Modeling Rare Events

Modeling Rare Events Chapter 4 Lecture 15 Yiren Ding Shanghai Qibao Dwight High School April 24, 2016 Yiren Ding Modeling Rare Events 1 / 48

Outline 1 The Poisson Process Three Properties Stronger Property 2 Exponential Distribution Exponential CDF Memoryless Property 3 Clustering Effect Exponential Density Clustering Effect Explained 4 Simulating Poisson Process Uniform Distribution Poisson Process Algorithm 5 Merging and Splitting of Poisson Process Mozart-Requiem Lacrimosa Yiren Ding Modeling Rare Events 2 / 48

The Poisson Process Three Properties Definition 1 (The Poisson Process). Consider the occurrence of a rare event as customers arriving at a facility. Suppose the number of potential customers is unlimited, and the customers act independently. The arrival is a Poisson process if A the customers arrive one at a time B the numbers of arrivals during non-overlapping time intervals are independent of one another C the probability mass function of the number of arrivals in any time interval (s, s + t) depends only on the length t of the interval and not on its position s on the time axis Examples of Poisson process include: - the emission of particles from a radioactive source - the occurrence of serious earthquakes, fires, power outages - the arrival of urgent calls to an emergency center Yiren Ding Modeling Rare Events 3 / 48

The Poisson Process Stronger Property Theorem 1 (Stronger Version of Property C). In a Poisson process, the number of arrivals during any given time interval has a Poisson distribution of which the expected value is proportional to the duration of the interval, i.e., P(k arrivals during a given time interval of duration t) αt (αt)k = e k! for k = 0, 1,... (1) where α is the arrival intensity representing the expected number of arrivals during a given time interval of unit length. Proof. First we show that the number of arrivals during any time interval of unit length has a Poisson distribution Po(α). To do that we partition the time interval of length 1 into n subintervals of equal length t = 1 n, where n is sufficiently large. Yiren Ding Modeling Rare Events 4 / 48

The Poisson Process Stronger Property Theorem 1 proof The numbers of arrivals in the n subintervals are i.i.d. by properties B and C; hence, denote by p the probability that precisely one customer will arrive in each of the n subintervals. Each subinterval can be viewed as an independent Bernoulli trial with success probability p, so for a fixed n, the number of arrivals follows a binomial distribution B(n, p). (It is Bernoulli because the probability of having two or more customers arriving is negligible for large n.) As n, we have p 0, (since np = α is fixed). It follows that the number of arrivals during an interval of unit length has a Poisson distribution with the expected value np = α. Therefore, p = α/n = α t. This means that the probability that precisely one customer will arrive in any time interval is proportional to the duration of the time interval by a factor of α. Yiren Ding Modeling Rare Events 5 / 48

The Poisson Process Stronger Property Theorem 1 proof We can repeat the same process with arbitrary time interval of duration t with n = t t, and p = α t. If follows by similar argument that the number of arrivals during an interval of length t has a Poisson distribution with expected value λ = np = t α t = αt. t Thus we may replace condition C with this stronger version. Finally note that condition B is critical for the Poisson process and cannot be satisfied unless the population of customers is very large. This explains why a Poisson process can be used to describe the emission of a particle by a radioactive source with many atoms, which act independently and decay with a very small probability. Yiren Ding Modeling Rare Events 6 / 48

Exponential Distribution Exponential CDF Relationship with the exponential distribution It is a shame to have studied the Poisson distribution without knowing its continuous counterpart: the exponential distribution. In a Poisson arrival process, the number of arrivals during a given time interval is a discrete random variable, but the time between two successive arrivals can take on any positive value and is thus a so-called continuous random variable. Denote by T the random variable representing the waiting time between two successive arrivals in a Poisson process with arrival intensity α. Then T has the cumulative distribution function, P(T t) = 1 e αt, for t 0. (2) It should not surprise you that E(T ) = 1 α. (Why?) Yiren Ding Modeling Rare Events 7 / 48

Exponential Distribution Exponential CDF Effect of α on Exponential CDF Yiren Ding Modeling Rare Events 8 / 48

