Instructions

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.

Setup

We can use the runif function in R to choose random numbers from a uniform distribution in the interval [0, 1].

B <- runif(10000, min = 0, max = 1)
C <- runif(10000, min = 0, max = 1)

We can see that both B and C are proper uniform probability distributions by creating histograms of each random variable.

hist(B)

hist(C)

Find the probability that B + C < 1/2

By creating a histogram of B + C we can see that although the two variables are uniform distributions their sum is not a uniform distribution. So we can find the probability that their sum would be less than 1/2 using simulation.

X <- B + C
hist(X, main = "Histogram of B + C")

a <- sum(X < .5) / 10000
a
## [1] 0.1243

The probability that B + C < 1/2 = 0.1243.

Find the probability that BC < 1/2

We can see in the histogram below that B times C is a right skewed distribution, so once again, we can find the probability that BC < 1/2 using simulation.

X <- B * C
hist(X, main = "Histogram of B x C")

b <- sum(X < .5) / 10000
b
## [1] 0.8421

The probability that B * C < 1/2 = 0.8421.

Find the probability that |B − C| < 1/2

The absolute value of the difference between B and C is also not a uniform or normal distribution, so we will use simulation again to determine the probability that |B − C| < 1/2.

X <- abs(B - C)
hist(X, main = "Histogram of |B − C|")

c <- sum(X < .5) / 10000
c
## [1] 0.7535

The probability that |B − C| < 1/2 = 0.7535.

Find the probability that max{B,C} < 1/2

We can also use simulation to find the probability that the max of B and C is less than 1/2.

X <- pmax(B,C)
hist(X, main = "Histogram of max{B,C}")

d <- sum(X < .5) / 10000
d
## [1] 0.2456

The probability that the max{B,C} < 1/2 = 0.2456.

Find the probability that min{B,C} < 1/2

Finally, we will use simulation to find the probability that the min of B and C is less than 1/2.

X <- pmin(B,C)
hist(X, main = "Histogram of min{B,C}")

e <- sum(X < .5) / 10000
e
## [1] 0.7438

The probability that the min{B,C} < 1/2 = 0.7438.