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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: iating yields dFY (c) = fY (c) = dc 1 √ 3c 1 √ 6c 0 0≤c&lt;1 1≤c&lt;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 &gt; 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 &gt; 0 is given in Section 3.6.1. So suppose a &lt; 0. Then FY (c) = P {aX + b ≤ c} = P {aX ≤ c − b} = P X ≥ c−b = 1 − FX c−b . a a Diﬀerentiating...
View Full Document

## 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.

Ask a homework question - tutors are online