Inverse Transform Simulations

Transcription

1 Inverse Transform Simulations (a) Using the Inverse Transform Method, write R codes to draw 100,000 observations from the following distributions (b) Check our simulations with built-in R functions both graphicall as well as using the Kolmogorov Smirnov test. Histograms and results of the Kolmogorov-Smirnov test are given below. All code is presented together at the end. Our Eponential( 2) a R's Eponential( 2) b > ks.test( a, "pep", 2) One-sample Kolmogorov-Smirnov test data a D = , p-value = Page 1 of 13

2 Our Gamma( alpha=3, beta=3) R's Gamma( alpha=3, beta=3) > ks.test(, "pgamma", 3, 3) One-sample Kolmogorov-Smirnov test data D = , p-value = Page 2 of 13

3 Our Normal( 2, 3) R's Normal( 2, 3) > ks.test(, "pnorm", 2, 3) One-sample Kolmogorov-Smirnov test data D = , p-value = Page 3 of 13

4 Our Cauch( 1, 2) 0e+00 1e+05 2e+05 3e+05 4e+05 R's Cauch( 1, 2) 0e+00 1e+05 2e+05 3e+05 4e+05 > ks.test(, "pcauch", 1, 2) One-sample Kolmogorov-Smirnov test data D = , p-value = Warning message In ks.test(, "pcauch", 1, 2) cannot compute correct p-values with ties Page 4 of 13

5 Our Poisson( 3) R's Poisson( 3) >ks.test(, ) Two-sample Kolmogorov-Smirnov test > data and D = , p-value = Warning message In ks.test(, ) cannot compute correct p-values with ties Page 5 of 13

6 Our Binomial( 5, 0.3) R's Binomial( 5, 0.3) > ks.test(, ) Two-sample Kolmogorov-Smirnov test data and D = 8e-04, p-value = 1 Warning message In ks.test(, ) cannot compute correct p-values with ties ks.test(, ) Page 6 of 13

7 Our NegativeBinomial( r=3, p=0.5) R's NegativeBinomial( r=3, p=0.5) > ks.test(, ) Two-sample Kolmogorov-Smirnov test data and D = 0.003, p-value = Warning message In ks.test(, ) cannot compute correct p-values with ties Page 7 of 13

8 Our Multinomial( size=3, p1=.2, p2=.4, p3=.3, p4=.04, p5=.06) R's Multinomial( size=3, p1=.2, p2=.4, p3=.3, p4=.04, p5=.06) > ks.test(, ) Two-sample Kolmogorov-Smirnov test data and D = 3e-04, p-value = 1 Warning message In ks.test(, ) cannot compute correct p-values with ties Page 8 of 13

9 0e+00 6e+04 0e+00 6e+04 Our Dirichlet( alpha1=0.2, alpha2=0.3, alpha3=0.4) R's Dirichlet( alpha1=0.2, alpha2=0.3, alpha3=0.4) > ks.test(, ) Two-sample Kolmogorov-Smirnov test data and D = , p-value = Page 9 of 13

10 R Codes ### eponential ### ################### par(mfrow=c(2,1)) a<-c(-log(runif(100000))/2) hist(a, main="our Eponential( 2)") b<-rep(100000,2) hist(b, main="r's Eponential( 2)") ks.test( a, "pep", 2) ### Gamma ### ############# =numeric(100000) for(i in ) [i]=-1/3*sum(log(runif(3,0,1))) hist(, main="our Gamma( alpha=3, beta=3)") <-rgamma(100000,3,3) hist(, main="r's Gamma( alpha=3, beta=3)") ks.test(, "pgamma", 3, 3) ### Normal ### ############## =numeric(100000) for(i in ) [i]=2+3*sqrt(-2*log(runif(1)))*cos(2*pi*runif(1)) hist(, main="our Normal( 2, 3)") <-rnorm(100000,2,3) hist(, main="r's Normal( 2, 3)") ks.test(, "pnorm", 2, 3) ### Cauch ### ############### =numeric(100000) for(i in ) [i]=1+2*tan(pi*(runif(1)-1/2)) Page 10 of 13

