Simulation and Random Number Generation

Summary Discrete Time vs Discrete Event Simulation Random number generation Generating a random sequence Generating random variates from a Uniform distribution Testing the quality of the random number generator Some probability results Evaluating integrals using Monte-Carlo simulation Generating random numbers from various distributions Generating discrete random variates from a given pmf Generating continuous random variates from a given distribution

Discrete Time vs. Discrete Event Simulation Solve the finite difference equation Given the initial conditions x0, x-1 and the input function uk, k=0,1,… we can simulate the output! This is a time driven simulation. For the systems we are interested in, this method has two main problems.

Issues with Time-Driven Simulation System Input System Dynamics u1(t) t u2(t) Δt t x(t) t Inefficient: At most steps nothing happens Accuracy: Event order in each interval

Time Driven vs. Event Driven Simulation Models Time Driven Dynamics Event Driven Dynamics In this case, time is divided in intervals of length Δt, and the state is updated at every step State is updated only at the occurrence of a discrete event

Pseudo-Random Number Generation It is difficult to construct a truly random number generator. Random number generators consist of a sequence of deterministically generated numbers that “look” random. A deterministic sequence can never be truly random. This property, though seemingly problematic, is actually desirable because it is possible to reproduce the results of an experiment! Multiplicative Congruential Method xk+1 = a xk modulo m The constants a and m should follow the following criteria For any initial seed x0, the resultant sequence should look like random For any initial seed, the period (number of variables that can be generated before repetition) should be large. The value can be computed efficiently on a digital computer.

Pseudo-Random Number Generation To fit the above requirements, m should be chosen to be a large prime number. Good choices for parameters are m= 231-1 and a=75 for 32-bit word machines m= 235-1 and a=55 for 36-bit word machines Mixed Congruential Method xk+1 = (a xk + c) modulo m A more general form: where n is a positive constant xk/m maps the random variates in the interval [0,1)

Quality of the Random Number Generator (RNG) The random variate uk=xk/m should be uniformly distributed in the interval [0,1). Divide the interval [0,1) into k subintervals and count the number of variates that fall within each interval. Run the chi-square test. The sequence of random variates should be uncorrelated Define the auto-correlation coefficient with lag k > 0 Ck and verify that E[Ck] approached 0 for all k=1,2,… where ui is the random variate in the interval [0,1).

Some Probability Review/Results Probability mass functions Cumulative mass functions X 1 2 3 Pr{X=1}=p1 Pr{X=2}=p2 Pr{X=3}=p3 1 X 1 2 3 p1 p1+ p2 p1+ p2 + p3

Some Probability Results/Review Probability density functions Cumulative distribution x fX(x) x1 Pr{x=x1}= 0! x FX(x) 1

Independence of Random Variables Joint cumulative probability distribution. Independent Random Variables. For discrete variables For continuous variables

Conditional Probability Conditional probability for two events. Bayes’ Rule. Total Probability

Expectation and Variance Expected value of a random variable Expected value of a function of a random variable Variance of a random variable

Covariance of two random variables The covariance between two random variables is defined as The covariance between two independent random variables is 0 (why?)

Markov Inequality If the random variable X takes only non-negative values, then for any value a > 0. Note that for any value 0<a < E[X], the above inequality says nothing!

Chebyshev’s Inequality If X is a random variable having mean μ and variance σ2, then for any value k > 0. σ μ x fX(x) kσ kσ This may be a very conservative bound! Note that for X~N(0,σ2), for k=2, from Chebyshev’s inequality Pr{.}<0.25, when in fact the actual value is Pr{.}< 0.05

The Law of Large Numbers Weak: Let X1, X2, … be a sequence of independent and identically distributed random variables having mean μ. Then, for any ε >0 Strong: with probability 1,

The Central Limit Theorem Let X1, X2, …Xn be a sequence of independent random variables with mean μi and variance σι2. Then, define the random variable X, Let Then, the distribution of X approaches the normal distribution as n increases, and if Xi are continuous, then

Chi-Square Test Let k be the number of subintervals, thus pi=1/k, i=1,…,k. Let Ni be the number of samples in each subinterval. Note that E[Ni]=Npi where N is the total number of samples Null Hypothesis H0: The probability that the observed random variates are indeed uniformly distributed in (0,1). Let T be Define p-value= PH0{T>t} indicate the probability of observing a value t assuming H0 is correct. For large N, T is approximated by a chi-square distribution with k-1 degrees of freedom, thus we can use this approximation to evaluate the p-value The H0 is accepted if p-value is greater than 0.05 or 0.01

Monte-Carlo Approach to Evaluating Integrals Suppose that you want to estimate θ, however it is rather difficult to analytically evaluate the integral. Suppose also that you don’t want to use numerical integration. Let U be a uniformly distributed random variable in the interval [0,1) and ui are random variates of the same distribution. Consider the following estimator. Strong Law of Large Numbers Also note that:

