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.
Proof of Proper Probability Distribution (Uniform Continuous Probability Distribution)
B and C defines a coordinate of A point in the unit square [0,1] as point (B,C).
B and C are subsets of real numbers that are randomly drawn and independent; and B and C are continuous between intervals 0 and 1.
Both B and C are equally likely to fall in any subset E of the unit interval [0,1].
Find the probability that:
P((B + C)) < 1/2
For a Uniform Continuous Probability Distribution - The CDF for B and C are -
P(B < 1/2) = \(\int_{0}^{1_2}\) x dB = 1/2
P(C < 1/2) = \(\int_{0}^{1_2}\) x dC = 1/2
Since max area of B and C = 1 + 1 = 2 units, min area = 0
p((B+C)<1/2) < 1/4 of the area for combined area of B and C.
x = punif (0.5 , min = 0, max = 2 )
x## [1] 0.25Maximum area for BC when B = 1 and C = 1 is 1. Miniimum area is 0. P(BC < 1/2) < 1/2
x = punif (0.5 , min = 0, max = 1 )
x## [1] 0.5maximum area for |B - C| is 1 when either B or C = 1 and the other equal to 0.
minimum area is 0 when both are 0. p(|B - C| < 1/2) < 1/2
x = punif (0.5 , min = 0, max = 1 )
x## [1] 0.5maximum area for function max{B,C} = 1 when either or both B and C = 1. minimum area is 0 when both are 0. p(max{B,c} < 1/2) < 1/2
x = punif (0.5 , min = 0, max = 1 )
x## [1] 0.5maximum area for function min{B,C} = 1 when either or both B and C = 1. minimum area is 0 when both are 0. p(max{B,c} < 1/2) < 1/2
x = punif (0.5 , min = 0, max = 1 )
x## [1] 0.5