offer for each event.

– We want E(𝑁𝑁), but 𝑁𝑁𝑖𝑖s are dependent even if they have marginal pmf’s that are geometric distributions.

– How to break the dependence?

3 4 1

25

Page 11: .edu Exponential Distribution and Poisson Processmetin/Probability/Folios/expoPois.pdfΒ Β· which is the Markov property. Continuity property – Right-continuous evolutionP lim 𝑑𝑑↓0

utda

llas.edu /~m

etinPage 11

Amount of Time until a Coupon for EachContinuous-time Analog

Suppose emails are sent at the rate of 1 per day. Number of emails 𝑁𝑁 𝑑𝑑 ∼ π‘ƒπ‘ƒπ‘œπ‘œ(1𝑑𝑑) with interevent time π‘Œπ‘Œπ‘–π‘– ∼ 𝐸𝐸π‘₯π‘₯πΈπΈπ‘œπ‘œ(1). 𝑁𝑁𝑖𝑖 𝑑𝑑 is the number of coupons of type 𝑖𝑖 obtained by time 𝑑𝑑. 𝑁𝑁𝑖𝑖 𝑑𝑑 ∼ π‘ƒπ‘ƒπ‘œπ‘œ(1𝐸𝐸𝑖𝑖𝑑𝑑) 𝑁𝑁𝑖𝑖 𝑑𝑑 is obtained from 𝑁𝑁(𝑑𝑑) via thinning so π‘΅π‘΅πŸπŸ βŠ₯ π‘΅π‘΅πŸπŸ βŠ₯ β‹― βŠ₯ π‘΅π‘΅πŸ“πŸ“ 𝑋𝑋𝑖𝑖 is the interevent time for 𝑁𝑁𝑖𝑖 𝑑𝑑 process. 𝑋𝑋𝑖𝑖 ∼ 𝐸𝐸π‘₯π‘₯πΈπΈπ‘œπ‘œ(1𝐸𝐸𝑖𝑖)

– 𝑋𝑋𝑖𝑖 is the time between two emails both containing coupon 𝑖𝑖 Amount of time until a coupon for each event

𝑋𝑋 = max 𝑋𝑋1,𝑋𝑋2,𝑋𝑋3,𝑋𝑋4,𝑋𝑋5 for π‘Ώπ‘ΏπŸπŸ βŠ₯ π‘Ώπ‘ΏπŸπŸ βŠ₯ β‹― βŠ₯ π‘Ώπ‘ΏπŸ“πŸ“. Relating the number of emails to the time of the emails.

– How many emails are received in 𝑋𝑋 amount of time? How much time does it take to receive 𝑁𝑁 emails?𝑁𝑁(𝑑𝑑 = 𝑋𝑋) or 𝑋𝑋 = βˆ‘π‘–π‘–=1𝑁𝑁 π‘Œπ‘Œπ‘–π‘–

– Proceeding in either way: E 𝑁𝑁 𝑋𝑋 = 1E 𝑋𝑋 = E(𝑋𝑋) or E 𝑋𝑋 = E π‘Œπ‘Œπ‘–π‘– E 𝑁𝑁 = 1E(𝑁𝑁), so E 𝑁𝑁 = E 𝑋𝑋 . To finish, we need E 𝑋𝑋 . First equality is from independence,

P 𝑋𝑋 ≀ 𝑑𝑑 = �𝑖𝑖=1

5

P 𝑋𝑋𝑖𝑖 ≀ 𝑑𝑑 =�𝑖𝑖=1

5

1 βˆ’ π‘’π‘’βˆ’π‘π‘π‘–π‘–π‘‘π‘‘

E 𝑋𝑋 = ∫0∞

P 𝑋𝑋β‰₯ 𝑑𝑑 𝑑𝑑𝑑𝑑 = οΏ½0

∞1βˆ’οΏ½

𝑖𝑖=1

5

1βˆ’ π‘’π‘’βˆ’π‘π‘π‘–π‘–π‘‘π‘‘ 𝑑𝑑𝑑𝑑

– For the number of event types approaching infinity, E 𝑋𝑋 β†’ ∞. For a single event type, E(𝑋𝑋) = 1.

