MAS1302 Computational Probability and Statistics

Transcription

1 MAS1302 Computational Probability and Statistics April 23, 2008

2 3. Simulating continuous random behaviour 3.1 The Continuous Uniform U(0,1) Distribution We have already used this random variable a great deal: the Uniform distribution on the interval (0,1). This is denoted by U(0,1) and denotes the random variable which has probability density function (p.d.f.): { 1 0 < x < 1 f (x) = 0 otherwise

3 3.1 The U(0,1) Distribution f (x) x Up to now, we have used U(0,1) random variables to simulate from many discrete distributions, and taken the U(0,1) random variables for granted. In this chapter on continuous random behaviour, we will start by thinking how we simulate from this most simple of continuous distributions.

4 Example Many Casio calculators provide sequences of random numbers using the function SHIFT RAN# = : A typical sequnce of these might be: 0.909, 0.304, 0.966, 0.837, 0.151,... These are in fact, U(0,1) random numbers recorded to 3 decimal places. So how does the calculator generate them?

5 Solution (Slide 1 of 2) The key is in the accuracy to which the numbers are generated! Generating U(0,1) random variables precisely would be very difficult! (Impossible?!?) However we can get three decimal places by generating a random integer x from the set {0,1,2,...,999}, with all outcomes equally likely, and then put u = x/1000. Then u is a simulation from a U(0,1) random variable recorded to three decimals.

6 Solution (Slide 2 of 2) The simulation of x is easy, e.g. using a congruential generator with m = 1000 and maximum period, as we saw in Section 2.4. Some calculators give the Uniform random numbers to 10 decimal places rather than 3. This could be achieved with a congruential generator with m = 10 10, so simulating x from the set of integers: {0,1,2,..., }. Setting u = x/10 10 would give a pseudo random number from the U(0,1) distribution, recorded to 10 decimals.

7 As we will see, R will take care of all the details for us, and will supply many U(0,1) random variables to a good degree of accuracy very quickly. We have already seen how continuous U(0,1) random variables can be used to generate discrete probability distributions (Sections ). Soon we will see how U(0,1) random variables can be used to generate other continuous probability distributions. First, however, we will see how simulation from U(0,1) random variables can sometimes be useful in its own right.

8 3.2 Monte Carlo Methods Introduction The term Monte Carlo is used to decribe any simulation study which involves random numbers. The name is a reference to the famous Monte Carlo Casino in Monaco, where repetition of random events is the order of the day! What is a simulation study? For our purposes, a simulation study is any study where we study the properties of a system using random numbers. We have already seen several simulation studies in Section 2.11, when we verified some rules for probability distributions, and checked how well some data fitted a probability model.

9 Often, simulation studies involve estimating the probability of an event (e.g. in Example , the probabilities of each outcome were estimated). In general, suppose we do an experiment which has the event A as one possible outcome. We would like to estimate the probability of A, denoted by P(A). Then by repeatedly simulating the experiment, it is simple to estimate P(A), the probability of the event A, using P F (A), where: P F (A) = No. of times A occurs Number of times experiment simulated. Here P F (A) is so-called because it is a Frequency estimate of P(A).

10 Why does this work? Well suppose we denote the number of times we simulate the experiment by n, then P F (A) has the following important property: as n, then P F (A) P(A). This means that the larger our number of simulations, the more accurate will be our estimate of P(A).

11 Example Suppose U 1 and U 2 are two independent U(0,1) random variables. Let Y = U 1 + U 2. Then Y is a random variable on the interval (0,2). I.e. Y is a random variable which can take any value in the range (0,2). Question 1: What is P(Y < 1.5), the probability that Y is less than 1.5? Question 2: (harder) What is the probability density function (p.d.f.) of Y? Question 3: What is the cumulative distribution function (c.d.f.) of Y?

12 Monte Carlo solution to Question 1 (Slide 1 of 2) Let s formulate Question 1 as a Monte Carlo problem: Here A is the event {Y < 1.5}. We want P(A). Suppose we simulate n observations from the random variable Y, and we observe the number of occasions, r, that A happens, i.e. r = #{Y < 1.5}. Then our Monte Carlo estimate of P(A) is: P F (A) = r/n.

13 Monte Carlo solution to Question 1 (Slide 2 of 2) This was carried out in R with n = 1000, and we obtained r = 861. Hence we estimate: P F (A) = 861/1000 = 0.861, where P F (A) is our estimate of P(A) = P(Y < 1.5).

14 In this case, it is possible to answer Question 1 mathematically ( analytically ). This solution will be exactly right, so it will enable us to see how well we did with the Monte Carlo solution (see Problems Class).

