FangQuant › Codes

### R-Code@June 06, 2016

submitted 2016-06-06 16:48:27

1.Normal Distribution

set.seed(1)
cord.x<-c(0,seq(0,5,0.1),5)
cord.y<-c(0,dnorm(seq(0,5,0.1),0,5),0)
curve(dnorm(x,0,5),xlim=c(-15,15),main='Normal Distribution,PDF',col="darkgreen",xlab="",ylab="Density",type="l",lwd=2,cex=2,cex.axis=.8)
polygon(cord.x,cord.y,col='skyblue')
runs<-1000
xs<-rnorm(runs,mean=0,sd=5)
sum(xs>=0&xs<=5)/runs


2.Binomial Distribution

runs<-1000
one.trial<-function(){
sum(sample(c(0,1),10,replace=T))>5
}
sum(replicate(runs,one.trial()))/runs
pbinom(q=5,size=10,prob=0.5,lower.tail=FALSE)


3.Estimate pi

runs<-10000
xs<-runif(runs,min=-5,max=5)
ys<-runif(runs,min=-5,max=5)
in.circle<-xs^2+ys^2<=5^2
(sum(in.circle)/runs)*4
plot(xs,ys,pch='.',col=ifelse(in.circle,"blue","orange"),xlab='',ylab='',asp=1,main=paste("MC Estimate of Pi =",(sum(in.circle)/runs)*4))


4.A/B Test&Beta distribution

#It certainly looks like B is the winner,but we'd really like to know how likely this is.We could of course run a single tailed t-test,that would require that we assume that these are Normal distributions(which isn't a terrible approximation in this case).However we can also solve this via a Monte Carlo simulation! We're going to take 100,000 samples from A and 100,000 samples from B and see how often A ends up being larger than B.#

x=seq(0,1,length=100)
y=dbeta(x,25,75)