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.

Solution

B and C are proper probability distributions because they have 100% chance of being continuously distributed in between [0,1] while both of them combined, that is, B+C, have a 100% chance of falling between [0,2]. Probability distribution can be simulated with large number of trials so as to ensure accuracy as much as possible. Therefore, I would try to simulate this using R:

#creating a 100, 000 pair for both B and C
B <- runif(100000)
C <- runif(100000)

Note that the point (B,C) is then chosen at random in the unit square. Find the probability that:

(a) B + C < 1/2.

t = 0
for(k in 1:length(B))
{
  if((B[k]+C[k]) < 0.5)
  {
    t = t + 1;
  }
}

lB <- length(B)
print(paste("The Probabilty B+C < 1/2 =", round(t/lB, 3)))
## [1] "The Probabilty B+C < 1/2 = 0.126"

(b) BC < 1/2.

t = 0
for(k in 1:length(B))
{
  if((B[k]*C[k]) < 0.5)
  {
    t = t + 1;
  }
}

print(paste("The Probabilty that BC < 1/2 =", round(t/lB, 3)))
## [1] "The Probabilty that BC < 1/2 = 0.845"

(c) |B − C| < 1/2.

t = 0
for(k in 1:length(B))
{
  if(abs((B[k]-C[k])) < 0.5)
  {
    t = t + 1;
  }
}

print(paste("The Probabilty that |B-C| < 1/2 =", round(t/lB, 4)))
## [1] "The Probabilty that |B-C| < 1/2 = 0.7507"

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

t = 0
for(k in 1:length(B))
{
  if(max(c(B[k],C[k])) < 0.5)
  {
    t = t + 1;
  }
}

print(paste("The Probabilty that max(B,C) < 1/2 =", round(t/lB, 4)))
## [1] "The Probabilty that max(B,C) < 1/2 = 0.2502"

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

t = 0
for(k in 1:length(B))
{
  if(min(c(B[k],C[k])) < 0.5)
  {
    t = t + 1;
  }
}

print(paste("The Probabilty that min(B,C) < 1/2 =", round(t/lB, 4)))
## [1] "The Probabilty that min(B,C) < 1/2 = 0.7483"