15 Analytical solution to Question 1 (Slide 1 of 5) Recall Y = U 1 + U 2 where U 1 and U 2 are independent U(0,1) random variables. The distribution of Y is on the interval (0,2), but it is not uniform, and it is not obvious what it is. However, we can answer Question 1 by thinking geometrically!

16 Analytical solution to Question 1 (Slide 2 of 5) In the plot of U 2 against U 1 below, the shaded region is completely identified with the event A = {Y < 1.5}. U U 1 Given that the total area of the square is 1, the area of the shaded region is the probability we want!

17 Analytical solution to Question 1 (Slide 3 of 5) It is clear from the diagram that the answer is: P(A) = 7/8 = N.B. This answer is exactly right, so we can look back and see how well our Monte Carlo estimate P F (A) compared with the true value of P(A). We got as our estimate of Not bad!

18 Analytical solution to Question 1 (Slide 4 of 5) Note that the area of the shaded region was easy to calculate using common sense arguments, but it could also have been expressed as an integral. Consider the line which forms the upper boundary in the sketch. It has the equation 1 0 < u 1 1/2, u 2 = f (u 1 ) = 1.5 u 1 1/2 < u 2 < 1, 0 otherwise.

19 Analytical solution to Question 1 (Slide 5 of 5) But note that the probability we want is actually the area under the curve, given by the integral 1 0 f (u 1 )du 1 = 7/8 = Check you can do the integration here!

20 An analytical solution to Question 2 in Example is beyond the scope of this course, but if you fancy a challenge, it is not *too* hard to guess the answer! When you have the answer to Question 2 (soon), you *should* be able to answer Question 3! The solution to Question 1 leads us on to think more formally about using Monte Carlo methods as a method for evaluating integrals...

21 3.3 Approximation of integrals Suppose we have a complicated function f defined on the interval (0, 1), and also suppose that f (x) (0, 1). 1 f(x) We would like to evaluate 1 0 f (x)dx, but the function may be too complicated to integrate.

22 We can find an approximate answer by noting that the integral is equal to the area under the curve, and using Monte Carlo methods. We design an experiment which would work in general, even if the function was defined on a general range (a,b), and if f (x) (0,c), for any positive value c.

23 We generate a random data point from a simulation grid. Let A = {the data point lies below the curve}. Then so that P(A) = area under curve area of simulation grid, b a f (x)dx = [area under curve] = P(A)[area of simulation grid].

24 Example Consider the complicated function we saw earlier, defined on (0,1), and with f (x) (0,1): 1 f(x) Estimate 1 0 f (x)dx using Monte Carlo methods.

25 Solution (Slide 1 of 3) 1. We simulate n data points from the simulation grid; here this is the unit square (0, 1) (0, 1) f(x)

26 Solution (Slide 2 of 3) 2. Each of the coordinates x and y are generated using a U(0,1) random variable. Note the fact that A = {data point (x,y) lies below the curve} = {y < f (x)}. 3. If r points lie below the curve, then the proportion r/n = P F (A) is an estimate of P(A), and thus 1 0 f (x)dx = P(A)[area of simulation grid] = P(A) 1 = P(A) P F (A) = r/n.

27 Solution (Slide 3 of 3) 4. In our case, from the figure above we estimate: 1 0 f (x)dx = r/n = 14/20 = 0.7.

28 Example It is well known that there is no possible analytical expression for values of the Normal cumulative distribution function. For example, suppose we are interested in the standard Normal distribution with mean 0 and variance 1. Consider a random variable Z with this distribution, i.e. Z N(0, 1). Now suppose we want to evaluate P(0 Z 1).

29 Example (continued) It is very easy to write this probability down as an integral: 1 0 φ(x)dx = π exp( x 2 /2)dx. Here φ(x) is the standard Normal probability density function evaluated at x. The problem is that it is impossible to evaluate this integral, because indefinite integrals of the form exp(ax 2 )dx cannot be solved. It is easy, however, to formulate this as a Monte Carlo problem...

30 Monte Carlo Solution (Slide 1 of 3) The sketch of the Normal probability density function (p.d.f.) below illustrates the required probability P(0 Z 1) as the shaded area. Once again, the simulation grid is the unit square. p.d.f x

31 Monte Carlo Solution (Slide 2 of 3) As before, we simulate X and Y as independent U(0,1) random variables, so that we obtain an observation (x,y) each time. The event A is that the point (x,y) lies in the shaded area. This is true if y < φ(x), so here A = {y < φ(x)}. If we simulate n pairs (x,y), and set r = #{y < φ(x)}, then our estimate of the required integral is r/n.

32 Monte Carlo Solution (Slide 3 of 3) This was carried out in R with n = 1000, and we obtained r = 346. Hence we estimate: π exp( x 2 /2)dx 346/1000 = The true value of this integral is known to be (to three decimals, obtained using numerical methods). So once again we see that Monte Carlo simulation has given us a fairly accurate estimate!