Monte-Carlo Approach to Evaluating Integrals Use Monte-Carlo simulation to estimate the following integral. Let y=(x-a)/(b-a), then the integral becomes What if Use the substitution y= 1/(1+x), What if

Example: Estimate the value of π. (-1,-1) (1,-1) (1,1) (-1,1) Let X, Y, be independent random variables uniformly distributed in the interval [-1,1] The probability that a point (X,Y) falls in the circle is given by SOLUTION Generate N pairs of uniformly distributed random variates (u1,u2) in the interval [0,1). Transform them to become uniform over the interval [-1,1), using (2u1-1,2u2-1). Form the ratio of the number of points that fall in the circle over N

Discrete Random Variates Suppose we would like to generate a sequence of discrete random variates according to a probability mass function x 1 u X Inverse Transform Method

Discrete Random Variate Algorithm (D-RNG-1) Generate u=U(0,1) If u<p0, set X=x0, return; If u<p0+p1, set X=x1, return; … Set X=xn, return; Recall the requirement for efficient implementation, thus the above search algorithm should be made as efficient as possible! Example: Suppose that X{0,1,…,n} and p0= p1 =…= pn = 1/(n+1), then

Discrete Random Variate Algorithm Assume p0= 0.1, p1 = 0.2, p2 = 0.4, p3 = 0.3. What is an efficient RNG? D-RNG-1: Version 1 Generate u=U(0,1) If u<0.1, set X=x0, return; If u<0.3, set X=x1, return; If u<0.7, set X=x2, return; Set X=x3, return; D-RNG-1: Version 2 Generate u=U(0,1) If u<0.4, set X=x2, return; If u<0.7, set X=x3, return; If u<0.9, set X=x1, return; Set X=x0, return; More Efficient

Geometric Random Variables Let p the probability of success and q=1-p the probability of failure, then X is the time of the first success with pmf Using the previous discrete random variate algorithm, X=j if As a result:

Poisson and Binomial Distributions Poisson Distribution with rate λ. Note: The binomial distribution (n,p) gives the number of successes in n trials given that the probability of success is p. Note:

Accept/Reject Method (D-RNG-AR) Suppose that we would like to generate random variates from a pmf {pj, j≥0} and we have an efficient way to generate variates from a pmf {qj, j≥0}. Let a constant c such that In this case, use the following algorithm D-RNG-AR: Generate a random variate Y from pmf {qj, j≥0}. Generate u=U(0,1) If u< pY/(cqY), set X=Y and return; Else repeat from Step 1.

Accept/Reject Method (D-RNG-AR) Show that the D-RNG-AR algorithm generates a random variate with the required pmf {pj, j≥0}. 1 pi/cqi qi Therefore

D-RNG-AR Example Determine an algorithm for generating random variates for a random variable that take values 1,2,..,10 with probabilities 0.11, 0.12, 0.09, 0.08, 0.12, 0.10, 0.09, 0.09, 0.10, 0.10 respectively. D-RNG-1: Generate u=U(0,1) k=1; while(u >cdf(k)) k=k+1; x(i)=k; D-RNG-AR: u1=U(0,1), u2=U(0,1) Y=floor(10*u1 + 1); while(u2 > p(Y)/c) u1= U(0,1); u2=U(0,1); Y=floor(10*rand + 1); y(i)=Y;

The Composition Approach Suppose that we have an efficient way to generate variates from two pmfs {qj, j≥0} and {rj, j≥0} Suppose that we would like to generate random variates for a random variable having pmf, a (0,1). Let X1 have pmf {qj, j≥0} and X2 have pmf {rj, j≥0} and define Algorithm D-RNG-C: Generate u=U(0,1) If u <= a generate X1 Else generate X2

Continuous Random Variates Inverse Transform Method Suppose we would like to generate a sequence of continuous random variates having density function FX(x) Algorithm C-RNG-1: Let U be a random variable uniformly distributed in the interval (0,1). For any continuous distribution function, the random variate X is given by x 1 FX(x) u X

Example: Exponentially Distributed Random Variable Suppose we would like to generate a sequence of random variates having density function Solution Find the cumulative distribution Let a uniformly distributed random variable u Equivalently, since 1-u is also uniformly distributed in (0,1)

Convolution Techniques and the Erlang Distribution Suppose the random variable X is the sum of a number of independent identically distributed random variables Algorithm C-RNG-Cv: Generate Y1,…,Yn from the given distribution X=Y1+Y2+…+Yn. An example of such random variable is the Erlang with order n which is the sum of n iid exponential random variables with rate λ.

Accept/Reject Method (C-RNG-AR) Suppose that we would like to generate random variates from a pdf fX(x) and we have an efficient way to generate variates from a pdf gX(x). Let a constant c such that In this case, use the following algorithm C-RNG-AR: Generate a random variate Y from density gX(x). Generate u=U(0,1) If u< fX(Y)/(cgX(Y)), set X=Y and return; Else repeat from Step 1.

