TELECOMMUNICATIONS ENGINEERING

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1 TELECOMMUNICATIONS ENGINEERING STATISTICS COMPUTER LAB SESSION # 3. PROBABILITY MODELS AIM: Introduction to most common discrete and continuous probability models. Characterization, graphical representation. Solving real problems by simulation in MATLAB/Octave. In general, to generate continuous random variables it is usual to apply the inverse transform method of the distribution function (if inverse exists). For discrete r.v. s, it is possible to generate them using a boolean. In MATLAB/Octave, there exist several functions implemented in MATLAB/Octave in order to generate frequently used probability models. The next table summarizes some of the most common functions used in MATLAB/Octave to generate probability models of discrete and continuous random variables: Function Description Syntax normrnd random numbers N(µ, σ) normrnd(mu,sigma,m,n) randn random numbers N(0, 1) randn(m,n) exprnd random numbers Exp(λ) exprnd(1/lambda,m,n) binornd random numbers Bin(n, p) binornd(n,p,m,n) poissrnd random numbers Poiss(λ) poissrnd(lambda,m,n) where m and n, are respectively the numbers of rows and columns generated. 1. Continuous case Normal distribution Recall that Normal distribution has density function: X N (µ, σ) : f(x) = 1 σ 1 2π exp 2σ 2 (x µ)2, donde x R, σ > 0 y µ R. 1. Create a function in MATLAB/Octave which returns the values of the density function of a r.v. distributed as N (µ, σ). %% create the function fd_normal.m function y = fd_normal(x, mu, sigma) for i=1:length(x) y(i) = exp(-0.5*((x(i)-mu)/sigma)^2) / (sigma*sqrt(2*pi)); end %% Telecommunications Engineering - Estadística ( ), LAB 3. PROBABILITY MODELS 1

2 2. Use the function created in previous exercise to represent graphically the Normal r.v., for different values of the parameters of the distribution: a) Fix σ and change µ Consider 3 Normal distributions with constant standard deviation (σ = 1) and different means (µ =-1,0,1). Analyze the shape and the position of the distribution in terms of model parameters. b) Fix µ and change σ Consider 3 Normal distributions with constant mean (µ = 0) and different standard deviation (σ =0.3,0.5,1.2). Analyze the shape and the position of the distribution in terms of model parameters. % a) >> x = -5:0.01:5; >> y1 = fd_normal(x,-1,1); % y1 is Normal with mean -1 and s.d. 1 >> y2 = fd_normal(x, 0,1); % y2 is Normal with mean 0 and s.d. 1 >> y3 = fd_normal(x, 1,1); % y2 is Normal with mean +1 and s.d. 1 % b) >> y4 = fd_normal(x, 0, 0.3); % y4 is Normal with mean 0 and s.d. 0.3 >> y5 = fd_normal(x, 0, 0.5); % y5 is Normal with mean 0 and s.d. 0.5 >> y6 = fd_normal(x, 0, 1.2); % y6 is Normal with mean 0 and s.d. 1.2 Once we have the 6 normal distributed r.v. s, we can compare them graphically using the commands: subplot 1 ; plot and hold on, hold off. % GRAPHIC for CASE a) >> subplot(1,2,1) % 1 row, 2 columns, graph n o 1 >> hold on % holds the current plot so that % subsequent graphing commands add to the existing graph >> plot(x,y1, b ) % graph x/y1 in blue >> plot(x,y2, g ) % graph x/y2 in green >> plot(x,y3, r ) % graph x/y2 in red >> hold off % returns to the default % GRAPHIC for CASE b) >> subplot(1,2,2) % 1 row, 2 columns, graph n o 2 >> hold on >> plot(x,y4, b ) >> plot(x,y5, g ) >> plot(x,y6, r ) >> hold off 1 subplot allow us to create several graphs in the same panel. For example, subplot(1,2,i), implies we plot in one row and two columns, graphs i = 1 and 2 Telecommunications Engineering - Estadística ( ), LAB 3. PROBABILITY MODELS 2