Exponential Distribution Exponential CDF Intuitive proof of (2) We show that P(T > t) = e αt, where α is the arrival intensity. Partition the positive side of the t-axis by 0, t, 2 t, 3 t,..., where t > 0 is fixed. (It helps to consider t as seconds.) Denote by X i the number of arrivals in the ith time interval. By Theorem 1, each X i follows the Poisson distribution Po(α t). For any t > 0, let n = t t, and since lim t t t 0 t = t, P(T > t) = lim P(T > n t) = lim [P(X i = 0)] n t 0 n = lim t 0 (e α t ) t t = e αt. It follows that the CDF of the exponential distribution is P(T t) = 1 e αt. Yiren Ding Modeling Rare Events 9 / 48

Exponential Distribution Memoryless Property Theorem 2 (Memoryless Property). The exponential distribution has an amazing memoryless property: that is, for every fixed point in time, the waiting period from that point until the first arrival after that point has the same exponential distribution as the inter-arrival times, regardless of how long it has been since the last client arrived before that point in time. Symbolically, this is P(T > s + t T > s) = P(T > t) for all t > 0. (3) Proof. From (2) we have P(T > t) = e αt. Using the definition of conditional probability we have P(T > s + t T > s) = P(T > s + t) P(T > s) = e αt = P(T > t). = e α(s+t) e αs Yiren Ding Modeling Rare Events 10 / 48

Exponential Distribution Memoryless Property Example 1. Out in front of Central Station, multiple-passenger taxicabs wait until either they have acquired four passengers or a period of ten minutes have passed since the first passenger stepped into the cab. Passengers arrive according to a Poisson process with an average of one passenger every three minutes. (a) You are the first passenger to get into a cab. What is the probability that you will have to wait ten minutes before the cab gets underway? (b) You were the first passenger to get into a cab and you have been waiting there for five minutes. In the meantime, two other passengers have entered the cab. What is the probability that you will have to wait another five minutes before the cab gets underway? Solution. First note that the arrival intensity of this Poisson process is α = 1 3, the expected number of arrivals per minute. Yiren Ding Modeling Rare Events 11 / 48

Exponential Distribution Memoryless Property Example 1 solution To answer part (a), we use Theorem 1 to conclude that the number of arrivals during ten minutes is Poisson distributed with expected value 10α = 10/3. Hence, P(you must wait ten minutes) = P(0, 1, or 2 passengers arrive within the next ten minutes) = e (10/3) (10/3) (10/3)1 (10/3) (10/3)2 + e + e 1! 2! = 0.3528. To answer part (b), we use the memoryless property of the Poisson process. The waiting period before the arrival of the next passenger is exponentially distributed with an expected value of 1/α = 3 minutes, regardless of how long you have waited! Thus the desired probability is e 5α = e 5/3 = 0.1889. Yiren Ding Modeling Rare Events 12 / 48

Clustering Effect Clustering Effect of Arrival Times Figure above gives simulated arrival times in the time interval (0, 45) for a Poisson process with arrival intensity α = 1. It is clear that customer arrival times reveal a strong tendency to cluster. To see why, we need to learn more about the exact shape of the exponential distribution. Yiren Ding Modeling Rare Events 13 / 48

Clustering Effect Exponential Density Exponential Density Let T denote the inter-arrival time. We know that T is a continuous random variable with a CDF P(T t) = 1 e αt, α > 0. Unlike a discrete random variable, which has a probability mass function, a continuous variable does not have a mass because P(T = t) = 0 for any t R +. (We will prove that later.) Assume that we can only measure inter-arrival time with accuracy up to seconds (unit 1); thus we have a new discrete random variable T with the following probability mass function: P(T = t) = P(t 1 < T t), for t = 1, 2,... It make perfect sense to define it this way because, for example, a measurement of 5 seconds may refer to any of the values in the interval (4, 5] due to rounding errors. Yiren Ding Modeling Rare Events 14 / 48

Clustering Effect Exponential Density Exponential Density Since the event {T t} is the union of events {t 1 < T t} and {T t 1}, and using the linearization e x 1 + x for small x, P(T = t) = P(t 1 < T t) = P(T t) P(T t 1) = (1 e αt ) (1 e α(t 1) ) = (e α 1)e αt αe αt Let T (n) denote inter-arrival time measured up to 1 n of a second. This discrete random variable will have the probability mass function P(T (n) = t) = (e α/n 1)e αt 1 n αe αt. A curious reader should already see that αe αt is common in both expressions. This is the exponential density function f (t). Yiren Ding Modeling Rare Events 15 / 48