Discrete-time:Number of emailsto obtain coupon 𝑖𝑖

Continuous-time:Amount of timeto obtain coupon 𝑖𝑖

3 4 1

25

𝑋𝑋3 𝑋𝑋4𝑋𝑋1𝑋𝑋5 𝑋𝑋2

Page 12: .edu Exponential Distribution and Poisson Processmetin/Probability/Folios/expoPois.pdfΒ Β· which is the Markov property. Continuity property – Right-continuous evolutionP lim 𝑑𝑑↓0

utda

llas.edu /~m

etinPage 12

Distribution of Event Times Conditioned on Number of Events

Recall from order statistics: {𝑋𝑋1,𝑋𝑋2, … ,𝑋𝑋𝑛𝑛} is an iid sequence, where 𝑋𝑋𝑖𝑖 ∼ 𝑓𝑓𝑋𝑋, and π‘Œπ‘Œπ‘–π‘– is the 𝑖𝑖th smallest element in the sequence. The joint pdf of π‘Œπ‘Œ = [π‘Œπ‘Œ1,π‘Œπ‘Œ2, … ,π‘Œπ‘Œπ‘›π‘›] is, for 𝑦𝑦1 ≀ 𝑦𝑦2 ≀ β‹― ≀ 𝑦𝑦𝑛𝑛,

π‘“π‘“π‘Œπ‘Œ(𝑦𝑦1,𝑦𝑦2, … ,𝑦𝑦𝑛𝑛) = 𝑛𝑛!�𝑖𝑖=1

𝑛𝑛

𝑓𝑓𝑋𝑋(𝑦𝑦𝑖𝑖)

Ex: What is the joint pdf of the order statistics vector π‘Œπ‘Œ if each of underlying random variable 𝑋𝑋𝑖𝑖 ∼ 𝐸𝐸(0,𝑑𝑑)? – The pdf of 𝑋𝑋𝑖𝑖 is 𝑓𝑓𝑋𝑋(π‘₯π‘₯) = πΌπΌπœ†πœ†βˆˆ(0,𝑑𝑑] /𝑑𝑑. Then the joint pdf of the order statistics vector π‘Œπ‘Œ is, for 0 < 𝑦𝑦1 < 𝑦𝑦2 < β‹― < 𝑦𝑦𝑛𝑛 < 𝑑𝑑,

π‘“π‘“π‘Œπ‘Œ 𝑦𝑦1,𝑦𝑦2, … ,𝑦𝑦𝑛𝑛 = 𝑛𝑛!�𝑖𝑖=1

𝑛𝑛1𝑑𝑑

=𝑛𝑛!𝑑𝑑𝑛𝑛

Regardless of the rate of Poisson, the conditional joint pdf of ([𝐸𝐸1,𝐸𝐸2, … , 𝐸𝐸𝑛𝑛]|𝑁𝑁(𝑑𝑑) = 𝑛𝑛) is

𝑓𝑓[𝑆𝑆1 ,𝑆𝑆2 ,…,𝑆𝑆𝑛𝑛]|𝑁𝑁(𝑑𝑑)=𝑛𝑛 𝑠𝑠1,𝑠𝑠2, … , 𝑠𝑠𝑛𝑛 =𝑛𝑛!𝑑𝑑𝑛𝑛

for 0 < 𝑠𝑠1 < β‹― < 𝑠𝑠𝑛𝑛 < 𝑑𝑑.

– First note [𝐸𝐸1 = 𝑠𝑠1,𝐸𝐸2 = 𝑠𝑠2, … , 𝐸𝐸𝑛𝑛 = 𝑠𝑠𝑛𝑛,𝑁𝑁(𝑑𝑑) = 𝑛𝑛] iff [𝑋𝑋1 = 𝑠𝑠1,𝑋𝑋2 = 𝑠𝑠2 βˆ’ 𝑠𝑠1, … ,𝑋𝑋𝑛𝑛 = 𝑠𝑠𝑛𝑛 βˆ’ π‘ π‘ π‘›π‘›βˆ’1,𝑋𝑋𝑛𝑛+1 β‰₯ 𝑑𝑑 βˆ’ 𝑠𝑠𝑛𝑛].

