Question

#3.7

distribution. 0 and check that the mode of the generated samples is close to the (check the histogram). theoretical mode mass function 3.5 A discrete random variable X has probability 3 4 AtB.8 HUS 2 X p(x) 0.1 0.2 0.2 0.2 0.3 a random sample of size Use the inverse transform method to generate 1000 from the distribution of X. Construct a relative frequency table and compare the empirical with the theoretical probabilities. Repeat using the R sample function. 3.6 Prove that the accepted variates generated by the acceptance-rejection sampling algorithm are a random sample from the target density fx 3.7 Write a function to generate a random sample of size n from the Beta(a, b) distribution by the acceptance-rejection method. Generate a random sample of size 1000 from the Beta(3,2) distribution. Graph the histogram of the sample with the theoretical Beta(3,2) density perimposed. su-

Answer 1

3.7) Acceptance-Rejection Algorithm for continuous random variables .

1. Generate a RV distributed as
7
.

2. Generate (independent from )

3. If
fy (Y) cgy (Y)
, then set
X Y(accept)
; otherwise go back to 1, (“reject”).

Take
Y~U(1)
. Then
gy 1;0y<1
.

The Beta density is

(3+2) fx (X) (3)r(2)(1 -x)1 x (x) 12x2(1-x); 0 < x <1

To find the maximum value of ,
x (x) 12x2(1 - x) gy (x)
is found by differentiating,

(1 -x) 0 - x) = 0 dx 2x(1 x)2 - x2 0 2(1 x) x 0 x = 2/3

Thus,

$c=12(2/3)^2(1-2/3)=16/9$

The Histogram and theoretical density are plotted below.

Histogram of Beta Distribution 0.0 0.2 0.4 0.6 0.8 1.0 X 0'0 O L Density

The complete R code below.

N <- 1000
c <- 16/9
n <- 1
X <- array(dim = N)
f<- function (x)
{
12*x^2*(1-x)
}
while(n<=N)
{
Y <- runif(1)
U <- runif(1)
if(U < f(Y)/c)
{
X[n] <- Y
n <- n+1
}
}

plot(1:1)
dev.new()
hist(X,prob = TRUE, col = "skyblue", main = "Histogram of Beta Distribution")
curve(dbeta(x,3,2),add=TRUE,lwd=2,col="blue")