Thus dierentiating fy yields fy c 4 1 fx g 1 c g

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Unformatted text preview: iating yields dFY (c) = fY (c) = dc 1 √ 3c 1 √ 6c 0 0≤c<1 1≤c<4 else. By LOTUS, 3 E [Y ] = E [(X − 1)2 ] = 0 1 (u − 1)2 du = 1 3 Example 3.8.4 Suppose X is a continuous-type random variable. (a) Describe the distribution of −X in terms of fX . (b) More generally, describe the distribution of aX + b in terms of fX , for constants a and b with a = 0. (This generalizes Section 3.6.1, which covers only the case a > 0.) Solution: (a) Let Y = −X, or equivalently, Y = g (X ) where g (u) = −u. We shall find the pdf of Y after first finding the CDF. For any constant c, FY (c) = P {Y ≤ c} = P {−X ≤ c} = P {X ≥ −c} = 1 − FX (−c). Differentiating with respect to c yields fY (c) = fX (−c). Geometrically, the graph of fY is obtained by reflecting the graph of fX about the vertical axis. (b) Suppose now that Y = aX + b. The pdf of Y in case a > 0 is given in Section 3.6.1. So suppose a < 0. Then FY (c) = P {aX + b ≤ c} = P {aX ≤ c − b} = P X ≥ c−b = 1 − FX c−b . a a Differentiating...
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This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Institute of Technology.

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