– Hence,𝑓𝑓𝑆𝑆1 ,𝑆𝑆2,…,𝑆𝑆𝑛𝑛 ,𝑁𝑁(𝑑𝑑) 𝑠𝑠1,𝑠𝑠2, … , 𝑠𝑠𝑛𝑛,𝑛𝑛 = 𝑓𝑓𝑋𝑋 𝑠𝑠1 𝑓𝑓𝑋𝑋 𝑠𝑠2βˆ’ 𝑠𝑠1 … 𝑓𝑓𝑋𝑋 𝑠𝑠𝑛𝑛 βˆ’ π‘ π‘ π‘›π‘›βˆ’1 1βˆ’πΉπΉπ‘‹π‘‹ π‘‘π‘‘βˆ’ 𝑠𝑠𝑛𝑛

= πœ†πœ† π‘’π‘’βˆ’πœ†πœ†π‘ π‘ 1πœ†πœ†π‘’π‘’βˆ’πœ†πœ† 𝑠𝑠2βˆ’π‘ π‘ 1 β€¦πœ†πœ†π‘’π‘’βˆ’πœ†πœ† π‘ π‘ π‘›π‘›βˆ’π‘ π‘ π‘›π‘›βˆ’1 π‘’π‘’βˆ’πœ†πœ† π‘‘π‘‘βˆ’π‘ π‘ π‘›π‘› = πœ†πœ†π‘›π‘›π‘’π‘’βˆ’πœ†πœ†π‘‘π‘‘.– Then going forward mechanically

𝑓𝑓[𝑆𝑆1 ,𝑆𝑆2,…,𝑆𝑆𝑛𝑛]|𝑁𝑁(𝑑𝑑)=𝑛𝑛 𝑠𝑠1,𝑠𝑠2, … , 𝑠𝑠𝑛𝑛 =𝑓𝑓𝑆𝑆1 ,𝑆𝑆2 ,…,𝑆𝑆𝑛𝑛 ,𝑁𝑁 𝑑𝑑 𝑠𝑠1,𝑠𝑠2, … , 𝑠𝑠𝑛𝑛,𝑛𝑛

P 𝑁𝑁 𝑑𝑑 = 𝑛𝑛=

πœ†πœ†π‘›π‘›π‘’π‘’βˆ’πœ†πœ†π‘‘π‘‘

(πœ†πœ†π‘‘π‘‘)π‘›π‘›π‘’π‘’βˆ’πœ†πœ†π‘‘π‘‘/𝑛𝑛!=𝑛𝑛!𝑑𝑑𝑛𝑛

.

From the last two examples, the conditional joint pdf of ([𝐸𝐸1,𝐸𝐸2, … , 𝐸𝐸𝑛𝑛]|𝑁𝑁(𝑑𝑑) = 𝑛𝑛) and the pdf of order statistics [π‘Œπ‘Œ1,π‘Œπ‘Œ2, … ,π‘Œπ‘Œπ‘›π‘›] with an underlying iid 𝑋𝑋𝑖𝑖 ∼ 𝐸𝐸𝑖𝑖(0,𝑑𝑑] have the same distribution.

𝐸𝐸𝑖𝑖 𝑁𝑁 𝑑𝑑 = 𝑛𝑛 = π‘Œπ‘Œπ‘–π‘–(𝐸𝐸1(0, 𝑑𝑑], … ,𝐸𝐸𝑛𝑛(0, 𝑑𝑑]) in distribution

Page 13: .edu Exponential Distribution and Poisson Processmetin/Probability/Folios/expoPois.pdfΒ Β· which is the Markov property. Continuity property – Right-continuous evolutionP lim 𝑑𝑑↓0

utda

llas.edu /~m

etinPage 13Expected Total Waiting Time for a Bus

Ex: A DART bus departs from UTD to grocery markets every 𝑇𝑇 minutes. o Passengers arrive according to Poisson process 𝑁𝑁(𝑑𝑑) with rate πœ†πœ† and wait for the departure.o E.g.,, the 𝑖𝑖th passenger that arrives over (0,𝑑𝑑] arrives at 𝐸𝐸𝑖𝑖 and waits for 𝑇𝑇 βˆ’ 𝐸𝐸𝑖𝑖. o The expected total waiting time is

E �𝑖𝑖=1

𝑁𝑁 𝑇𝑇

𝑇𝑇 βˆ’ 𝐸𝐸𝑖𝑖 = E E �𝑖𝑖=1

𝑁𝑁 𝑇𝑇

𝑇𝑇 βˆ’ 𝐸𝐸𝑖𝑖 |𝑁𝑁 𝑇𝑇 = 𝑛𝑛 = �𝑛𝑛=0

∞

E �𝑖𝑖=1

𝑁𝑁 𝑇𝑇

𝑇𝑇 βˆ’ 𝐸𝐸𝑖𝑖 |𝑁𝑁 𝑇𝑇 = 𝑛𝑛(πœ†πœ†π‘‡π‘‡)π‘›π‘›π‘’π‘’βˆ’πœ†πœ†π‘‡π‘‡

𝑛𝑛!