11 lim = c(min(), ma()) hist(, lim=lim, main="our Cauch( 1, 2)") = rcauch(100000, 1, 2) hist(, lim=lim, main="r's Cauch( 1, 2)") ks.test(, "pcauch", 1, 2) ### Poisson (3) ### ################### lambda = 3 B = = numeric( B) for( i in 1B ) u = numeric( 0) u[1] = runif( 1 ) while( prod( u ) >= ep( -lambda ) ) u = c( u, runif( 1 ) ) [ i ] = length( u) - 1 hist(, main="our Poisson( 3)") = rpois( B, 3) hist(, main="r's Poisson( 3)") ks.test(, ) ### Binomial( 5, 0.3) ### ######################### size = 5 #number of trials p = 0.3 #prob of success B = #number of observations = numeric( B) #initialize random vector for( i in 1B) counter = 0 for( j in 1size) if( runif( 1) < p) counter = counter + 1 [i] = counter hist(, main="our Binomial( 5, 0.3)") = rbinom( n=b, size=size, prob=p) Page 11 of 13

12 hist(, main="r's Binomial( 5, 0.3)") ks.test(, ) rm( size, p ) ### Negative Binomial( r=3, p=0.5) ### ###################################### r = 3 p = 0.5 B = #number of observations = numeric( B) for( i in 1B) u = runif( 3) [i] = sum( floor( log( u)/ log( 1 - p) ) ) hist(, main="our NegativeBinomial( r=3, p=0.5)") = rnbinom( n=b, size=r, prob=p) hist(, main="r's NegativeBinomial( r=3, p=0.5)") ks.test(, ) ### Multinomial( size=3, prob=c(.2,.4,.3,.04,.06)) ### ########################################################### B = #number of random vectors/observations to draw size = 3 #number of "items"/"balls" to distribute among the "urns" prob =c(.2,.4,.3,.04,.06) #probabilities for each urn prob = prob/ sum( prob) #normalize proportions into probabilities breaks = numeric(1) #initialize breaks for( i in 1length(prob) ) #break the unit line into lengths proportional to probabilities in "prob" breaks[i+1] = sum( prob[1i] ) = matri( 0, nrow=length(prob), ncol=b ) #initialize matri of random number vectors; each column is a trial. for( i in 1B) for( j in 1size) u = runif( 1) if( an( breaks == u ) ) #check to see if u landed eactl on one of the breaks. if( u == 0 ) winner = 1 #ball goes into first urn if u == 0 if( u == 1 ) winner = length( prob) #similarl, ball goes into last urn if u == 1 Page 12 of 13

13 else #otherwise flip a coin to decide over to which side of the break the ball will fall adjustment = runif( 1, -0.5, 0.5) while( adjustment == 0) adjustment = runif( 1, -0.5, 0.5) #if adjustment==0, then we have another tie. we don't want that, so keep flipping coin until there is no tie. winner = floor( which( breaks == u) + adjustment) #when <= adjustment < 0, ball falls to the left of the break; when 0 < adjustment < 0.5 ball goes to right of the break. else winner = min( which( breaks > u ) ) - 1 #<- algorithm when u does not land eactl on a break [ winner, i] = [ winner, i] + 1 #save the result into the matri hist(, main="our Multinomial( size=3, p1=.2, p2=.4, p3=.3, p4=.04, p5=.06)") = rmultinom( B, size=size, prob=prob ) hist(, main="r's Multinomial( size=3, p1=.2, p2=.4, p3=.3, p4=.04, p5=.06)") ks.test(, ) ### Dirichlet( alpha1=0.2, alpha2=0.3, alpha3=0.4) ### ###################################################### alpha1 = 0.2 alpha2 = 0.3 alpha3 = 0.4 alpha = c( alpha1, alpha2, alpha3) B = = matri( nrow=b, ncol=length(alpha) ) for( i in 1B) = numeric( length( alpha)) for( j in 1length( alpha) ) [j] = rgamma( 1, shape=alpha[j], 1) #total = sum( ) [i,] = / sum( ) hist(, main="our Dirichlet( alpha1=0.2, alpha2=0.3, alpha3=0.4)") = rdirichlet(b, alpha= alpha) hist(, main="r's Dirichlet( alpha1=0.2, alpha2=0.3, alpha3=0.4)") ks.test(, ) par( mfrow= c(1,1)) Page 13 of 13