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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. CONTINUOUSTYPE 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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 Spring '08
 Zahrn
 The Land

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