Clustering Effect Exponential Density Exponential Density From physics, you know that Density = In probability, there is a similar version: Mass Volume. Probability Density = Probability Mass Interval Length. Indeed, we see that P(T (n) = t) lim = αe αt = f (t). n 1/n A formal definition of probability density will be presented later. Yiren Ding Modeling Rare Events 16 / 48

Clustering Effect Exponential Density Probability Density Visualized Yiren Ding Modeling Rare Events 17 / 48

Clustering Effect Exponential Density Probability Density Visualized Yiren Ding Modeling Rare Events 18 / 48

Clustering Effect Exponential Density Probability Density Visualized Yiren Ding Modeling Rare Events 19 / 48

Clustering Effect Exponential Density Probability Density Visualized Yiren Ding Modeling Rare Events 20 / 48

Clustering Effect Exponential Density Probability Density Visualized Yiren Ding Modeling Rare Events 21 / 48

Clustering Effect Exponential Density Probability Density Visualized Yiren Ding Modeling Rare Events 22 / 48

Clustering Effect Exponential Density Probability Density Visualized Yiren Ding Modeling Rare Events 23 / 48

Clustering Effect Exponential Density Probability Density Visualized Yiren Ding Modeling Rare Events 24 / 48

Clustering Effect Exponential Density Probability Density Visualized Yiren Ding Modeling Rare Events 25 / 48

Clustering Effect Exponential Density Probability Density Visualized Yiren Ding Modeling Rare Events 26 / 48

Clustering Effect Exponential Density Probability Density Visualized Yiren Ding Modeling Rare Events 27 / 48

Clustering Effect Exponential Density Probability Density Visualized Yiren Ding Modeling Rare Events 28 / 48

Clustering Effect Exponential Density Probability Density Visualized Note that the heights of the bars in the previous graphs do not represent the probability; instead, the areas of the bars do! This begins to resemble a Riemann sum. Indeed, we have P(T t) = t 0 f (x) dx = Hence it should be intuitively clear that t 0 αe αx dx = ( e αx + C) t 0 = 1 e αt. P(s < T t) = t s f (x) dx. The value of f (t) at a single point is not a probability! Yiren Ding Modeling Rare Events 29 / 48

Clustering Effect Exponential Density Effect of α on Exponential Density Yiren Ding Modeling Rare Events 30 / 48

Clustering Effect Clustering Effect Explained Clustering Effect Explained Denote by F (t) = 1 e αt the cumulative distribution function. Have you noticed the remarkable fact that the probability density function is simply the derivative of the CDF? That is, F (t) = d dt (1 e αt ) = αe αt = f (t). By the definition of the derivative (hope you haven t forgotten), This implies that f (t) = lim t 0 F (t + t) F (t). t P(t < T t + t) f (t) t, for t small. Yiren Ding Modeling Rare Events 31 / 48

Clustering Effect Clustering Effect Explained Clustering Effect Explained From its shape, we know that f (t) is largest when t = 0. Hence for a fixed, small value t, the point at which the probability of T being close to t, P(t < T t + t), is largest is at t = 0. This explains the clustering effect because a relatively shorter inter-arrival times occur more frequently than longer ones! Suppose that each day there is a 1% chance of someone being murdered in Shanghai. The occurrence of murder can be modeled by a Poisson process with arrival intensity 3.65 murders per year. Over a period of, say twenty years, the probability of having nine or more murders in a 12-month period is approximately 60%. This contrasts with the probability of nine or more murders in a given 12-month period of only 0.0127. Yiren Ding Modeling Rare Events 32 / 48

Clustering Effect Clustering Effect Explained Example 2. In a given city, traffic accidents occur according to a Poisson process with an average of α = 10 accidents per week. In a certain week, seven accidents have occurred. What is the probability that exactly one accident has occurred on each day of that week? Can you explain beforehand why this probability must be small? Solution. Define random variables X 1, X 2,..., X 7 where X i denote the number of accidents occurring in day i, and let X = 7 i=1 X i. The probability we are looking for is P(X 1 = 1,..., X 7 = 1 X = 7). (4) By properties B and C of the Poisson process and Theorem 1, the X i s are i.i.d. Poisson random variables with expected value α/7, and X is Poisson distributed with expected value α. Yiren Ding Modeling Rare Events 33 / 48

