This is somewhat like throwing two six-sided dice. For two six sided dice, each dice will be discretely range from 1 to 6 and the combination of the two dice will range discretely from 2 to 12. What makes the two six sided dice proper probabilty distributions is the each die has a 100% chance of rolling [1,2,3,4,5,6] and the pair of dice has a 100% chance of falling in the range of [2,3,4,5,6,7,8,9,10,11,12]
What makes B and C a proper probabilty distrubution is that both B and C have a 100% chance of falling in between [0,1] albeit continuously instead of discretely, and B+C combined have a 100% chance of falling between [0,2]
To demonstrate, I have used r to make 100,000 such pairs:
B <- runif(100000)
head(B)## [1] 0.2721655 0.6335266 0.5737134 0.6398595 0.4187566 0.2797840tail(B)## [1] 0.1583106 0.4648774 0.6142615 0.1678318 0.3278460 0.7622023C <- runif(100000)
head(C)## [1] 0.04676928 0.05308840 0.58126611 0.67874371 0.37893887 0.21403610tail(C)## [1] 0.9249220 0.8270274 0.6886993 0.8071693 0.1750513 0.6560202head(B+C)## [1] 0.3189347 0.6866150 1.1549796 1.3186032 0.7976955 0.4938201tail(B+C)## [1] 1.0832327 1.2919048 1.3029608 0.9750011 0.5028973 1.4182224j = 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 = 0.1248"The probabilty is 12.5%. For this to be true B and C must be < 1/2, which is 0.5x0.5 = 0.25 and C or B must be < 1/2-[Compliment] the range where it can be true is [0,0.5]. The overall probabilty is 0.25x0.5=0.125 which is what we see in the simulation.
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 = 0.84797"The probabilty is about 85%. In this, there is a 50% that both B and C are < \(\frac{1}{\sqrt{2}}\) =0.7071 , this means the odds of both being >\(\frac{1}{\sqrt{2}}\) = 0.7071, which is the condition for BC>\(\frac{1}{2}\), is (1-\(\frac{1}{\sqrt{2}}\))/2=0.1464466. The overall probabilty is 1 - 0.1464466 =0.8535534, which is what we see in the simulation.
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 = 0.74907"The closer one number get to another, the closer their difference appoaches zero. So in the case where B(<1/2) and C(<1/2), or B(>1/2) and C(>1/2) are going to have a diffence that is <1/2. This accounts for 50% of all cases. The table below for (d) and (e) may help illustrate this. The remaining 25% comes from cases where either B or C is >1/2 and C or B is <1/2. For |B-C| < 1/2 , then B or C must be (.75,0.5), and C or B (0.5,0.25). This accounts for 1/2 of the cases where one random variable is <1/2 and the other is <1/2. This means the probabilty for the remaining cases is 1/2x1/2=1/4=0.25. Such that 0.5+0.25=0.75, which is supported by the simulation.
Both (d) and (e) can best be described by the following table:
| B or C | C < 1/2 | C > 1/2 |
|---|---|---|
| B < 1/2 | B(<1/2)&C(<1/2) | B(<1/2)&C(>1/2) |
| B > 1/2 | B(>1/2)&C(<1/2) | B(>1/2)&C(>1/2) |
There is only 25% chance that the max of either B or C is < 1/2. However there is a 75% chance that the minimum of either B or C is < 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 = 0.25012"The simulation for max{B,C} supports the table.
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 = 0.75132"The simulation for min{B,C} supports the table.