Random Processes DS GA 1002 Probability and Statistics for Data - PowerPoint PPT Presentation

Poisson process N ( t n ) ( x 1 , . . . , x n ) p � N ( t 1 ) ,..., � � � N ( t 1 ) = x 1 , . . . , � � = P N ( t n ) = x n � � N ( t 1 ) = x 1 , � � N ( t 2 ) − � N ( t 1 ) = x 2 − x 1 , . . . , � N ( t n ) − � = P N ( t n − 1 ) = x n − x n − 1

Poisson process N ( t n ) ( x 1 , . . . , x n ) p � N ( t 1 ) ,..., � � � N ( t 1 ) = x 1 , . . . , � � = P N ( t n ) = x n � � N ( t 1 ) = x 1 , � � N ( t 2 ) − � N ( t 1 ) = x 2 − x 1 , . . . , � N ( t n ) − � = P N ( t n − 1 ) = x n − x n − 1 � � � � � � � N ( t 2 ) − � � N ( t n ) − � � = P N ( t 1 ) = x 1 N ( t 1 ) = x 2 − x 1 N ( t n − 1 ) = x n − x n − 1 P . . . P

Poisson process N ( t n ) ( x 1 , . . . , x n ) p � N ( t 1 ) ,..., � � � N ( t 1 ) = x 1 , . . . , � � = P N ( t n ) = x n � � N ( t 1 ) = x 1 , � � N ( t 2 ) − � N ( t 1 ) = x 2 − x 1 , . . . , � N ( t n ) − � = P N ( t n − 1 ) = x n − x n − 1 � � � � � � � N ( t 2 ) − � � N ( t n ) − � � = P N ( t 1 ) = x 1 N ( t 1 ) = x 2 − x 1 N ( t n − 1 ) = x n − x n − 1 P . . . P = p ( λ t 1 , x 1 ) p ( λ ( t 2 − t 1 ) , x 2 − x 1 ) . . . p ( λ ( t n − t n − 1 ) , x n − x n − 1 )

Poisson process ( λ = 0 . 2) 1.0 0.8 0.6 0.4 0.2 0.0 0 2 4 6 8 10 t

Poisson process ( λ = 0 . 2) 1.0 0.8 0.6 0.4 0.2 0.0 0 2 4 6 8 10 t

Poisson process ( λ = 0 . 2) 1.0 0.8 0.6 0.4 0.2 0.0 0 2 4 6 8 10 t

Poisson process ( λ = 1) 1.0 0.8 0.6 0.4 0.2 0.0 0 2 4 6 8 10 t

Poisson process ( λ = 1) 1.0 0.8 0.6 0.4 0.2 0.0 0 2 4 6 8 10 t

Poisson process ( λ = 1) 1.0 0.8 0.6 0.4 0.2 0.0 0 2 4 6 8 10 t

Poisson process ( λ = 2) 1.0 0.8 0.6 0.4 0.2 0.0 0 2 4 6 8 10 t

Poisson process ( λ = 2) 1.0 0.8 0.6 0.4 0.2 0.0 0 2 4 6 8 10 t

Poisson process ( λ = 2) 1.0 0.8 0.6 0.4 0.2 0.0 0 2 4 6 8 10 t

Call-center data ◮ Example: Data from a call center in Israel ◮ We compare the histogram of the number of calls received in an interval of 4 hours over 2 months and the pmf of a Poisson random variable fitted to the data

Call-center data 0.14 Real data 0.12 Poisson distribution 0.10 0.08 0.06 0.04 0.02 0.00 0 5 10 15 20 25 30 35 40 Number of calls

Poisson process Distribution of interarrival times? F T ( t )

Poisson process Distribution of interarrival times? F T ( t ) := P ( T ≤ t )

Poisson process Distribution of interarrival times? F T ( t ) := P ( T ≤ t ) = 1 − P ( T > t )

Poisson process Distribution of interarrival times? F T ( t ) := P ( T ≤ t ) = 1 − P ( T > t ) = 1 − P ( no events in an interval of length t )

Poisson process Distribution of interarrival times? F T ( t ) := P ( T ≤ t ) = 1 − P ( T > t ) = 1 − P ( no events in an interval of length t ) = 1 − e − λ t

Poisson process Distribution of interarrival times? F T ( t ) := P ( T ≤ t ) = 1 − P ( T > t ) = 1 − P ( no events in an interval of length t ) = 1 − e − λ t f T ( t ) = λ e − λ t Iid exponential sequence (allows to simulate Poisson process!)