Clustering Effect Clustering Effect Explained Example 2 solution By the definition of conditional probability, (4) is (P(X 1 ) = 1) (P(X 7 ) = 1) P(X = 7) = ( ) e α 7 7 α 7 e α α 7 7! = 7! 7 7 = 0.0162. This is indeed a small probability. The tendency of Poisson arrivals to cluster explains why this probability is so small. Incidentally, this is the same as the probability of independently drawing seven random numbers and have exactly one random number in each of the seven intervals (0, 1 7 ), ( 1 7, 2 7 ),..., ( 6 7, 1). Thus there is a close relationship between the Poisson arrival process and the uniform distribution, which can be used to simulate the Poisson process in higher-dimensions. (e.g. Stars in the sky) Yiren Ding Modeling Rare Events 34 / 48

Clustering Effect Clustering Effect Explained Randomly Generated Points in 2D Yiren Ding Modeling Rare Events 35 / 48

Clustering Effect Clustering Effect Explained Randomly Generated Points in 3D Yiren Ding Modeling Rare Events 36 / 48

Simulating Poisson Process Uniform Distribution Uniform Distribution A discrete random variable X is said to have a uniform distribution if the probability of all values in the range of X is identical. For example, rolling a fair die yields a uniform distribution, P(i) = 1, i = 1,..., 6. 6 Likewise, we can extend this to the continuous case: A continuous random variable X is said to have a uniform density over the interval (a, b) if its probability density function is given by f (x) = 1 b a for a < x < b (5) By Axiom 2, P(X (a, b)) = 1, which is a mass over the whole interval (b a). By uniformity, the density must be 1/(b a). Yiren Ding Modeling Rare Events 37 / 48

Simulating Poisson Process Uniform Distribution Uniform Distribution Yiren Ding Modeling Rare Events 38 / 48

Simulating Poisson Process Poisson Process Algorithm Example 3. Let the random variable be given by T = 1 α ln(u), where U is a random number between (0, 1) and α a positive number. Show that T is exponentially distributed. That is, P(T t) = 1 e αt. First note that T is a positive random variable. For any t > 0, P(T t) = P ( 1α ) ln(u) t = P(ln(U) αt) = P(U e αt ) = 1 P(U e αt ). Since U is uniform in (0, 1), P(U e αt ) = e αt P(T t) = 1 e αt. 0 1 1 dx = e αt, so Hence T is exponentially distributed with expected value 1 α. Yiren Ding Modeling Rare Events 39 / 48

Simulating Poisson Process Poisson Process Algorithm Simulating a Poisson Process Example 3 gives a perfect algorithm to simulate Poisson arrivals. (1) Generate a random number u between 0 and 1. (2) Take t = 1 α ln(u) as the inter-arrival time. The following Python code generates 20 arrivals with α = 1. import matplotlib.pyplot as plt import numpy as np from math import log n, alpha, x, X = 20, 1, 0, [] for i in range(n): u = np.random.random() x = x - (1/alpha)*log(u) X.append(x) plt.plot(x, len(x) * [0], "x") plt.show() Yiren Ding Modeling Rare Events 40 / 48

Simulating Poisson Process Poisson Process Algorithm Simulation Results Yiren Ding Modeling Rare Events 41 / 48

Simulating Poisson Process Poisson Process Algorithm Power of Simulation The previous algorithm is extremely powerful in approximating solutions to problems are not easy to solve analytically. Consider the problem we encountered before: Assume that for each day there is a 1% probability of someone being murdered. What is the probability that somewhere within a time frame of ten years there will be one 12-month period containing nine or more murders? What about twenty years? It is reasonable to model the occurrence of murder by a Poisson process with an arrival intensity α = p/ t = 3.65 murders per year. This problem is ridiculous to solve using analytical method, but using Python we found that the former probability is around 36%, and the latter probablity is around 60%. Sometimes in real life, having an approximation is good enough! Yiren Ding Modeling Rare Events 42 / 48

Simulating Poisson Process Poisson Process Algorithm Python Code import numpy as np from math import log numtrials, count = 10000, 0 def PoissonProcess(alpha, time): x, X = 0, [] while x <= time: u = np.random.random() x = x - (1/alpha)*log(u) X.append(x) return X for num in range(numtrials): dupe = False X = PoissonProcess(3.65, 10) for i in range(len(x)-8): if X[i+8]-X[i] <= 1: dupe = True if dupe == True: count += 1 prob = count / numtrials print(prob) Yiren Ding Modeling Rare Events 43 / 48