3 % You can also >> subplot(1,2,1) >> plot(x,y1, b,x,y2, g,x,y3, r ) >> subplot(1,2,2) >> plot(x,y4, b,x,y5, g,x,y6, r ) mu = -1, sigma = 1 mu = 0,sigma = 0.3 mu = 0, sigma = 1 mu = 0, sigma = mu = 1, sigma = mu = 0,sigma = Note: Another alternative way to represent the graph for a Normal distribution (µ = 0 y σ = 2,5) for x ( 10, 10) may be: >> fplot( fd_normal(x, 0, 2.5), [-10 10], g ) MATLAB s command disttool in the Command Window, we can plot the density functions for more probability models selection Function Type/PDF. 3. Generate in MATLAB/Octave random numbers from a Normal distribution. In MATLAB/Octave s it is possible to generate random numbers from a Normal distribution (µ, σ), using build-in functions as: >> x= normrnd(mu,sigma,m,n); Command randn, generates random number from a standard Normal distribution (i.e., with mean µ = 0 and standard deviation σ = 1). From randn, it is also possible to generate random numbers N (µ, σ). Let Z N (0, 1), we can obtain X N (µ, σ), given that: X = Z σ + µ Telecommunications Engineering - Estadística ( ), LAB 3. PROBABILITY MODELS 3

4 >> z= randn(m,n); >> x= z*sigma + mu; We can check that the generated numbers are normally distributed by N (µ, σ) by its histogram: >> m=1000; n=1; >> sigma=0.75; mu=1; >> z=randn(m,n); x=z*sigma + mu; >> hist(x) Cumulative distribution function (CDF). In MATLAB/Octave, command normcdf(x,mu,sigma) gives the probability p = P (X x) of a Normal distribution of parameters µ and σ. Represent the cumulative distribution function for values of x [ 3, 3], given that X is standard normal. >> x = [-3:0.01:3]; >> mu=0; sigma=1; >> p = normcdf(x,mu,sigma); % p = normcdf(x) gives % the CDF of a % N(0,1) % Its inverse is x = norminv(p,mu,sigma) >> plot(x,p) >> grid on Telecommunications Engineering - Estadística ( ), LAB 3. PROBABILITY MODELS 4

5 2. Discrete case Binomial distribution The probability function p(x) of a Binomial distribution Bin(n, p) is: ( ) n X Bin(n, p) : p(x) = p x (1 p) n x where x = 0, 1,..., n y 0 p < 1. x where n is the number of trials and the parameter p the probabolity of success. 1. Create a MATLAB/Octave s function to represent graphically the probability function of a Binomial distribution. %% create fp_binomial.m function y=fp_binomial(x,n,p) for i=1:length(x) y(i)=(nchoosek(n,x(i)))*(p^x(i))*((1-p)^(n-x(i))) ; end %% Note: function nchoosek(n,x) computes ( n x), however, for large values of n, the result is not exact. Therefore, for large values of n, we can use Gamma function, Γ: Γ(p) = + Properties of Gamma function (Γ): a) Γ(1) = 0! = 1 0 e x x p 1 dx, where p > 0 b) Γ(p) = (p 1)Γ(p 1), p > 0 and therefore Γ(p) = (p 1)!, p N. c) ( ) n x = Γ(n+1) d) Γ( 1 2 ) = π. Γ(x+1) Γ(n x+1) 2. Create in MATLAB/Octave a function to represent probability functions from a Binomial distribution for large values of n. %% crete fp_binomialn.m %% using Prop. 3 of Gamma function function y = fp_binomialn(x,n,p) for i=1:length(x) y(i)=(gamma(n+1)*(p^x(i))*((1-p)^(n-x(i))))/ (gamma(x(i)+1)*gamma(n-x(i)+1)); end %% Telecommunications Engineering - Estadística ( ), LAB 3. PROBABILITY MODELS 5

6 3. Use the function created in the previous section, and represent graphically the probability function of a Binomial r.v. for: a) Fix p and change n create 3 Binomials: Bin(5,0.2), Bin(10,0.2), Bin(20,0.2) b) Fix n and change p create 3 Binomials: Bin(100,0.1), Bin(100,0.5), Bin(100,0.8) % Case a): >> x5=0:5 % create a sequence between 0 and 5 >> y5=fp_binomial(x5,5,0.2) % call to fp_binomial.m >> sum(y5) % check if sum is =1 >> x10=0:10; >> y10=fp_binomial(x10,10,0.2); >> x20=0:20; >> y20=fp_binomial(x20,20,0.2); % Caso b): >> x100 = 0:100; >> y1 = fp_binomialn(x100, 100, 0.1); % call to >> y2 = fp_binomialn(x100, 100, 0.5); % fp_binomialn.m >> y3 = fp_binomialn(x100, 100, 0.8); We can compare both graphs: Telecommunications Engineering - Estadística ( ), LAB 3. PROBABILITY MODELS 6