Call-center data ◮ Example: Data from a call center in Israel ◮ We compare the histogram of the inter-arrival times between calls occurring between 8 pm and midnight over two days and the pdf of an exponential random variable fitted to the data

Call center 0.9 Exponential distribution 0.8 Real data 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0 1 2 3 4 5 6 7 8 9 Interarrival times (s)

Generating a Poisson process To sample from a Poisson random process with parameter λ we: 1. Generate independent samples from an exponential random variable with parameter λ t 1 , t 2 , t 3 , . . . 2. Set the events of the Poisson process to occur at t 1 , t 2 , t 3 , . . .

Mean and autocovariance � � � X ( t ) E

Mean and autocovariance � � � X ( t ) = λ t E

Mean and autocovariance � � � X ( t ) = λ t E R � X ( t 1 , t 2 ) = λ min { t 1 , t 2 }

Mean and autocovariance � � � X ( t ) = λ t E R � X ( t 1 , t 2 ) = λ min { t 1 , t 2 } The process is neither strictly nor wide-sense stationary

Earthquakes ◮ Earthquakes in San Francisco follow a Poisson process with parameter 0.3 earthquakes/year ◮ Probability of no earthquakes in the next 10 years and then at least 1 over the following 20 years?

Earthquakes � � X ( 10 ) = 0 , � � X ( 30 ) ≥ 1 P

Earthquakes � � � � X ( 10 ) = 0 , � � X ( 10 ) = 0 , � � X ( 30 ) − � X ( 30 ) ≥ 1 = P X ( 10 ) ≥ 1 P

Earthquakes � � � � X ( 10 ) = 0 , � � X ( 10 ) = 0 , � � X ( 30 ) − � X ( 30 ) ≥ 1 = P X ( 10 ) ≥ 1 P � � � � � X ( 30 ) − � � = P X ( 10 ) = 0 X ( 10 ) ≥ 1 P

Earthquakes � � � � X ( 10 ) = 0 , � � X ( 10 ) = 0 , � � X ( 30 ) − � X ( 30 ) ≥ 1 = P X ( 10 ) ≥ 1 P � � � � � X ( 30 ) − � � = P X ( 10 ) = 0 X ( 10 ) ≥ 1 P � � � � �� � X ( 30 ) − � � = P X ( 10 ) = 0 1 − P X ( 10 ) = 0

Earthquakes � � � � X ( 10 ) = 0 , � � X ( 10 ) = 0 , � � X ( 30 ) − � X ( 30 ) ≥ 1 = P X ( 10 ) ≥ 1 P � � � � � X ( 30 ) − � � = P X ( 10 ) = 0 X ( 10 ) ≥ 1 P � � � � �� � X ( 30 ) − � � = P X ( 10 ) = 0 1 − P X ( 10 ) = 0 = e − 3 � 1 − e − 6 � = 4 . 97 10 − 2

Random walk Process that evolves by taking steps in random directions Step sequence � Z is iid � with probability 1 + 1 � 2 S ( i ) = with probability 1 − 1 2 We define a random walk � X as � 0 for i = 0 � X ( i ) := � i j = 1 � S ( j ) for i = 1 , 2 , . . .

Random walk 6 4 2 0 2 4 6 0 2 4 6 8 10 12 14 i

Random walk 6 4 2 0 2 4 6 0 2 4 6 8 10 12 14 i

Random walk 6 4 2 0 2 4 6 0 2 4 6 8 10 12 14 i

First-order pmf X ( i ) ( x ) ? p � Distribution of number of positive steps S + ? Negative steps: S − = i − S + X ( i ) ( x ) p �

First-order pmf X ( i ) ( x ) ? p � Distribution of number of positive steps S + ? Negative steps: S − = i − S +   i �  �  X ( i ) ( x ) = P S ( i ) = x p � j = 0

First-order pmf X ( i ) ( x ) ? p � Distribution of number of positive steps S + ? Negative steps: S − = i − S +   i �  �  X ( i ) ( x ) = P S ( i ) = x p � j = 0 = P ( S + − S − = x )

First-order pmf X ( i ) ( x ) ? p � Distribution of number of positive steps S + ? Negative steps: S − = i − S +   i �  �  X ( i ) ( x ) = P S ( i ) = x p � j = 0 = P ( S + − S − = x ) = P ( 2 S + − i = x )

First-order pmf X ( i ) ( x ) ? p � Distribution of number of positive steps S + ? Negative steps: S − = i − S +   i �  �  X ( i ) ( x ) = P S ( i ) = x p � j = 0 = P ( S + − S − = x ) = P ( 2 S + − i = x ) � � S + = i + x = P 2

First-order pmf X ( i ) ( x ) ? p � Distribution of number of positive steps S + ? Negative steps: S − = i − S +   i �  �  X ( i ) ( x ) = P S ( i ) = x p � j = 0 = P ( S + − S − = x ) = P ( 2 S + − i = x ) � � S + = i + x = P 2 � i � 1 if i + x = is an integer between 0 and i i + x 2 i 2 2