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Unformatted text preview: Math 302.102 Fall 2010 Solutions to Assignment #6 1. The definition of conditional probability implies that P { X > t + s  X > t } = P { X > t + s,X > t } P { X > t } = P { X > t + s } P { X > t } since the only way for both { X > t + s } and { X > t } to happen is if { X > t + s } happens. Since X ∼ Exp( λ ), we find that if a > 0, then P { X > a } = Z ∞ a λe λx d x = e λx ∞ a = e λa . Therefore, P { X > t + s  X > t } = P { X > t + s } P { X > t } = e λ ( t + s ) e λt = e λs = P { X > s } . Equivalently, Bayes’ rule implies that P { X > t + s  X > t } = P { X > t  X > t + s } P { X > t + s } P { X > t } = P { X > t + s } P { X > t } since P { X > t  X > t + s } = 1; that is, if know X is at least t + s , then we know with certainty that X is at least t . We find, as above, P { X > t + s  X > t } = P { X > t + s } P { X > t } = e λ ( t + s ) e λt = e λs = P { X > s } . 2. If X ∼ Unif(0 , 1), then f ( x ) = ( 1 , ≤ x ≤ 1 , , otherwise . (a) We find μ = E ( X ) = Z ∞∞ xf ( x )d x = Z 1 x · 1d x = 1 2 x 2 1 = 1 2 and therefore σ 2 = var( X ) = Z ∞∞ ( x μ ) 2 f ( x )d x = Z 1 ( x 1 / 2) 2 · 1d x = 1 3 ( x 1 / 2) 3 1 = (1 / 2) 3 ( 1 / 2) 3 3 = 1 12 . Equivalently, var( X ) can be determined by first computing E ( X 2 ) = Z ∞∞ x 2 f ( x )d x = Z 1 x 2 · 1d x = 1 3 x 3 1 = 1 3 and noting that σ 2 = var( X ) = E ( X 2 ) [ E ( X )] 2 = 1 3 1 2 2 = 1 12 . (b) We begin by noting that P { μ 2 σ < X < μ + 2 σ } = P 1 2 1 √ 3 < X < 1 2 + 1 √ 3 = Z 1 2 + 1 √ 3 1 2 1 √ 3 f ( x )d x. Observe that 1 2 1 √ 3 < 0 and 1 2 + 1 √ 3 > 1 . Thus, since f ( x ) = 1 only when 0 ≤ x ≤ 1, we see that Z 1 2 + 1 √ 3 1 2 1 √ 3 f ( x )d x = Z 1 2 1 √ 3 f ( x )d x + Z 1 f ( x )d x + Z 1 2 + 1 √ 3 1 f ( x )d x = Z 1 2 1 √ 3 0d x + Z 1 1d x + Z 1 2 + 1...
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This note was uploaded on 04/07/2011 for the course MATH 302 taught by Professor Israel during the Spring '08 term at UBC.
 Spring '08
 ISRAEL
 Conditional Probability, Probability

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