7 >> subplot(1,2,1) >> plot(x5,y5,., x10,y10, +, x20, y20, * ); >> legend( n=5, p=0.2, n=10, p=0.2, n=20, p=0.2 ) >> subplot(1,2,2) >> plot(x100, y1,., x100, y2, +, x100, y3, * ); >> legend( n = 100, p = 0.1, n = 100, p = 0.5, n = 100, p = 0.8 ); n=5, p=0.2 n=10, p=0.2 n=20, p= n = 100, p = 0.1 n = 100, p = 0.5 n = 100, p = By definition, we know tha Bin(n, p) is the sum of n independent Bernoulli r.v. s: Binom(n, p) = Bern(p) Bern(p) }{{} n times Create a function in MATLAB/Octave to generate random numbers from the Binomial distribution. %% create function na_binomial.m function y= na_binomial(n_dat, n, p) % n_dat is % n o of binomial data % to obtain prob = rand(n,n_dat); % simulate n trials success = prob < p; % if it holds, then we obtain a success y =sum(success); % sum all success %% In Command Window: Telecommunications Engineering - Estadística ( ), LAB 3. PROBABILITY MODELS 7

8 % Obtain 10 binomial numbers from the sum of the n o % of successes from n trials with probability of success p >> y = na_binomial(10,n,p) Numerically, we can verified the properties of the mean and variance of the simulated Binomial: E[X] = n p Var[X] = n p q >> n=100; p=0.1; >> y = na_binomial(10,n,p) >> mean(y) % aprox. 10 >> var(y) % aprox Generation of random variables As noted in previous practical, the inverse transform method allow us to generate continuous random variables from its distribution function if it admits inverse. Weibull distribution: Let X be a Weibull 2 random variable, such that X Weibull(α, β), with distribution function: { 0 x > 0 F X (x) = 1 e (x/β)α 0 x where α and β are known parameters. Give MATLAB/Octave s code to generate Weibull distributed r.v s. From inverse transform method, considering u = F X (x), 1 e (x/β)α = u e (x/β)α = 1 u (x/β) α = ln(1 u) x/β = [ ln(1 u)] 1 α x = β[ ln(1 u)] 1 α 2 NOTA: this distribution is very common in reliability, for instance, to study the lifetimes of components untill it fails. Telecommunications Engineering - Estadística ( ), LAB 3. PROBABILITY MODELS 8

9 And therefore: >> u = rand(n,1); >> x = beta*(-log(1-u))^(1/alpha); 3. Q function The Q function is defined as the complement of the distribution function of a standard normal N (0, 1), i.e. as: Q(x) = P (X > x) with X N (0, 1) Suppose a Q function with upper bounds (cs1) and (cs2) and lower bound (ci) given by Q(x) 1 x2 e 2 x 0 (cs1) 2 Q(x) < 1 e x2 2 x > 0 (cs2) 2πx Q(x) > 1 ( 1 1 ) 2πx x 2 e x2 2 para todo x > 1 (ci) a) MATLAB/Octave s function normcdf(x,mu,sigma) returns the CDF of a N (µ, σ). How would you compute in MATLAB/Octave Q function? >> Q = 1 - normcdf(x,0,1) b) Write a MATLAB/Octave s code to represent graphically the Q function in the interval (0, 5] with upper bounds cs1 and cs2 and lower bounds ci. The obtained result must be similar to figure 1. Note: Represent x (0, 5] as x=[0.01:0.01:5]. For the lower bound ci, observe that there is a vertical asymptote in x = 1, therefore, define xi=[1.01:0.01:5] only for this bound. Use logarithmic scale to represent the bounds. Telecommunications Engineering - Estadística ( ), LAB 3. PROBABILITY MODELS 9