Merging and Splitting of Poisson Process Merging and Splitting of Poisson Process Consider a call center that receive emergency phone calls from two different companies A and B. Calls from company A and calls from company B are independent Poisson processes with arrival intensities α A and α B, respectively. Claim 1: The merging of these two arrival processes gives a Poisson process with arrival intensity α A + α B. Now consider the occurrence of any magnitude of earthquake is a Poisson process with arrival intensity α. There is the probability p of any earthquake being a high-magnitude one, and 1 p being a low-magnitude one. Claim 2: The occurrences of high-magnitude and low-magnitude earthquakes are two independent Poisson processes with arrival intensities αp, and α(1 p), respectively! Yiren Ding Modeling Rare Events 44 / 48

Merging and Splitting of Poisson Process Example 4. A piece of radioactive material emits alpha-particles according to a Poisson process with an intensity of 0.84 particle per second. A counter detects each emitted particle, independently, with probability of 0.95. In a 10-second period the number of detected particles is 12. What is the probability that more than 15 particles were emitted in that period? By Claim 2, the emission of undetected particle is an independent Poisson process with arrival intensity 0.84 0.05 = 0.0402. Denote by X number of undetected particles in the 10-second period. Then X Po(λ) where λ = 10 0.0402 = 0.402. The desired probability is then P(X > 3) = 1 3 k=0 0.402 (0.402)k e k! = 0.00079. Yiren Ding Modeling Rare Events 45 / 48

Merging and Splitting of Poisson Process Mozart-Requiem Lacrimosa Example 5 (Variant of Coupon Collector Problem). In the casino game of craps two dice are repeatedly rolled. What is the probability that a total of 2, 3, 4, 5, 6, 8, 9, 10, 11, and 12 will be rolled before rolling a seven? Beautiful Solution. A beautiful trick is to imagine the rolling of the two dice follow a Poisson process with arrival intensity α = 1. By Claim 2, this Poisson process can be split into 11 independent Poisson processes with arrival intensities αp j = p j, where p j is the probability of rolling a total of j in rolling of two dice. Fix t, and suppose that a roll of 7 occurs for the first time at time t. By independence of Poisson processes, the probability of rolling a total of 2, 3, 4, 5, 6, 8, 9, 10, 11 and 12 showing up in (0, t) given that the first appearance of 7 is at time t is f (t) = (1 e p 2t ) (1 e p 6t )(1 e p 8t ) (1 e p 12t ). Yiren Ding Modeling Rare Events 46 / 48

Merging and Splitting of Poisson Process Mozart-Requiem Lacrimosa Mozart-Requiem Lacrimosa The probability that 7 will occur for the first time in the very small interval (t, t + t) is approximately equal to p 7 e p 7t t, since p 7 e p 7t is the exponential density of rolling a 7. By the Law of Conditional Probability, the desired probability is just a Riemann sum that converges to an improper integral, lim t 0 f (i t)p 7 e p7i t t = i=0 0 f (t)p 7 e p 7t dt. An improper integral f (x) dx is defined as lim f (x) dx. 0 a 0 Plugging the various values of p j and using numerical integration give the value of 0.005258 for the desired probability. Uncontrollable tears of joy fall loudly on the keyboard. a Yiren Ding Modeling Rare Events 47 / 48

Merging and Splitting of Poisson Process Mozart-Requiem Lacrimosa Mozart-Requiem Lacrimosa In some casino, the payout of this craps is $175 for each $1 bet. This corresponds to a house edge of 1 (1 0.005258) 174 0.005258 = 7.99%. To cry harder, consider another game: You win if any total is rolled twice before a seven. A win pays 2 for 1. It is left as a homework exercise to show that the win probability is, 1 0 (e p 2t + p 2 te p 2t ) (e p 6t + p 6 te p 6t ) (e p 8t + p 8 te p 8t ) (e p 12t + p 12 te p 12t )p 7 e p 7t dt = 0.47107 This corresponds to a house edge of 1 (1 0.47107) 1 0.47107 = 5.79%. Yiren Ding Modeling Rare Events 48 / 48