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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 continuoustype 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.
 Spring '08
 Zahrn
 The Land

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