Accept/Reject Method (C-RNG-AR) The C-RNG-AR is similar to the D-RNG-AR algorithm except the comparison step where rather than comparing the two probabilities we compare the values of the density functions. Theorem The random variates generated by the Accept/Reject method have density fX(x). The number of iterations of the algorithm that are needed is a geometric random variable with mean c Note: The constant c is important since is implies the number of iterations needed before a number is accepted, therefore it is required that it is selected so that it has its minimum value.

C-RNG-AR Example Use the C-RNG-AR method to generate random variates X that are normally distributed with mean 0 and variance 1, N(0,1). First consider the pdf of the absolute value of |X|. We know how to generate exponentially distributed random variates Y with rate λ=1. Determine c such that it is equal to the maximum of the ratio

C-RNG-AR Example C-RNG-AR for N(0,1): u1=U(0,1), u2=U(0,1); Y= -log(u1); while(u2 > exp(-0.5(Y-1)*(Y-1))) u1= U(0,1); u2=U(0,1); u3= U(0,1); If u3 < 0.5 X=Y; Else X= -Y; Suppose we would like Z~N(μ, σ2), then

Generating a Homogeneous Poisson Processes A homogenous Poisson process is a sequence of points (events) where the inter-even times are exponentially distributed with rate λ (The Poisson process will be studied in detail during later classes) Let ti denote the ith point of a Poisson process, then the algorithm for generating the first N points of the sequence {ti, i=1,2,…,N} is given by Algorithm Poisson-λ: k=0, t(k)=0; While k<N k= k+1; Generate u=U(0,1) t(k)= t(k-1) – log(u)/lambda; Return t.

Generating a Non-Homogeneous Poisson Processes Suppose that the process is non-homogeneous i.e., the rate varies with time, i.e., λ(t) ≤λ, for all t<T. Let ti denote the ith point of a Poisson process, and τ the actual time, then the algorithm for generating the first N points of the sequence {ti, i=1,2,…,N} is given by Algorithm Thinning Poisson-λ: k=0, t(k)=0, tau= 0; While k<N Generate u1=U(0,1); tau= tau – log(u1)/lambda; Generate u2= U(0,1); If(lambda(tau)\lambda < u2) k= k+1, t(k)= tau; Return t.

Generating a Non-Homogeneous Poisson Processes Again, suppose that the process is non-homogeneous i.e., the rate varies with time, i.e., λ(t) ≤λ, for all t<T but now we would like to generate all points ti directly, without thinning. Assuming that we are at point ti, then the question that we need to answer is what is the cdf of Si where Si is the time until the next event Thus, to simulate this process, we start from t0 and generate S1 from FS1 to go to t1=t0+S1. Then, from t1, we generate S2 from FS2 to go to t2=t1+S2 and so on.

Example of Non-Homogeneous Poisson Processes Suppose that λ(t)= 1/(t+α), t ≥0, for some positive constant a. Generate variates from this non-homogeneous Poisson process. First, let us determine the rate of the cdf Inverting this yields

Example of Non-Homogeneous Poisson Processes Inverse Transform Thus we start from t0=0 …

The Composition Approach Suppose that we have an efficient way to generate variates from cdfs G1(x),…, Gn(x). Suppose that we would like to generate random variates for a random variable having cdf Algorithm C-RNG-C: Generate u=U(0,1) If u<p1, get X from G1(x), return; If u<p1+p2, get X from G2(x), return; …

Polar Method for Generating Normal Random Variates Let X and Y be independent normally distributed random variables with zero mean and variance 1. Then the joint density function is given by X Y Then make a variable change θ R The new joint density function is Exponential with rate 1/2 Uniform in the interval [0,2π]

Polar Method for Generating Normal Random Variates Algorithm C-RNG-N1: Generate u1=U(0,1), u2=U(0,1); R= -2*log(u1); W= 2*pi*u2; X= sqrt(R) cos(W); Y= sqrt(R) sin(W); Generates 2 independent RVs But, sine and cosine evaluations are inefficient! Algorithm C-RNG-N2: Generate u1=U(0,1), u2=U(0,1); Set V1= 2*u1-1, V2= 2*u2-1; S=V1*V1+V2*V2; If S > 1, Go to 1 R= sqrt(-2*log(S)/S); X= R*V1; Y= R*V2; (-1,1) (-1,-1) (1,1) (1,-1) (V1,V2)

Simulation of Discrete Event Systems INITIALIZE CLOCK STRUCTURE RNG STATE TIME EVENT CALENDAR e1 t1 e2 t2 … Update Time t’=t1 Update State x’=f(x,e1) Add New Feasible Events Delete Infeasible Events

Verification of a Simulation Program Standard debugging techniques Debug “modules” or subroutines Create simple special cases, where you know what to expect as an output from each of the modules Often choosing carefully the system parameters, the simulation model can be evaluated analytically. Create a trace which keeps track of the state variables, the event list and other variables.