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Unformatted text preview: ) = 1 for all x 0. Observe also that P [ X 1] = F X (1) = 1 3 + 0 = 1 3 while, on the other hand, P [ X < x ] = F X ( x ) = 0 for any x < 0. This implies that P [ X =1] = 1 3 What is happening here is that the function F X has a discontinuity at1. Let us calculate P [  X1 3  < 1]: P X1 3 < 1 = P 2 3 < X < 4 3 = P X < 4 3 P X 2 3 = 1F X 2 3 = 1 1 3 + 2 3 1 3 2 1...
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 Spring '02
 Sengupta
 Probability

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