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.
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] 0.6908698 0.3928684 0.1078697 0.7738888 0.1766512 0.1848385tail(B)## [1] 0.45791471 0.52997784 0.87696845 0.66126171 0.90010342 0.02680453C <- runif(100000)
head(C)## [1] 0.7829510 0.7792084 0.8440596 0.4678556 0.1533400 0.6504237tail(C)## [1] 0.1792427 0.5371778 0.2676244 0.6793426 0.2577355 0.6636308head(B+C)## [1] 1.4738208 1.1720768 0.9519293 1.2417444 0.3299912 0.8352622tail(B+C)## [1] 0.6371574 1.0671556 1.1445928 1.3406043 1.1578389 0.6904353Note that the point (B,C) is then chosen at random in the unit square. Find the probability that
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 = 0.12478"The probabilty is 12.5%. For this to be true B and C must be \(< \frac{1}{2}\), which is \(0.5*0.5 = 0.25\) and C or B must be \(< \frac{1}{2} - [Compliment]\) 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.
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 = 0.84606"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 = 0.75097"The closer one number get to another, the closer 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 \(1/2*1/2 = 1/4 = 0.25\). Such that \(0.5+0.25 =0.75\) , which is supported by the simulation.
Both (d) and (e) can best be discribed 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 \(< \frac{1}{2}\). However there is a 75% chance that the minimum of either B or C is \(< \frac{1}{2}\).
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 = 0.25047"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 = 0.75024"The simulation for min{B,C} supports the table.