To double check the answer note that if 1 i 6 then p g

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Unformatted text preview: (−c). Thus, FY (c) = FX (c) − FX (−c) c ≥ 0 0 c ≤ 0; Differentiating to get the pdf yields: fY (c) = fX (c) + fX (−c) c ≥ 0 0 c < 0; Basically, for each c > 0, there are two terms in the expression for fY (c) because there are two ways for Y to be c–either X = c or X = −c. A geometric interpretation is given in Figure 3.19. Figure 3.19(a) pictures a possible pdf for X. Figure 3.19(b) shows a decomposition of the probability mass into a part on the negative line and part on the positive line. Figure 3.19(c) shows the result of reflecting the probability mass on the negative line to the positive line. Figure 3.19(d) shows the pdf of |Y |, obtained by adding the two functions in Figure 3.19(c). Another way to think of this geometrically would be to fold a picture of the pdf of X in half along the vertical axis, so that fX (c) and fX (−c) are lined up for each c, and then add these together to get fY (c) for c ≥ 0.. Example 3.8.9 Let X be an exponentially distributed random variable with parameter λ. Let...
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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 Tech.

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