P EXAM QUESTIONS OF THE WEEK
S. Broverman, 2005
Week of January 23/06
\ and ] have a joint distribution on the two-dimensional region " B " , # C #, and the pdf of the joint distribution is 0 B C - (constant) on the region. Find the probability T l\l l] l .
The solution can be found below.
Week of January 23/06 - Solution
Whenever a joint distribution has a constant density over the entire probability space W , the probability of any subregion E is equal to Area of E . Area of W The area of the full probability space is # % ) (the area of a rectangle with sides that are 2. The graph of the region defined by l\l l] l is the shaded region illustrated below To find this region, we first find the boundary l\l l] l , which is the combination of the two lines C B and C B. We then determine "which sides" of the lines represent the inequality. Alternatively, we consider the relationship between and B C in the four quadrants: (i) 1st quadrant, B ! C ! so lBl B lCl C and the inequality becomes B C , (ii) 2nd quadrant, B ! C ! so lBl B lCl C the inequality becomes B C , (iii) 3rd quadrant lBl B lCl C the inequality becomes B C (or B C ), and (iv) 4th quadrant lBl B lCl C the inequality becomes B C (or B C ) .
The area of the shaded region is ) (Area of blank region) , and the area of the blank region is 2 (two triangles, each with base 2 and height 1), so the area of the shaded region is 6. The probability is T l\l l] l ' . ) Note that since the joint pdf is the constant - , it must be true that " " " - Area of Total Region of Probability #% ) . The probability would be the double integral of the joint pdf over the shaded region. Since the joint pdf is constant at " , the integral over the shaded region would be ) " (Area of shaded region) . )