Assignment 5

Fundamentals of Computational Mathematics

CUNY MSDS DATA 605, Fall 2018

Rose Koh

09/23/2018

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.

Find the probability of a,b,c,d,e:

First, let’s create the two numbers B,C at random from the inverval [0,1] with uniform density.

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

B <- as.data.frame(B.nums)
C <- as.data.frame(C.nums)

To prove that B, C are proper probability distributions:

1. We must prove that All the probabilities must be between 0 and 1 inclusive.

summary.B <- B %>% 
  mutate(condition.1 <- ifelse(B > 0 && B <= 1,TRUE,FALSE)) %>% # create `condition.1` column that fills as True if 0 < B <= 1, otherwise False.
  setNames(c("B", "condition.1")) %>%                           # rename columns
  group_by(condition.1) %>%                                     # group by the condition.1 column
  tally()                                                       # table summary.

summary.C <- C %>% 
  mutate(condition.1 <- ifelse(B > 0 && B <= 1,TRUE,FALSE)) %>%
  setNames(c("C", "condition.1")) %>%
  group_by(condition.1) %>% 
  tally()

sum.b <- sum(B)
sum.c <- sum(C)

paste("Total value of B: " , nrow(B) )

## [1] "Total value of B:  10000"

paste("Number of B values that is between 0 and 1 inclusive: ", summary.B[2])

## [1] "Number of B values that is between 0 and 1 inclusive:  10000"

paste("Total value of C: " , nrow(C) )

## [1] "Total value of C:  10000"

paste("Number of C values that is between 0 and 1 inclusive: ", summary.C[2])

## [1] "Number of C values that is between 0 and 1 inclusive:  10000"

2. We must prove the sum of the probabilities of the outcomes must be 1.

outcome.b <- B %>% 
  mutate(prob.b <- B.nums/sum.b) %>%
  setNames(c("B.nums", "prob.b")) %>% 
  summarise(sum(prob.b)) 

outcome.c <- C %>% 
  mutate(prob.c <- C.nums/sum.c) %>%
  setNames(c("C.nums", "prob.c")) %>% 
  summarise(sum(prob.c))

paste("The sum of probabilities of the outcome is: ", "For B:", outcome.b[1,1], "and For C:", outcome.c[1,1])

## [1] "The sum of probabilities of the outcome is:  For B: 1 and For C: 1"

By satisfying both conditions, It has proven that B and C have proper probabilities.

1. B + C < 1/2

P.a <- sum((B.nums+C.nums) < 1/2)/length(B.nums)
P.a

## [1] 0.1226

2. BC < 1/2

P.b <- sum((B.nums*C.nums) < 1/2)/length(B.nums)
P.b

## [1] 0.8485

3. |B − C| < 1/2

P.c <- sum(abs((B.nums-C.nums)) < 1/2)/length(B.nums)
P.c

## [1] 0.7478

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

P.d <- sum(pmax(B, C) < 1/2)/nrow(B)
P.d

## [1] 0.2506

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

P.e <- sum(pmin(B, C) < 1/2)/nrow(B)
P.e

## [1] 0.7553