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English Pages 721 [1] Year 2014
(a) From the above PDF we can determine the value of c by integrating the PDF and setting it equal to 1, yielding Z 2 cx dx = 2c = 1. (2) 0
Therefore c = 1/2. R1 (b) P[0 X 1] = 0
x 2
dx = 1/4.
(c) P[ 1/2 X 1/2] =
R 1/2 0
x 2
dx = 1/16.
(d) The CDF of X is found by integrating the PDF from 0 to x. 8 > x < 0, Z x
0 : 1 x > 2.
(3)
Problem 4.3.2 Solution From the CDF, we can find the PDF by direct di↵erentiation. The CDF and corresponding PDF are 8 > x < 1,
: 1 x > 1, ( 1/2 1 x 1, fX (x) = (2) 0 otherwise.
Problem 4.3.3 Solution We find the PDF by taking the derivative of FU(u) on each piece that FU(u) is
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