33 Remarks 1 The accuracy of the estimate is dependent on the value of n. Remember P F (A) P(A) as n. 2 We will use this method in Assignment 2 to estimate the circular constant π. 3 Later we will use Monte Carlo methods with non uniform random variables. 4 Notice that in Example Question 1 the outcome of interest was a probability, whereas in Examples and it was an integral.

34 3.4 Non Uniform Continuous Random Variables In order to simulate many real life systems we will need to generate non uniform continuous random variables. As with the various discrete random variables we generated in Section 2, we will use the continuous Uniform U(0,1) distribution as our starting point. We will use a method called the Inversion Method, also known as the Table Look-Up Method, to transform a U(0,1) random variable to any other continuous distribution we desire. This works as long as we can evaluate the cumulative distribution function (c.d.f.) of the continuous distribution in question.

35 Theorem: The Inversion Method Suppose we wish to simulate a continuous random variable X with cumulative distribution function F(x). (Remember the notation is: F(x) = P(X x) ). Now let U be a U(0,1) random variable. Then X = F 1 (U) has the required distribution. Proof: P(X x) = P(F 1 (U) x) = P(U F(x)) = F(x).

36 Example Suppose X is a random variable with an exponential distribution with mean 2. How do we simulate from X?

37 Solution (Slide 1 of 3) The Exponential distribution with mean (1/λ) has c.d.f. F(x) = 1 e λx. We simulate a value u from a U(0,1) distribution, and set F(x) = 1 e λx = u. Now rearranging this to make x a function of u, we get: x = 1 λ log(1 u) = F 1 (u). (1) So given any simulated value u, we use Equation (1) with λ = 1/2 (giving mean equal to 2, as required), to obtain x, our simulation from X, where X Exp(1/2) as required.

38 Solution (Slide 2 of 3) Putting λ = 1/2, equation (1) becomes: x = 2log(1 u). So, for example, if we generated u = 0.785, then we would obtain: x = F 1 (u) = 2log( ) = It is easiest to see why this works by visualising the process...

39 Solution (Slide 3 of 3) The inversion method for the Exp(0.5) distribution with u = 0.785: u = F(x) 1 F(x) = 1 e x/ x = F (u) 10

40 Summary So, in order to simulate a value x from a random variable X, we need to take the following two steps: 1 Simulate a random number u from the U(0,1) distribution. 2 Set x = F 1 (u), where F is the c.d.f. of X, and F 1 is therefore the inverse c.d.f. This is all O.K., but what happens if we can t write down the function F 1?

41 E.g. Suppose Y is a random variable with a Normal N(µ,σ 2 ) distribution. How do we simulate from Y? This time we have a problem, because we can t write down the c.d.f. Φ(x) for the Normal distribution, so we can t invert it explicitly to obtain the inverse c.d.f. Φ 1 (u). Fortunately tables for Φ 1 (u) exist (e.g. see Neave s tables), but these are only for the standard Normal distribution, i.e. for the random variable Z, where Z N(0,1).

42 So starting with a simulated observation u from the Uniform U(0,1) distribution, we can look up our simulated observation z from our standard Normal random variable Z N(0,1). However, remember we wanted an observation y from Y N(µ,σ 2 ). This is now easy to generate if we remember that Y µ σ N(0,1). We put so that Y µ σ = Z, Y = µ + σz N(µ,σ 2 ). (2) By plugging our simulated observation z from Z into Equation (1) we immediately obtain our simulated value y from Y.

43 Example A. Convert the simulated value u = from U(0,1) to an observation from Y N(100,15 2 ). B. Convert the simulated value u = from U(0,1) to an observation from Y N(36,2 2 ).

44 Solution (Slide 1 of 2) A. We take the following steps: 1 Look up z = Φ 1 (u) to obtain z = Φ 1 (0.904) = Calculate y = µ + σz = =

45 Solution (Slide 2 of 2) B. We take the following steps: 1 Look up z = Φ 1 (u) to obtain z = Φ 1 (0.414) = Calculate y = µ + σz = = N.B. Check you can repeat this process! Use the following observations: a) u = 0.687; b) u =

46 3.5 Exponential Variables and Poisson Processes In Example we saw how to simulate psuedo random numbers from the Exponential Distribution with parameter λ (mean 1/λ). The Exponential Distribution is important in statistics, often as a model for the inter arrival times between events. For example, consider a queue for a cashier at a bank. Now suppose that the chance of a customer arriving at any instant in time is constant over time. Then the distribution of the time interval between arrivals is a random variable having the Exponential Distribution.

47 Now suppose we start counting at some arbitrary time zero, then the time to the first event, E 1, and the time interval between the (i 1) th and the i th events, E i, are all independent Exponential random variables with some parameter λ: E i exp(λ), independently, i = 1,2,3,... The p.d.f. of the E i is f (x) = λe λx, and the mean inter arrival time is 1/λ.

