Random Number Generation

Write a program to generate 10 normal random numbers using method of Acceptance Rejection

# Normal Random Numbers using Acceptance Rejection Sampling

set.seed(24)
# Specify the number of random numbers to be generated
size=10
#First randon numbers are generated from Uniform distribution to generate #g(y). Here as some samples will be rejected in AR method we generate 3 times 
u=runif(3*size)
# the random numbers from the covering distribution is generated
exp=-log(u)
#Second set of random numbers to check the condition is generated
u2=runif(3*size)
#function to evalauate the upper limit for uniform random number
const=function(y){
  return(exp(-1*(y-1)^2/2))
}
n=1
i=1
#modz represents the list of ABSOLUTE standard normal ramdon numbers
modz=c()
while (n<=size){
  if(u2[i]<=const(exp[i])){
    modz=append(modz,exp[i])
    n=n+1
  }
  i=i+1
}
#Generate binary random number which assumes -1 0r +1  with equal probability
u3=runif(size)
#s is the random numbers of the random variable  assuming -1 0r +1  with equal probability
s=c()
for (i in 1:size){
  if (u3[i]<=0.5){
    s=append(s,-1)
  }
  else{
    s=append(s,1)
  }
}
#multiply absolute standard normal random number with -1 0r +1
z=modz*s
z

##  [1] -1.22903780 -1.49213880 -0.35066028 -0.65604959  0.08289935 -1.27391050
##  [7]  0.26942252  0.22110736  0.50271054 -0.99226790

Write a Program in r to generate 1000 standard normal random numbers and plot the histogram of the same

set.seed(24)
# Specify the number of random numbers to be generated
size=10
#First randon numbers are generated from Uniform distribution to generate #g(y). Here as some samples will be rejected in AR method we generate 3 times 
u=runif(3*size)
# the random numbers from the covering distribution is generated
exp=-log(u)
#Second set of random numbers to check the condition is generated
u2=runif(3*size)
#function to evalauate the upper limit for uniform random number
const=function(y){
  return(exp(-1*(y-1)^2/2))
}
n=1
i=1
#modz represents the list of ABSOLUTE standard normal ramdon numbers
modz=c()
while (n<=size){
  if(u2[i]<=const(exp[i])){
    modz=append(modz,exp[i])
    n=n+1
  }
  i=i+1
}
#Generate binary random number which assumes -1 0r +1  with equal probability
u3=runif(size)
#s is the random numbers of the random variable  assuming -1 0r +1  with equal probability
s=c()
for (i in 1:size){
  if (u3[i]<=0.5){
    s=append(s,-1)
  }
  else{
    s=append(s,1)
  }
}
#multiply absolute standard normal random number with -1 0r +1
z=modz*s

Now we shall consider the p-p plot of the generated random numbers to confirm whether the distribution of the random number is normal.

par(mfrow=c(2,1))
library(stats)
ProbDist=pnorm(sort(z))
plot(ppoints(length(z)), ProbDist,xlab = 'Observed Probability',main = 'P P Plot',ylab = 'Expected Probability',lwd=1,type = 'l',ylim =c(0,1))
abline(0,1)
hist(z,freq = FALSE)

ProbDist=pnorm(sort(z))
plot(ppoints(length(z)), ProbDist,xlab = 'Observed Probability',main = 'P P Plot',ylab = 'Expected Probability',lwd=1,type = 'l',ylim =c(0,1))
abline(0,1)

x=seq(-3,3,0.05)

par(mar = c(3, 3, 1, 1))
plot(x,pnorm(x),type='l')

Chi Sqaure test for goodness of fit

Testing of goodness of fit is used to evaluate whether the assumed distribution fits the given data.

Discrete Distribution

90 observations on a random variable \(X\) assuming values \(0,1,2,3\) are respectively \(20,30, 15\) and \(25\). Can we conclude that data is uniformly distributed.

# Observed frequencies
observed <- c(20, 30, 15, 25)

# Expected probabilities (equal proportions)
expected <- rep(1/4, 4)

# Perform the chi-square goodness of fit test
result <- chisq.test(observed, p = expected)

# Print the test result
print(result)

## 
##  Chi-squared test for given probabilities
## 
## data:  observed
## X-squared = 5.5556, df = 3, p-value = 0.1354

Since p-value is greater than \(0.05\), we cannot reject the null hypothesis.

Five coins are tossed and number of heads noted. The experiment is repeated 128 times and the following distribution is obtained:

## 
##  0  1  2  3  4  5 
##  6 20 24 42 24 12

Fit a Binomial distribution. Also test the goodness of fit.

Answer

To fit a binomial distribution first we have to estimate the parameters of the distribution. To estimate the parameter \(n\), we assign the maximum value taken by the random variable. Hence \(\hat{n}=6\) here. Now we shall estimate \(p\) using method of moments. Hence we equate sample mean to population. On simplification we get \(\hat{p}=\dfrac{\bar{x}}{\hat{n}}\)

values=seq(0,5)
frequ=c(6,20,24,42,24,12)
nhat=max(values)
nhat

## [1] 5

# Now we shall find the sample mean
mean_x=weighted.mean(values,frequ)
mean_x

## [1] 2.734375

phat=mean_x/nhat
phat

## [1] 0.546875

Now we will evaluate thee expected probabilities using dbinom() function.

ExpProb=c()
for (i in 0:5){
  ExpProb=append(ExpProb,dbinom(i,nhat,phat))
}
ExpProb

## [1] 0.01910250 0.11527368 0.27824682 0.33581513 0.20264706 0.04891481

chisq.test(frequ,p=ExpProb)

## Warning in chisq.test(frequ, p = ExpProb): Chi-squared approximation may be
## incorrect

## 
##  Chi-squared test for given probabilities
## 
## data:  frequ
## X-squared = 16.249, df = 5, p-value = 0.006169

Random Number Generation

Department of Statistics

2023-08-01

Chi Sqaure test for goodness of fit

Discrete Distribution