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: (a) B + C < 1/2
We know that B and C are bounded by the closed interval [0,1]. We also know that B and C are independent, hence the following is true:
Lets let B=x and C=y \[ f(B,C)=f(x,y)=f(x)f(y) \]
We define our probability as follows: \[ P(B+C<1/2)=p(x+y<1/2)\\ \] We can define x in terms of y as \[ y<1/2-x \]
We define our double integral as follows: \[ =\int _{ 0 }^{ 1/2 } \int _{ 0 }^{ 1/2 -x }f(x,y)dxdy \] The inner integral correpsonds to the y variable and the outer intgereal correpsonds to the x variable.When performing double intgration, you want to work from inside to outside. http://tutorial.math.lamar.edu/Classes/CalcIII/DoubleIntegrals.aspx
f(x,y)=1
\[ =\int _{ 0 }^{ 1/2 } \int _{ 0 }^{ 1/2 -x }1dxdy \]
Evaluate the innerintegral first and integrate with respect to y and apply fundamental theorem of calculus part 1.Proceed to evaluate the outer integral.
\[ =\int _{ 0 }^{ 1/2 } ((\frac{1}{2}-x)-0)dx\\ =\int _{ 0 }^{ 1/2 } (\frac{1}{2}-x)dx\\ =((\frac{1}{2} )(\frac{1}{2})-\frac{(\frac{1}{2})^{2}}{2})-(0-0)\\ =\frac{1}{4}-\frac{1}{8}\\ =\frac{1}{8}\\ p(B+C<1/2)=\frac{1}{8} \]
If we were to let B=1, then C would be <1/2 vice versa . We can use integration to solve this problem, however we do not need to use double integration.
\[ BC<\frac{1}{2} \]
Lets set up our problem \[ xy<\frac{1}{2} \]
IF we follow a similar process as part a, then we can re-write y as follows: \[ y<\frac{1}{2x} \]
We set up our integrals as follows: \[ =\frac{1}{2}+\int _{ 1/2 }^{ 1 } \int _{ 0 }^{ 1/2x }f(x,y)dydx \] 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)
\[ =\frac{1}{2}+\int _{ 1/2 }^{ 1}(\frac{1}{2x})dx \]
\[ =\frac{1}{2}+\frac{1}{2}[ln(1)-ln(\frac{1}{2})] \] \[ =\frac{1}{2}+\frac{1}{2}[0-ln(\frac{1}{2})]\\ =\frac{1}{2}+\frac{1}{2}[ln(\frac{1}{2})] \] 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. The intersections of this region with the unit square forms two right triangles. We can simply use the formula for the area of a right trangle. We then subtract that doubled right triangle area from 1.
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 \[ |x-y|<1/2\\ y=x+1/2\\ y=x-1/2 \]
If you plot these two lines on a graph, you would see that they intersect the unit square at the points (0,1/2),(1/2,0), and (1,1/2). The area of interest excludes two right triangles close to the edges each with length 1/2 and base 1/2. This is why it is easier to simply subtract the areas of those right triangles from 1 in order to find the area in question.
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.
\[ 1-P(B\le\frac{1}{2})P(C\le\frac{1}{2})=(\frac{1}{2})(\frac{1}{2})=\frac{1}{4} \]
Of course we can also use techniques of calculus to solve this problem.
\[ Z=min(B,C)<1/2 \] Since in part D, we computed the max of BC<1/2, then the min of BC being less than 1/2 should simply be the compliment but lets see if our work matches that hypothesis
\[ P(B\ge\frac{1}{2},C\ge\frac{1}{2})\\ =1-[1-P(B\le\frac{1}{2})][1-P(C\le\frac{1}{2})]\\ =1-[1-\frac{1}{2}][1-\frac{1}{2}]\\ =1-[\frac{1}{2}][\frac{1}{2}]\\ =1-\frac{1}{4}\\ =\frac{3}{4} \]