Recall from section 25 that the geometric

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Unformatted text preview: = fX (u)du −∞ 1 A(1 − u2 )du = −1 = A u− u3 3 1 = −1 4A , 3 so A = 3 . 4 The support of fX (the set on which it is not zero) is the interval [−1, 1]. To find P {0.5 < X < 1.5}, we only need to integrate over the portion of the interval [0.5, 1.5] in the support of fX . That is, we only need to integrate over [0.5, 1] : 1.5 P {0.5 < X < 1.5} = fX (u)du 0.5 1 = 0.5 = 3 4 3(1 − u2 ) du 4 u− u3 3 1 = 0 .5 3 4 2 11 − 3 24 = 5 . 32 Because the support of fX is the interval [−1, 1], we can immediately write down the partial answer: 0 c ≤ −1 ? −1 < c < 1 FX (c) = 1 c ≥ 1, where the question mark represents what hasn’t yet been determined, namely, the value of FX (c) for −1 < c < 1. So let −1 < c < 1. Then FX (c) = P {X ≤ c} c 3(1 − u2 ) = du 4 −1 3 u3 c u− = 4 3 −1 = 2 + 3c − c3 . 4 This allows us to give the complete expression for FX : 0 c ≤ −1 2+3c−c3 FX (c) = −1 < c < 1 4 1 c ≥ 1. 3.3. UNIFORM DISTRIBUTION 77 ∞ −∞ ufX (u)du, The mean, µX , is zero, because µX = over R is zero. (For example, ufX (u) = 9 32 and ufX (u) is an odd function so its integral and uf...
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