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 that
B and C are have the boundary interval [0,1]and are independent and thus:
Lets let B=x and C=y \[ f(B,C)=f(b)f(c) \]
We define our probability as follows: \[ P(B+C<1/2)=p(b+c<1/2)\\ \] We can define b in terms of c as \[ c<1/2-b \]
We define our double integral as follows: \[ =\int _{ 0 }^{ 1/2 } \int _{ 0 }^{ 1/2 -b }f(b,c)dbdc \]
f(b,c)=1
\[ =\int _{ 0 }^{ 1/2 } \int _{ 0 }^{ 1/2 -b }1dbdc \]
Following standard calculus, we evaluate the inner and outer intergrals with respect to be and c \[ =\int _{ 0 }^{ 1/2 } ((\frac{1}{2}-b)-0)db\\ =\int _{ 0 }^{ 1/2 } (\frac{1}{2}-b)db\\ =((\frac{1}{2} )(\frac{1}{2})-\frac{(\frac{1}{2})^{2}}{2})-(0-0)\\ =\frac{1}{4}-\frac{1}{8}\\ =\frac{1}{8}\\ p(B+C<1/2)=\frac{1}{8} \]
for this problem we dont need double intergration. it is solved as below
\[ BC<\frac{1}{2} \]
\[ xy<\frac{1}{2} \]
\[ y<\frac{1}{2b} \]
We set up our integrals as follows: \[ =\frac{1}{2}+\int _{ 1/2 }^{ 1 } \int _{ 0 }^{ 1/2b }f(b,c)dcdb \] We pretty much divide our region into two parts taking the area of x from 0 to y=1/2x using geometry and 1/2 to 1. The points of intersection would be (1/2,1) and (1,1/2)
\[ =\frac{1}{2}+\int _{ 1/2 }^{ 1}(\frac{1}{2b})db \]
\[ =\frac{1}{2}+\frac{1}{2}[ln(1)-ln(\frac{1}{2})] \] \[ =\frac{1}{2}+\frac{1}{2}[0-ln(\frac{1}{2})]\\ =\frac{1}{2}+\frac{1}{2}[ln(\frac{1}{2})] \] If you punch this into a calculator, the probability is roughly .8
Hence, we have found the probability of BC<1/2
\[ |B-C|<1/2 \]
Using the definition of absolute value, we consider C to be the lines C=B+1/2 and C=B-1/2 and since the interesection forms a right triangle, we can use the triangle area formula to aid in this computation. The base has a unit length of 1/2 and the length is 1 unit
\[ 1-2A(triangle)\\ 1-2\frac{1}{2}bh\\ 1-2(\frac{1}{2}(\frac{1}{2})(\frac{1}{2}))\\ =\frac{3}{4} \]
This is what is happening geometrically \[ |B-C|<1/2\\ C=B+1/2\\ C=B-1/2 \]
we need to compute maximum of function items to compute \[ Z=max(B,C)<1/2 \]
We can re-write as \[ P(B\le\frac{1}{2},C\le\frac{1}{2}) \] Just by inference, if the max of the product BC are less than a half, then we can assume that B and C are both going to be less than or equal to a half.
\[ 1-P(B\le\frac{1}{2})P(C\le\frac{1}{2})=(\frac{1}{2})(\frac{1}{2})=\frac{1}{4} \]
\[ Z=min(B,C)<1/2 \]
\[ P(B\ge\frac{1}{2},C\ge\frac{1}{2})\\ =1-[1-P(B\le\frac{1}{2})][1-P(C\le\frac{1}{2})]\\ =1-[1-\frac{1}{2}][1-\frac{1}{2}]\\ =1-[\frac{1}{2}][\frac{1}{2}]\\ =1-\frac{1}{4}\\ =\frac{3}{4} \]