Dr. Hackney STA Solutions pg 47

# Dr. Hackney STA Solutions pg 47 - Second Edition 4-3 1 V 2...

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Second Edition 4-3 U 1 2 3 2 1 12 1 6 1 12 V 3 1 12 1 6 1 12 4 1 12 1 6 1 12 4.11 The support of the distribution of ( U, V ) is { ( u, v ) : u = 1 , 2 , . . . ; v = u + 1 , u + 2 , . . . } . This is not a cross-product set. Therefore, U and V are not independent. More simply, if we know U = u , then we know V > u . 4.12 One interpretation of “a stick is broken at random into three pieces” is this. Suppose the length of the stick is 1. Let X and Y denote the two points where the stick is broken. Let X and Y both have uniform(0 , 1) distributions, and assume X and Y are independent. Then the joint distribution of X and Y is uniform on the unit square. In order for the three pieces to form a triangle, the sum of the lengths of any two pieces must be greater than the length of the third. This will be true if and only if the length of each piece is less than 1 / 2. To calculate the probability of this, we need to identify the sample points ( x, y ) such that the length of each piece is less than 1 / 2. If y > x , this will be true if x < 1 / 2, y - x < 1 / 2 and 1 - y < 1 / 2.
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