10 >>x1= [1.01:0.01:5]; >>for i=1:length(x1) ci(i) = (1/x1(i)*sqrt(2*pi))*(1- (1/x1(i)^2))*exp(-0.5*x1(i)^2); end >>x = [0.01:0.01:5]; >>for i=1:length(x) cs1(i)=0.5*exp(-0.5*x(i)^2); cs2(i)=(1/(x(i)*sqrt(2*pi))) * exp(-0.5*x(i)^2); end >>q = 1- normcdf(x,0,1); >> plot(x, log(q), black ) >> hold on >> plot(x, log(cs1), r ) >> plot(x, log(cs2), g ) >> plot(x1, log(ci), m ) >> hold off 10 5 log(q) log(cs1) log(cs2) log(ci) Figura 1: Graphical representation of Q function with bounds in log scale, log(cs1), log(cs2) and log(ci). Vertical line represent the asymptte in x = Exercises of interest 1. Given Y distributed as Cauchy, which can be obtained as Y = tg (X) with X as U ( π 2, π 2 ), how would you use this information to generate a Cauchy? Telecommunications Engineering - Estadística ( ), LAB 3. PROBABILITY MODELS 10

11 If it is knows that there is no need to apply the inverse transform method, we can simply: >> x=unifrnd(-pi/2,pi/2, 100,1); % generate uniforms in the interval % (-pi/2,pi/2) >> y=tan(x); % apply tangent 2. Let be X N (0, σ = 3). If a circle with radius x of the defined random variable. Compute in terms of the Q function the probability that the generated circle has a greater or equal area to π. Evaluate the probability with MATLAB/Octave analitically and by simulation. Let C be a r.v. area of the circle, we have to compute P (C π) = P ( πx 2 π ) = P ( X 2 1 ) = P (X 1) + P (X 1) ( X 0 = P 1 0 ) ( X 0 + P 1 0 ) ( ) 1 = 2Q to evaluate the result in MATLAB/Octave: >> x=2*(1-normcdf(1/3)) by simulation: >> x=normrnd(0,3,1000,1); >> area=pi*x.>2; >> c=(area>=pi); >> prob=sum(c)/ Let X be a r.v. hours dedicated to perform an activity, with density function f (x) = { 1 4 (x + 1) 0 < x < 2 0 otherwise a) Calculate, analitically and using MATLAB/Octave, the probability that the time required to perform the activity is more than one hour and a half. b) If 10 activities are performed following a r.v. X, calculate analitically and using MATLAB/Octave, the probability of exactly in three of the activities, the time employed to perform each of them is more than one hour and a half. Telecommunications Engineering - Estadística ( ), LAB 3. PROBABILITY MODELS 11

12 a) Analitically we have: P (X > 1,5) = = x=1,5 f (x) dx = 2 x=1,5 ( 4 9 ) + 1 ( 2 3 ) (x + 1) dx = = = = 0,3438 [ x 2 ] 2 1, [x]2 1,5 To solve using MATLAB/Octave, we use the inverse transform method, so we calculate the distribution function: 0 x < 0 x 1 F (x) = 0 4 (y + 1) dy = 1 8 x x 0 x < x and the inverse: 1 8 x x = u x 2 + 2x 8u = 0 x = 2 ± u 2 From the definition of the r.v. X we only consider the positive value, and therefore to generate the random variable: u U (0, 1) x = u 2 The MATLAB/Octave s code is: >> u=rand(1000,1); >> x=(-2+sqrt(4+32*u))/2; >> prob=sum(x>1.5)/ b) Analitically, if the random variable Y is defined as number of activities among the 10 in which the employed time is more than one hour and a half, we have Y Bin ( n = 10, p = 11 32) and calculate P (Y = 3) = ( 10 3 ) ( ) 3 ( ) 21 7 = 0, Telecommunications Engineering - Estadística ( ), LAB 3. PROBABILITY MODELS 12