= �𝑛𝑛=0

∞

E �𝑖𝑖=1

𝑛𝑛

𝑇𝑇 βˆ’ π‘Œπ‘Œπ‘–π‘– 𝐸𝐸1 0, 𝑇𝑇 , … ,𝐸𝐸𝑛𝑛 0,𝑇𝑇 P 𝑁𝑁 𝑇𝑇 = 𝑛𝑛

= �𝑛𝑛=0

∞

�𝑖𝑖=1

𝑛𝑛

E 𝑇𝑇 βˆ’ π‘Œπ‘Œπ‘–π‘– (𝐸𝐸1(0, 𝑇𝑇], … ,𝐸𝐸𝑛𝑛(0,𝑇𝑇]) P(𝑁𝑁 𝑇𝑇 = 𝑛𝑛)

= �𝑛𝑛=0

∞

𝑛𝑛𝑇𝑇 βˆ’ E �𝑖𝑖=1

𝑛𝑛

π‘Œπ‘Œπ‘–π‘– 𝐸𝐸1 0, 𝑇𝑇 , … ,𝐸𝐸𝑛𝑛 0, 𝑇𝑇 P(𝑁𝑁 𝑇𝑇 = 𝑛𝑛)

= �𝑛𝑛=0

∞

𝑛𝑛𝑇𝑇 βˆ’ 𝑛𝑛𝑇𝑇2 P 𝑁𝑁 𝑇𝑇 = 𝑛𝑛 =

12 𝑇𝑇𝑇 𝑁𝑁 𝑇𝑇 =

𝟏𝟏𝟐𝟐 𝑻𝑻𝑻𝑻𝑻𝑻

πœ†πœ†π‘‡π‘‡

𝑇𝑇

Arrival time 𝐸𝐸𝑖𝑖|𝑁𝑁(𝑑𝑑) = π‘Œπ‘Œπ‘–π‘–(𝐸𝐸1(0,𝑇𝑇], … ,𝐸𝐸𝑛𝑛(0,𝑇𝑇])

Waiting time

Area of the triangleis the expected total

waiting time

Sum of order statistics = Sum of sample

E �𝑖𝑖=1

𝑛𝑛

π‘Œπ‘Œπ‘–π‘– = E �𝑖𝑖=1

𝑛𝑛

𝐸𝐸𝑖𝑖

= 𝑛𝑛𝑇 𝐸𝐸𝑖𝑖 = 𝑛𝑛𝑇𝑇2

Page 14: .edu Exponential Distribution and Poisson Processmetin/Probability/Folios/expoPois.pdfΒ Β· which is the Markov property. Continuity property – Right-continuous evolutionP lim 𝑑𝑑↓0

utda

llas.edu /~m

etinPage 14

Number of calls arriving to a call centerin 15 minute time intervals

Generalizations

Nonhomogeneous Poisson process.– Non-stationary Stationary Increments– Time dependent event rate πœ†πœ†(π‘₯π‘₯) at time π‘₯π‘₯– The expected number of events over (0,𝑑𝑑] is

οΏ½0

π‘‘π‘‘πœ†πœ† π‘₯π‘₯ 𝑑𝑑π‘₯π‘₯

– The number of events during (0,𝑑𝑑] has

π‘ƒπ‘ƒπ‘œπ‘œ οΏ½0

π‘‘π‘‘πœ†πœ† π‘₯π‘₯ 𝑑𝑑π‘₯π‘₯

– Independent increments?– To model varying traffic intensity

Compound Poisson process: A Poisson process 𝑁𝑁(𝑑𝑑) and an iid sequence {π‘Œπ‘Œπ‘–π‘–} can be combined to obtain the compound Poisson process:

𝑋𝑋(𝑑𝑑) = �𝑖𝑖=1

𝑁𝑁 𝑑𝑑

π‘Œπ‘Œπ‘–π‘–

– Independent increments?– To model batch arrivals of customers to a store (or buying multiple units at once)

Earlymorning

Earlyevening

Page 15: .edu Exponential Distribution and Poisson Processmetin/Probability/Folios/expoPois.pdfΒ Β· which is the Markov property. Continuity property – Right-continuous evolutionP lim 𝑑𝑑↓0

utda

llas.edu /~m

etinPage 15Summary

Continuous-time Markov Process Poisson Process Thinning Conditioning on the Number of Events Generalizations


