Generate a continuous random variable

Drawing from exponential distribution with \(\lambda = 1\)

Use inverse of cdf

n<-1000
lambda = 1
y = runif(n) ## generate n numbers between 0 and 1
f.inv = function(x){-1/lambda*log(1-x)}
X.exp = f.inv(y)

Compare empirical density to true density using the dexp function in R

 ## empirical density
hist(X.exp,freq=FALSE, main  = "Inverse cdf density vs Truth")    ## empirical density

## true density
t<-c(1:100)/10 
lines(t,dexp(t,1),col = "red")

Compare the empirical cdf to the true cdf

plot(ecdf(X.exp),main = "rexp, Empirical cdf vs truth")                  ## empirical cdf using ecdf function
lines(t,pexp(t,1),col=2)    ## true cdf using pexp function

Compare to using the rexp function in R

X.exp2 = rexp(n,1)
hist(X.exp2,freq=FALSE, main = "rexp vs truth")
t<-c(1:100)/10 
lines(t,dexp(t,1),col = "red") 

plot(ecdf(X.exp2),main = "rexp, Empirical cdf vs truth")                  ## empirical cdf versus true cdf
lines(t,pexp(t,1),col=2)    ## true cdf using pexp function

Check probability-integral transformation: Drawing X from Exp(1)), computing Y=F(X), and show that Y~UNIF(0,1)

y<-pexp(X.exp,1) ### compute F(x)
hist(y,freq=FALSE)  ## look at pdf of Y
s<-c(0:100)/100
lines(s,dunif(s,0,1),col = "red") ### compare to uniform (0,1)

Drawing from a chi-square distribution with 1 degree of freedom

Generate 10000 from chi-square(1) distribution. Compare empirical density and cdf with true density and cdf.

n<-10000
X<-rchisq(n,1)
hist(X,freq=FALSE, main = "Chi-squared empirical cdf with true density")
t<-c(0:1200)/100
lines(t,dchisq(t,1),col = 2)  ## empirical cdf with true density

plot(ecdf(X), main = "Chi-squared empirical cdf with true cdf")
lines(t,pchisq(t,1),col=2)  ## empirical cdf with true cdf

Generate 10000 U from UNIF(0,1) and then generate V from chi-squared using the inverse cdf of a chi-squared distribution.

n <- 1000
U <- runif(n)
V = qchisq(U,1)
hist(V,  freq = FALSE,main = "Chi-squared using inverse cdf, empirical cdf with true density")
t<-c(0:2000)/100
lines(t,dchisq(t,1), col = "red")  ## empirical pdf with true density

plot(ecdf(X), main = "Chi-squared using inverse cdf, empirical cdf with true cdf")
t<-c(0:2000)/100
lines(t,pchisq(t,1),col=2)  ## empirical cdf with true cdf

Simulation to estimate the pmf for a transformed discrete random variable

X is the value shown when you roll a six sided die and \(Y = (X-2)^2\). Find f_Y(y).

Simulation code

## ++++++++++++++++
## simulation parameters
## ++++++++++++++++
num.sims<-10000
seed.use <- 124

set.seed(seed.use)

## ++++++++++++++++
## create place holders
## ++++++++++++++++
Y.vals <- NULL

## ++++++++++++++++
## run simulation
## ++++++++++++++++
for(i in 1:num.sims) {
  x<-sample(1:6,1,replace=TRUE)
  Y.vals[i] = (x-2)^2
}

Summarize results

## ++++++++++++++++
## calculate probability of all different birthday
## ++++++++++++++++
## simulation result
sum.results = data.frame(round(table(Y.vals)/num.sims,3))
colnames(sum.results) = c("Y","P(Y=y)")
rownames(sum.results) =NULL

pander(sum.results)
Y P(Y=y)
0 0.163
1 0.333
4 0.166
9 0.171
16 0.167