Ec 1e 0 1 ec 1 e 0c1 otherwise to nd the limit of

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Unformatted text preview: with respect to c, using the chain rule and the fact: Φ (s) = fY (c) = √1 24πc √ exp − ( c−2)2 6 √ + exp − ( c+2)2 6 √1 2π 2 exp(− s2 ), to obtain if c ≥ 0 0 (3.4) if c < 0. Example 3.8.3 Suppose X is uniformly distributed over the interval [0, 3], and Y = (X − 1)2 . Find the CDF, pdf, and expectation of Y. Solution. Since X ranges over the interval [0, 3], Y ranges over the interval [0, 4]. The expression for FY (c) is qualitatively different for 0 ≤ c ≤ 1 and 1 ≤ c ≤ 4, as seen in the following sketch: 4 4 c 1 c 1 1 3 1 3 In each case, FY (c) is equal to one third the length of the shaded interval. For 0 ≤ c ≤ 1, √ √ √ 2c 2 FY (c) = P {(X − 1) ≤ c} = P {1 − c ≤ X ≤ 1 + c} = . 3 For 1 ≤ c ≤ 4, 2 FY (c) = P {(X − 1) ≤ c} = P {0 ≤ X ≤ 1 + √ √ 1+ c c} = . 3 102 CHAPTER 3. CONTINUOUS-TYPE RANDOM VARIABLES Combining these observations yields: FY (c) = 0 c<0 0≤c<1 1≤c<4 c ≥ 4. √ 2c 3 √ 1+ c 3 1 Different...
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