13 The MATLAB/Octave s code is: >> u=rand(1000,10); >> x=(-2+sqrt(4+32*u))/2; >> conta=(x>1.5); >> suma=sum(conta,2); >> prob=sum(suma==3)/ Ex. FEB 2004 Ing. Tel. (C1a) (link). The aim of this problem is to analyze a communication channel. When the channel canal transmits a 1, the receptor recieves a r.v. which follow a Normal distribution with mean 1 and variance 0, 5. If the binary channel transmits a 2, the receptor recieves a Normal r.v. with mean 2 and variance 0, 5. Let P (1) be the probability if transmiting a 1. a) Given P (1) = 0, 75. What is the probability that a 1 has been transmited if the receptor has received a signal greater than 2? T 1 = transmit 1, P (T 1 ) = 0,75, R T 1 N(1, 0,5) T 2 = transmit 2, P (T 2 ) = 0,25, R T 2 N(2, 0,5) P (T 1 R > 2) = P (R > 2 T 1)P (T 1 ) = P (R > 2) P (R > 2 T 1 )P (T 1 ) = P (R > 2 T 1 )P (T 1 ) + P (R > 2 T 2 )P (T 2 ) Let Z 1 = R T 1 1 0,5 and Z 2 = R T 2 2 0,5, P (Z 1 > 1,41) 0,75 P (T 1 R > 2) = P (Z 1 > 1,41) 0,75 + P (Z 2 > 0) 0,25 0,0793 0,75 = 0,0793 0,75 + 0,5 0,25 = 0,3224 In MATLAB/Octave, it is possible to approximate the probability by: >> n=10000; >> u=rand(n,1); % simulate n random trials >> p=0.75; % is the probability to transmit a 1 % message transmition in the channel >> t=1*(u<=p)+2*(u>p); % message reception >> r=(t==1).*normrnd(1,sqrt(0.5),n,1)+(t==2).*normrnd(2,sqrt(0.5),n,1); >> r=2*(r>2)+1*(r<=2); Telecommunications Engineering - Estadística ( ), LAB 3. PROBABILITY MODELS 13

14 % create a crosstable with transmited/received >> a=crosstab(t,r) >> a = % a(1,1) a(1,2) % a(2,1) a(2,2) % to calculate the probability P(t==1 r>2), we calculate >> a(1,2)/(a(1,2)+a(2,2)) Ex. SEP 2007 Ing. Téc. Tel. (P2b). (link). The duration in days of a type of sensors follows a Weibull distribution with: ( ( ) ) t 1/2 F (t) = 1 exp α ( f (t) = 1/2 ( ) ) t 1/2 α 1/2 t 1/2 exp α con α > 0. a) Suppose a box with 60 unused sensors with duration time distributed as a Weibull with α = 1 4 days, which verifies that E [T ] = 1 2 days and V [T ] = 5 4 days2. Suppose you start using a sensor from the box and replace it with another one from the box whenever it fails. What is the probability that when the last sensor of the box has failed it has been passed 47 days? Use any of this values. x 1,64 1,64 1,96 1, Q (x) 0,05 0,95 0,025 0, b) Comment next MATLAB s code, what it the approximate value of prob. >> u=rand(60,1000); >> w=0.25*(-log(1-u)).^(1/0.5); >> s=sum(w); >> prob=sum(s<47)/1000 Telecommunications Engineering - Estadística ( ), LAB 3. PROBABILITY MODELS 14

15 a) Let T i a r.v. of time until i th sensor fails, calculate: P ( 60 ) T i 47 i=1 Given that the r.v. s T i verify Central Limit Theorem conditions, we have (60 12 ), if we standardize P ( Z < P ( 60 i=1 T i i=1 ) T i N = P i=1 T i < ) = P (Z < 1, 96) = 1 Q (1, 96) = 1 0, 025 = 0, , 66 b) Firstly, we have to generate a Weibull distribution. We can apply inverse transform method. Therefore ( ( ) ) t β F (t) = 1 exp = u α ( ) t β ln (1 u) = α t = α [ Ln (1 u)] 1/β First line generates a matrix with random numbers U (0, 1). In second line, we generate a matrix, with random number distributed as a Weibull with α = 1 4 days, (we are generating times when sensors are failing). Third line, applies the total sum of the 60 sensors. Finally, fourth line checks how many times out of 1000, the total duration is less than 47, by using he boolean condition (s < 47). Therefore, this code has the same answer as in a). So the result of prob should be very close to the theoretical value of 0, Telecommunications Engineering - Estadística ( ), LAB 3. PROBABILITY MODELS 15

16 5. Proposed exercises NOTE: These exercises are not part of the compulsory assignment. Their resolution is recommended for gaining a better understanding of the concepts introduced in this Lab Session. 1. Indicate MATLAB/Octave s code to solve the problem proposed in: Ex. SEP 2005 Ing. Téc. Tel. C3b (link). 2. Indicate MATLAB/Octave s code to solve the problem proposed in: Ex. JUN 2002 Ing. Inf. (P1b) (link). 3. Indicate MATLAB/Octave s code to solve the problem proposed in: Ex. SEP 2007 Ing. Téc. Tel. (C4a y C4b). (link). Telecommunications Engineering - Estadística ( ), LAB 3. PROBABILITY MODELS 16