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MATH 5010–001 SUMMER 2003 SOLUTIONS TO ASSIGNMENT 6 Throughout these exercises, we will encounter compute this once and for all. Lemma. For any k ≥ 0, Z ∞ R y k e−y dy a few times. So let us y k e−y dy = k!. 0 Remark. By now, you should recognize the integral as Γ(k + 1) and so the above lemma should be familiar. Here is the familar proof. R R Proof. Recall integration by parts: u dv = uv − v du. Apply this with u = y k and dv = e−y dy to get du = ky k−1 dy and v = −e−y . Thus, Z 0 ∞ k −y y e Z ∞ k −y dy = −y e +k 0 ∞ y 0 k−1 −y e Z dy = k ∞ 0 y k−1 e−y dy. R∞ That is, ak = 0 y k e−y dy satisfies the recursion, ak = kak−1 for all k ≥ 1. Apply this recursion to itself, ak = kak−1 = k(k − 1)ak−2 = · · · = k!a0 . R ∞ −y But a0 = 0 e dy = 1. This proves the lemma. Problems from pp. 228–231 R 2. First find C by setting Z 1=C f dx = 1. But ∞ 0 −x/2 xe Z dx = 4C ∞ 0 ye−y dy (y = x/2). Thanks to the Lemma, this equals 4C, which means that C = 14 . Consequently, 1 P {X ≥ 5} = 4 " Z ∞ 5 −x/2 xe Z dx = Z ∞ −y = −ye + 5/2 = ∞ 5/2 ∞ 5/2 ye−y dy (y = x/2) # e−y dy 5 −5/2 7 + e−5/2 = e−5/2 . e 2 2 6. (a) Change variable (y = x/2), and apply the Lemma: E(X) = R∞ 2 0 y 2 e−y dy = 4. 1 1 4 R∞ 0 x2 e−x/2 dx = R R1 6. (b) First, find c by setting f dx = 1. That is, 1 = c −1 (1 − x2 ) dx = R1 In particular, E(X) = 34 −1 x(1 − x2 ) dx = 0. 6. (c) E(X) = R∞ 5 5 x 4c 3 . So, c = 34 . dx = +∞. 21. Let X denote the height, in inches, of a randomly-selected 25-year old man. The first part asks for P {X ≥ 74}. Since X is normally distributed with mean µ = 61 and standard deviation σ = 2.5, Z = (X − µ)/σ is normally distributed with mean zero and standard deviation one. So use the normal table (p. 203) to obtain 74 − 71 P {X ≥ 74} = P Z ≥ = 1 − Φ(1.2) = 1 − 0.8849 = 0.1151. 2.5 In the second part, we are asked to find: P {X ≥ 77 | X ≥ 72} = 1 − Φ(2.4) 1 − 0.9918 P {X ≥ 77} = = = 0.023796. P {X ≥ 72} 1 − Φ(0.4) 1 − 0.6554 31. (a) If X is uniformly distributed on (0, A), then for any number a ∈ (0, A), Z Z Z 1 A 1 a 1 A E{|X − a|} = |x − a| dx = (a − x) dx + (x − a) dx A 0 A 0 A a (A − a)2 a2 + . = 2A 2A a Call this ψ(a) and minimize ψ by letting ψ 0 = 0 and check ψ 00 . Now ψ 0 (a) = A − A−a = A 2a−A A 2 A 0 00 A . Therefore, ψ = 0 when a = 2 . Also, ψ (a) = A ≥ 0 so the value of a = 2 is a minimum. Finally, A A (A/2)2 (A/2)2 + = . min E{|X − a|} = ψ = 2 2A 2A 4 a∈(0,A) 31. (b) In a similar way, Z ∞ Z a Z ∞ −λx −λx E{|X − a|} = |x − a|λe dx = λ (a − x)e dx + λ (x − a)e−λx dx 0 0 a Z a Z ∞ = λe−λa yeλy dy + λe−λa ze−λz dz. 0 0 I have changed variables by letting y = a − x and z = a − x. Change variables once more (w = λy, v = λz) to obtain: Z Z e−λa aλ w e−λa ∞ −v E{|X − a|} = we dw + ve dv λ λ 0 0 "Z # aλ e−λa = wew dw + 1 . λ 0 2 The Lemma was used in the last line. Now integrate by parts to see that R aλ w aλ wew |aλ − eaλ + 1. Therefore, 0 − 0 e dw = aλe R aλ 0 wew dw = e−λa aλeaλ − eaλ + 2 λ 1 2e−aλ =a− + . λ λ E{|X − a|} = Call this ψ(a) and minimize. Clearly, ψ 0 (a) = 1 − 2e−aλ and ψ 00 (a) = 2λe−aλ ≥ 0. So we have a minimum where ψ 0 = 0; i.e., e−aλ = 12 , which is at a = ln(2) λ . Thus, min E{|X − a|} = ψ a>0 3 ln(2) λ = ln 2 . λ