DATA 605 HW5

Choose independently two numbers B and C at random from the interval [0, 1] with uniform density. Prove that B and C are proper probability distributions. Note that the point (B,C) is then chosen at random in the unit square.

# Set seed v alue
set.seed(1)
n <- 1000000
# Generate two sereis of random variables B and C
B <- runif(n,0,1)
C <- runif(n,0,1)
# See first few values
head(B)

## [1] 0.2655087 0.3721239 0.5728534 0.9082078 0.2016819 0.8983897

head(C)

## [1] 0.14011775 0.69562066 0.72888445 0.09164734 0.06661200 0.61285721

To prove proper probability distributions we’ve to show that all all the values are greater than 0 and the empirical cumulative values lies between 0 to 1. Here all the values of variable B and C are greater than 0.

ecdf(B)

## Empirical CDF 
## Call: ecdf(B)
##  x[1:999880] = 1.5483e-07, 5.2433e-07, 1.0892e-06,  ...,      1,      1

ecdf(C)

## Empirical CDF 
## Call: ecdf(C)
##  x[1:999865] = 1.099e-07, 9.5111e-07, 1.522e-06,  ...,      1,      1

From the empirical cumulitive distribution we can see that the value of cumulutive distrubiotn functions for both B and C lies between 0 to 1 that is the total probability is 1.

From the above discussion we can say that randomly choosen two random numbers B and C are proper probabilijty distributions.

Find the probability that (a) B + C < 1/2.

Answer:

We find the sum of the proabilities of each B + C < 0.5.

sum(punif((B+C)<0.5, min=0, max=1)) / n

## [1] 0.125325

The probability that the the sum of B and C values falls below \(\frac{1}{2}\) is 0.125325. (b) BC < 1/2.

Answer:

We find the sum of the proabilities of each B * C < 0.5.

sum(punif((B*C)<0.5, min=0, max=1)) / n

## [1] 0.846724

The probability that the the product of B and C values falls below \(\frac{1}{2}\) is 0.846724.

3. |B − C| < 1/2.

Answer:

We find the sum of the proabilities of each |B − C| < 1/2.

sum(punif(abs(B-C)<0.5, min=0, max=1)) / n

## [1] 0.7502

The probability that the the absolute difference of B and C values falls below \(\frac{1}{2}\) is 0.7502.

4. max{B,C} < 1/2.

Answer: We find the sum of the proabilities of each max{B,C} < 1/2.

favor <- 0
for(i in 1:length(B)){
  if(max(B[i], C[i]) < 0.5){
    favor <- favor + 1
  }
}
favor/n

## [1] 0.250287

The probability that the maximum value of B and C falls below \(\frac{1}{2}\) is 0.250287.

5. min{B,C} < 1/2. Answer: We find the sum of the proabilities of each min{B,C} < 1/2.

favor <- 0
for(i in 1:length(B)){
  if(min(B[i], C[i]) < 0.5){
    favor <- favor + 1
  }
}
favor/n

## [1] 0.749991

The probability that the minimum value of B and C falls below \(\frac{1}{2}\) is 0.749991.