48 Now consider the number of events N occurring in a unit time interval. If the chance of an arrival at any instant in time is constant over time, then N Poisson(λ). That is to say, the distribution of the number of events in a unit time-interval is Poisson, with parameter (mean) equal to λ. Moreover, the distribution of the number of events in a time interval of length t 1 is Poisson(λt 1 ). Finally, the distribution of the number of events in a non overlapping time interval of length t 2 is indpendently distributed Poisson(λt 2 ).

49 Summary So, for any process where events occur so that the chance of an occurence remains constant through time, and the average number of events per unit time is λ, the following properties automatically hold: 1 The inter event times are independently Exponentially distributed with parameter λ, mean interval 1/λ. 2 The number of events in a time interval of length t is Poisson(λt), independently of the number of events in any other non overlapping time interval. Any process like this is called a Poisson Process in time. The fact that events occur at a constant rate means it is called a homogeneous Poisson Process.

50 Examples of Poisson Processes in Time (Slide 1 of 1) Goals scored in a football match.(may not be homogeneous e.g. the chance of a goal in any given instant may increase towards the end of the match). Crashes of a computer system. Emission of particles from a radioactive source. Instances of a river level being higher than a crucial safety level x. Instances of asteroid impact explosions greater than y megatonnes. (In the study of the history of the earth, such occurences are linked to mass extinctions of species.)

51 3.6 Simulation of Random Systems Now that we are able to simulate from any continuous random variable for which we can write down the c.d.f., we are in a position to simulate much more interesting/realistic examples than we have thus far. We will see that systems with quite complex behaviour can be simulated by building them from components which are simple random variables.

52 Example Queueing system Suppose we want to model a queue for a specialist service at a bank. A really simple model might be as follows (units are minutes): Inter-arrival times are distributed Exp(0.05), i.e. Exponential with mean 20. Service times are distributed N(25,5 2 ). Consider the following questions: 1 What is the distribution of the number of customers waiting at time t = 120 minutes? 2 In a 6-hour day, what is the mean customer waiting time? 3 In a 6-hour day, what is the distribution of the maximum customer waiting time?

53 Example Queueing system (continued) In principle we could find mathematical solutions to these problems. In practice this would be extremely difficult. However, very good approximate answers can be obtained by repeated simulations of six hours of queue behaviour. This is easily achieved by simulating a sequence of arrival times from the Exp(0.05) distribution, and a sequence of service times from the N(25,5 2 ) distribution, and then constructing a time line, on which we monitor the size of the queue. Let s see how this works by simulating the first hour of queue time...

54 Example Birth Death Process Consider a population of organisms (e.g. bacteria) evolving through time: The initial population size is denoted by N(0). Any individual splits ( birth ) with rate λ. Any individual dies with rate µ. Once again the rules are very simple to set up, but the evolution of the population size (N(t)) over time (t > 0) can be complex. It is easy to study this process by simulation, but very hard to do the maths.

55 Example Birth Death Process (continued) It is not immediately obvious how we can simulate from such a process, because any individual could split or die at any time. However, one solution is as follows: It can be shown that the time from one event at time t (birth or death) to the next event is a random variable E with distribution E Exp[(λ + µ)n(t)]. The next event is either a birth or a death with probabilities λ/(λ + µ) and µ/(λ + µ) respectively.

56 Solution (Slide 1 of 1) This gives rise to a 2 step simulation process from one event to the next: 1 Simulate the time to the next event from an Exp[(λ + µ)n(t)] random variable. 2 Simulate the type of event from a Bernoulli random variable with success probability λ/(λ + µ) corresponding to a birth, otherwise the event is a death. (We will be simulating birth death processes in Assignment 2!)

57 Example Progress of an epidemic Here we consider the evolution of the population of infected individuals as a birth-death process. Here the birth process corresponds to new individuals becoming infected. The death process corresponds to infected individuals being removed from the population (by being cured, or hospitalized, or dying). In order to stop the epidemic, it is necessary to get the death rate higher than the birth rate.

58 Case study: The SARS epidemic of 2003 SARS (Severe Acute Respiratory Syndrome) is a viral disease which has had one major epidemic (Nov Jul. 2003). Denote the rate of infections at time t by λ t. Denote the rate of removal from infectious state at time t by µ t.

59 Case study: The SARS epidemic of 2003 In the early stages (small t), we had λ t > µ t, and the epidemic spread. As time passed (large t), measures were taken to reduce infectivity, and hence achieve λ t < µ t. The epidemic was reduced and halted. Note that one reason this was achievable was because SARS is most infectious during the symptomatic period. It would be much more difficult for diseases such as influenza, which are most infectious pre-symptoms!