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ISYE6644 Midterm II solutions B-ALL ANSWERS CORRECT

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ISyE 6644 | Fall 2020 | Test #2b Solutions (Revised 11/6/20) This test is 120 minutes. There are 38 questions, each worth 2.5 points, and a \free" 39th question, worth 5 points (total = 100). In additon, later on, we will add 4 points to everyone’s score (up to a max of 100). But in exchange, you will not be allowed to whine about (and we will not consider) any grade scoring issues unless they involve more than 4 points. Let’s get some additional awyer stuff out of the way. . . You’re allowed the following items: • Pencil / pen and scratch paper. • A reasonable calculator. • Two cheat sheets (4 sides total). • Normal, t, and χ2 tables. (I will supply these.) But note that • You are not allowed to use Arena, even though I’m asking questions about it. • This test requires some sort of proctor. • If you encounter a ProctorTrack issue, contact us immediately (but don’t get an ulcer over it). Good luck! I want you to make this test sorry that it ever tried to mess around with you!!! 1. Suppose X and Y are continuous random variables with joint p.d.f. f(x; y) = cxy2; 0 ≤ x ≤ 2; 0 ≤ y ≤ 1, for some appropriate constant c. Find E[2X − 1]. (a) 1/6 (b) 5/3 (c) 5/2 (d) 11/32 (e) 12 Solution: First, we need to find c, so we set: 1 = Z0 1 Z0 2 f(x; y) dx dy = Z0 1 Z0 2 cxy2 dx dy = 2c 3 : Therefore, c = 3 2. Next, we find the marginal p.d.f. fX(x): fX(x) = Z0 1 3 2xy2 dy = x=2; 0 ≤ x ≤ 2: Finally, we compute E[X]: E[X] = Z0 2 xfX(x) dx = 4=3: Thus, E[2X − 1] = 2E[X] − 1 = 5=3. Thus, the answer is (b). 2 2. Suppose X and Y are discrete random variables with the following joint p.m.f. f(x; y) X = 2 X = 3 Pr(Y = y) Y = 0 0.2 0.5 0.7 Y = 1 0.0 0.3 0.3 Pr(X = x) 0.2 0.8 1.0 Find Var(X + 2Y ). (a) 0.16 (b) 0.21 (c) 0.49 (d) 1.243 (e) 3.14 Solution: We can compute the following quantities of interest: E[X] = X x xfX(x) = 2:8 E[X2] = X x x2fX(x) = 8 Var(X) = E[X2] − (E[X])2 = 0:16 E[Y ] = X y yfY (y) = 0:3 E[Y 2] = X y y2fY (y) = 0:3 Var(Y ) = E[X2] − (E[X])2 = 0:21 E[XY ] = X x Xy xyfX;Y (x; y) = 0:9 Cov(X; Y ) = E[XY ] − E[X]E[Y ] = 0:06 Therefore, Var(X + 2Y ) = Var(X) + 4Var(Y ) + 4Cov(X; Y ) = 1:24: Thus, the answer is (d). 2 3. If X and Y are independent Exp(λ = 1) random variables, find Cov(−3X; 4Y − 2). (a) −3 (b) −2 (c) 0 (d) 0.5 (e) 1 Solution: By independence, Cov(−3X; 4Y − 2) = −12Cov(X; Y ) = 0, i.e., choice (c)

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