Georgia Institute Of Technology ISYE 6644 f19-HW 3-Module 2b-solns-GRADED A+

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ISyE 6644 | HW3 | Fall 2019 (covers Module 2.10-2.16) 1. (Lesson 2.10: Conditional Expectation.) BONUS: Suppose that f(x; y) = 6x for 0 ≤ x ≤ y ≤ 1: Hint (you may already have seen this some... place): The marginal p.d.f. of X turns out to be fX(x) = 6x(1 − x) for 0 ≤ x ≤ 1: Find the conditional p.d.f. of Y given that X = x. (a) f(yjx) = 1−1x; 0 ≤ x ≤ y ≤ 1 (b) f(yjx) = 1−1x; 0 ≤ x ≤ 1 (c) f(yjx) = 1−1y ; 0 ≤ y ≤ 1 (d) f(xjy) = 1−1x; 0 ≤ x ≤ y ≤ 1 Solution: f(yjx) = f(x; y) fX(x) = 6x 6x(1 − x) = 1 1 − x; 0 ≤ x ≤ y ≤ 1: So the answer is (a).  2. (Lesson 2.10: Conditional Expectation.) BONUS: Again suppose that f(x; y) = 6x for 0 ≤ x ≤ y ≤ 1: Hint (you may already have seen this someplace): The marginal p.d.f. of X turns out to be fX(x) = 6x(1 − x) for 0 ≤ x ≤ 1: Find E[Y jX = x]. (a) E[Y jX = x] = 1=2; 0 ≤ x ≤ 1 (b) E[Y jX = x] = 1+2x; 0 ≤ x ≤ 1 (c) E[Y jX = x] = 1+2y ; 0 ≤ y ≤ 12 (d) E[XjY = y] = 1+2 y; 0 ≤ y ≤ 1 Solution: By the definition of conditional expectation, we have E[Y jX = x] = Z−1 1 yf(yjx) dy = Zx1 1 −y x dy = 1 +2 x; 0 ≤ x ≤ 1: So the answer is (b).  3. (Lesson 2.10: Conditional Expectation.) BONUS: Yet again suppose that f(x; y) = 6x for 0 ≤ x ≤ y ≤ 1: Hint (you may already have seen this someplace): The marginal p.d.f. of X turns out to be fX(x) = 6x(1 − x) for 0 ≤ x ≤ 1: Find EhE[Y jX]i. (a) 1/2 (b) 2/3 (c) 3/4 (d) 1 Solution: By the Law of the Unconscious Statistician and part (b), EhE[Y jX]i = Z0 1 E[Y jx]fX(x) dx = Z0 1 1 +2 x 6x(1 − x) dx = 3=4: So the answer is (c).  Let’s check this answer. First of all, the marginal p.d.f. of Y is fY (y) = Z0 y f(x; y) dx = Z0 y 6x dx = 3y2; for 0 ≤ y ≤ 1:3 Then E[Y ] = R01 yfY (y) dy = R01 3y3 dy = 3=4 [Show More]

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