Fry er Folios - University of Queensland313592/ff5_1...Fry er JUNE 2010 Folios 2 UQ LIBRARY Fryer Folios is published twice a year by The University of Queensland Library to illustrate

Fry er Folios - University of Queensland313592/ff5_1...Fry er JUNE 2010 Folios 2 UQ LIBRARY Fryer Folios is published twice a year by The University of Queensland Library to illustrate

Content Folios Installation and Administration Guide...Content Folios Installation and Administration Guide 1-1 Introduction Chapter 1.INTRODUCTION OVERVIEW Content Folios provides

Content Folios Installation and Administration Guide...Content Folios Installation and Administration Guide 1-1 Introduction Chapter 1.INTRODUCTION OVERVIEW Content Folios provides

Initial Program Approval Training for MODIFIED FOLIOSΒ Β· Initial Program Approval Training for MODIFIED FOLIOS . ... β€’ English for Speakers of Other Languages ... β€’ Request to

Initial Program Approval Training for MODIFIED FOLIOSΒ Β· Initial Program Approval Training for MODIFIED FOLIOS . ... β€’ English for Speakers of Other Languages ... β€’ Request to

Codex Buranus Folios 57-72

Codex Buranus Folios 57-72

Collaborator Folios

Collaborator Folios

Phys 401 Spring 2020 Lecture # 23 1 April, 2020€¦ · The probability to be in the left well vanishes at time 𝑑𝑑= πœ‹πœ‹/2βˆ†π’Š. It then increases to unity at time 𝑑𝑑=

Phys 401 Spring 2020 Lecture # 23 1 April, 2020€¦ · The probability to be in the left well vanishes at time 𝑑𝑑= πœ‹πœ‹/2βˆ†π’Š. It then increases to unity at time 𝑑𝑑=

Irving Berlin Folios

Irving Berlin Folios

VIVA CREATIVE LITHOGRAPHERS FOLIOS

VIVA CREATIVE LITHOGRAPHERS FOLIOS

FOLIOS ACADEMICOS.pdf

FOLIOS ACADEMICOS.pdf

Folios of "The Power of Crystals"

Folios of "The Power of Crystals"

Total = 06 Folios - sistemas.istpcsr.edu.pe

Total = 06 Folios - sistemas.istpcsr.edu.pe

Content Folios

Content Folios

Differential Deep Learning on Graphs and its Applications...2020/02/07 Β Β· A model of dynamics on graphs. 𝑑𝑑𝑑𝑑(𝑑𝑑) 𝑑𝑑𝑑𝑑. = 𝑓𝑓(𝑋𝑋𝑑𝑑,𝐺𝐺,πœƒπœƒ,𝑑𝑑)

Differential Deep Learning on Graphs and its Applications...2020/02/07 Β Β· A model of dynamics on graphs. 𝑑𝑑𝑑𝑑(𝑑𝑑) 𝑑𝑑𝑑𝑑. = 𝑓𝑓(𝑋𝑋𝑑𝑑,𝐺𝐺,πœƒπœƒ,𝑑𝑑)

New York State Common Core 6 Mathematics Curriculum...can evaluate the formula for 𝑑𝑑: 𝑑𝑑= 4(12) + 3, 𝑑𝑑= 48 + 3, 𝑑𝑑= 51. Students continue to evaluate given

New York State Common Core 6 Mathematics Curriculum...can evaluate the formula for 𝑑𝑑: 𝑑𝑑= 4(12) + 3, 𝑑𝑑= 48 + 3, 𝑑𝑑= 51. Students continue to evaluate given

de Evaluacion Inv Publica 01-2015.pdfΒ Β· NΓΊmero de folios Proponente 1: Trescientos veinticinco (325) folios (dos sobres: sobre 1, 314 folios, sobre 2, 11 folios) El dΓ­a Catorce

de Evaluacion Inv Publica 01-2015.pdfΒ Β· NΓΊmero de folios Proponente 1: Trescientos veinticinco (325) folios (dos sobres: sobre 1, 314 folios, sobre 2, 11 folios) El dΓ­a Catorce

Student e folios

Student e folios

FOLIOS Kaufman_ortograΓ­a perspectiva constructivista

FOLIOS Kaufman_ortograΓ­a perspectiva constructivista

Sharada Loose Folios

Sharada Loose Folios

Guitar Sheet Music and Mixed Folios - Alfred

Guitar Sheet Music and Mixed Folios - Alfred

Copyright Folios

Copyright Folios