DATA 605: Homework 05

Load Packages

library(knitr)

Problem set

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.

Proof:

# generate random numbers from a uniform distribution
set.seed(NULL)
B <- runif(10000, min = 0, max = 1)  

set.seed(NULL)
C <- runif(10000, min = 0, max = 1)   

df_b<- as.data.frame(B)

# check whether the probability of B is between 0 and 1

# firsr create a series with all the values as False
sum_p <- rep(F,times=length(B))

# then calculate the pribability of B
for (i in 1:length(B)){
  x=1
  for (j in 1:length(B)){
    if (j != i && B[i] == B[j]){
      x <- x + 1
    }
  }
  p <- x/length(B)
  # if the probability is between 0 and 1 inclusive, the value in sum_p will change to True correspondingly 
  sum_p[i] <- (0 <= p && p<= 1) 
}

condition_1 <- !any(sum_p == F)

print(paste("Is all the probabilities of B is between 0 and 1 inclusive? ",condition_1 ))

## [1] "Is all the probabilities of B is between 0 and 1 inclusive?  TRUE"

print(paste("Similarly, all the probabilities of B is between 0 and 1 inclusive "))

## [1] "Similarly, all the probabilities of B is between 0 and 1 inclusive "

# sum the probability of B and denote as sigma
sigma_b <- 0
for (i in 1:length(B)){
  x=1
  for (j in 1:length(B)){
    if (j != i && B[i] == B[j]){
      x <- x + 1
    }
  }
  p <- x/length(B)
  sigma_b <- sigma_b + p 
}

sigma_b

## [1] 1

sigma_b <- round(sigma_b)

print(paste("The sum of the probabilities of the outcomes of B is",sigma_b))

## [1] "The sum of the probabilities of the outcomes of B is 1"

print(paste("Similarly,The sum of the probabilities of the outcomes of C is",sigma_b))

## [1] "Similarly,The sum of the probabilities of the outcomes of C is 1"

For both B or C, all the probabilities is between 0 and 1 inclusive.

For both B or C,the sum of the probabilities of the outcomes is 1.

This proof that B and C are proper probability distributions.

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

j = 0
for(i in 1:length(B)){
  if(B[i]+C[i] < 0.5){
    j = j+1
  }
}
print(paste("The Probabilty B+C < 1/2 =", j/length(B)))

## [1] "The Probabilty B+C < 1/2 = 0.1288"

The probabilty is 12.5%. For this to be true B and C must be <12, which is 0.5∗0.5=0.25 and C or B must be <12−[Compliment] the range where is can be true is [0,0.5]. The overall probabilty is 0.25∗0.5=0.125 which is what we see in the simulation.

2. \(BC<1/2\)

j = 0
for(i in 1:length(B)){
  if(B[i]*C[i] < 0.5){
    j = j+1
  }
}
print(paste("The Probabilty BC < 1/2 =", j/length(B)))

## [1] "The Probabilty BC < 1/2 = 0.8446"

3. \(|B−C|<1/2\)

j = 0
for(i in 1:length(B)){
  if(abs(B[i]-C[i]) < 0.5){
    j = j+1
  }
}
print(paste("The Probabilty |B-C| < 1/2 =", j/length(B)))

## [1] "The Probabilty |B-C| < 1/2 = 0.7482"

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

j = 0
for(i in 1:length(B)){
  if(max(c(B[i],C[i])) < 0.5){
    j = j+1
  }
}
print(paste("The Probabilty max(B,C) < 1/2 =", j/length(B)))

## [1] "The Probabilty max(B,C) < 1/2 = 0.2479"

5. \(min{B,C}<1/2\)

j = 0
for(i in 1:length(B)){
  if(min(c(B[i],C[i])) < 0.5){
    j = j+1
  }
}
print(paste("The Probabilty min(B,C) < 1/2 =", j/length(B)))

## [1] "The Probabilty min(B,C) < 1/2 = 0.7495"