1. Computer simulation Sep. 9, 2013

2. QUIZ 2 Determine whether the following experiments have discrete or continuous out comes A fair die is tossed and the number of dots on the face noted. Identify the random experiment, the set of outcomes, and the probabilities of each possible outcome.

3. Introduction Introduce computer simulations (ComSim) Show how to use ComSim to provide counter examples (can’t be used to prove theorems) simulate the outcomes of a discrete random variable Give examples of typical ComSim used in probability Monte Carlo computer approaches are a broad class of computationa algorithms that rely on repeated random sampling to obtain numerical results; i.e., by running simulations many times over in order to calculate those same probabilities heuristically just like actually playing and recording your results in a real casino situation: hence the name. 3

4. Why Use Computer Simulations? To provide counterexamples to proposed theorems To build intuition by experimenting with random numbers To lend evidence to a conjecture (an unproven proposition). 4

5. Building intuition using ComSim If U 1 and U 2 are the outcomes of two experiments each is a number from 0 to 1. What are the probabilities of X = U 1 + U 2 ? The mathematical answer will be given in later lectures Assume X is equally likely to be in the interval [0,2]. Let’s check if our intuition is correct by carrying out a ComSim. Generate values U 1 and U 2 them sum them up to obtain X. Repeat this procedure M times. Build a histogram which gives the number of outcomes in each bin. 5

6. Building intuition using ComSim Assume the following M = 8 outcomes were obtained {1.7, 0.7, 1.2, 1.3, 1.8, 1.4, 0.6, 0.4} Choosing the four bins [0, 0.5], (0.5, 1], (1, 1.5], (1.5, 2] we get 6

7. Building intuition using ComSim It is clear that values of X are not equally likely. 7 M = 1000 More probable

8. Building intuition using ComSim The probabilities are higher near one because there are more ways to obtain these values. X = 2 can be obtained from U 1 = U 2 = 1, but X = 1 can be obtained from U 1 = U 2 = ½ or U 1 = ¼, U 2 = ¾, or U 1 = ¾, U 2 = ¼, etc. 8

9. Building intuition using ComSim The result can be extended to the addition of three of more experimental outcomes. Define X 3 = U 1 + U 2 + U 3 and X 4 = U 1 + U 2 + U 3 + U 4. The histogram appears more like a bell-shaped(Gaussian) curve Conjecture: as we add more outcomes we obtain Gaussian 9

10. ComSim of Random Phenomena A random variable (RV) X is a variable whose value is subject to variations due to chance. Discrete : the number of dots on a die X can take on the values in the set {1, 2, 3, 4, 5, 6} Continuous : the distance of a dart from the center of the dartboard radius r = 1. {r : 0 ≤ r ≤ 1} To determine various properties of X we perform a number of experiments (trials) that is denoted by M. Assume that X = {x 1, x 2, …,x N } with probabilities { p 1, p 2, …,p N }. 10

11. ComSim of Random Phenomena As an example if N = 3 we can generate M values of X by using the following code segment. 11 For a continuous RV X that is Gaussian we can use the code

12. Determining Characteristics of RV The probability of the outcomes in the discrete case and the PDF in the continuous case are complete description of a random phenomenon. Consider a discrete RV, the outcome of a coin toss. Let X be 0 if a tail is observed with probability p and let X be 1 if head is observed with probability 1 – p. 12 To determine the probability of head we could toss a coin a large number of times and estimate p. We can simulate this result by using ComSim and estimate p. However, it is not always correct. p is slightly large than the true value of 0.4 due to imperfection of the random number generator

13. Probability density function(PDF) estimation PDF can be estimated by first finding the histogram and then dividing the number of outcomes in each bin by M to obtain the probability. The to obtain the PDF p X ( x ) recall that the probability of X taking on a value in an interval is found as the area under the PDF and if a = x 0 – Δ x/2 and b = x 0 – Δ x/2,where Δ x is small, then Hence, we need only divide the estimated probability by the bin width Δ x. 13

14. Probability density function(PDF) estimation Applying this estimation procedure to the set of simulated outcomes that has Gaussian PDF we are able to obtain estimated PDF. 14

15. Probability of an interval To determine P [ a ≤ X ≤ b ] we need to generate M realizations of X, then count the number of outcomes that fall into the [ a, b ] interval and divide by M. If we let a = 2, and b = ∞, then we should obtain the value (using numerical integration) And therefore very few realizations can be expected to fall in this interval. 15

16. Determining Characteristics of RV Average(mean) value A mean value of transformed variable f(x) = x 2 16

17. Multiple random variables Consider an experiment the choice of a point in the square {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} according to some procedure. So it yields two RVs or the vector [ X 1, X 2 ] T. This procedure may or may not cause the value of x 1 to depend on the value of x 2. 17 No dependencyDependency There is a strong dependency, because if for example x 1 = 0.5, then x 2 would have to lie in the interval [0.25, 0.75].

18. Multiple Random Variables Consider the two random vectors, where U i is generated using rand. 18 Then the result of M = 1000 realizations are the scatter diagrams No dependencyDependency

19. Monte Carlo simulation Consider a circle inscribed in a unit square. Given that the circle and the square have a ratio of areas that is π/4, the value of π can be approximated using a Monte Carlo method: 1.Draw a square on the ground, then inscribe a circle within it. 2.Uniformly scatter some objects of uniform size (grains of rice or sand) over the square. 3.Count the number of objects inside the circle and the total number of objects. 4.The ratio of the two counts is an estimate of the ratio of the two areas, which is π/4. Multiply the result by 4 to estimate π. 19

20. Digital Communications In a phase-shift keyed (PSK) digital system a bit is communicated to receiver by sending 0 : s 0 (t) = Acos(2 π F 0 t + π) 1 : s 1 (t) = Acos(2 π F 0 t ) 20 The receiver

21. Digital Communications The input to the receiver is the noise corrupted signal where w(t) is the channel noise. The output of the multiplier will be (ignoring the noise) 21 Recall

22. Digital Communications After the lowpass filter, which filters out the Acos(2 π F 0 t ) part of the signal and sampler we have 22 To model the channel noise we assume that the actual value ξ of observed is, where W is a Gaussian RV

23. Digital Communications It is of interest to determine how the error depends on the signal amplitude A. If A is a large positive amplitude, the chance that the noise will cause an error or equivalently, ξ ≤ 0,should be small. The probability of error P e = P[A/2 + W ≤ 0] 23 Usually P e = 10 -7.