A geometric interpretation is given in figure 319

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Unformatted text preview: egative reals, so the support of Y is the nonnegative reals. Also, there is no value v such that P {Y = v } > 0. So Y is a continuous-type random variable and we will find its CDF. At this point we have: 0 c<0 FY (c) = ??? c ≥ 0; we must find FY (c) for c ≥ 0. This completes Step 1, “scoping the problem,” so we take a breath. Continuing, for c ≥ 0, √ √ P {X 2 ≤ c} = P {− c ≤ X ≤ c} √ = √ −c √ c = √ c fX (u)du = c √ −c exp(−|u|) du 2 exp(−u)du = 1 − e− 0 where the setup here is illustrated in Figure 3.14. So √ c , 100 CHAPTER 3. CONTINUOUS-TYPE RANDOM VARIABLES f (u) X g(u) c u c !c Figure 3.14: FX (c) is area of the shaded region. 0 √ c<0 1 − e− c c ≥ 0. FY (c) = Finally, taking the derivative of FY (using the chain rule of calculus) yields 0 √ fY (c) = e− c √ 2c c≤0 c > 0, Note that the value of a pdf at a single point can be changed without changing the integrals of the pdf. In this case we decided to let fY (0...
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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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