STAT 230 Final with Solution, Fall 2008

American University of Beirut STAT 230 Introduction to Probability and Random Variables Fall 2008-2009 Final Exam - solution Exercise 1 Let f(x, y) = 2 , 0 ≤ x ≤ y < 1, be the joint pdf of X and Y . Find P(X > Y 2). y = x x 6 - R y ................. 1 x = y2 P(X > Y 2) = � 1 0 � y y2 2 dxdy = � 1 0 2(y − y2)dy = 1/3 Exercise 2 Let X1, X2, X3 be independent random variables that represent lifetimes (in hours) of three key components of a device. Say their respective distributions are exponential with means 1000, 1500, and 2000. Let Y = min(X1, X2, X3). Find P(Y > 1000). P(Y > 1000) = P(min(X1, X2, X3) > 1000) = P(X1 > 1000) ∗ P(X2 > 1000) ∗ P(X3 > 1000) (cause X1, X2, X3 are independent) but if X �→ f(x) = 1 θe−x/θ, then P(X > y) = � ∞ y 1 θe−x/θdx = e−y/θ ; hence P(X1 > 1000) = e−1, P(X2 > 1000) = e−2/3 , and P(X3 > 1000) = e−0.5 and P(Y > 1000) = e−1e−2/3e−0.5 = e−13/6 Exercise 3 A consumer buys n light bulbs, each of which has a lifetime that has a normal dis- tribution with mean 800 hours, and a standard deviation of 100 hours. A light bulb is replaced by another as soon as it burns out. Assuming independence of the lifetimes, find the smallest n so that the succession of light bulbs produces light for at least 10000 hours with probability 0.8997. Let X1, X2, . . . , Xn be the lifetimes of the n bulbs used in succession. find n such that P(X1 + X2 + . . . + Xn > 10000) = 0.8997; but Xi �→ N(800, 1002), and X1 + X2 + . . . + Xn �→ N(800n, 1002n) hence P(X1 + X2 + . . . + Xn > 10000) = P( X1+X2+...+Xn−800n 100√n > 100−8n √n ) = P(Z > 100−8n √n ) = 0.8997. Form normal table, 100−8n √n = −1.28, and n = 14 by solving a quadratic equation. Exercise 4 Let X and Y be a couple of random variables with pdf f(x, y) = 1 x2y2 x ≥ 1 , y ≥ 1 a. find the joint pdf of U = XY and V = X/Y solving the system of equations yields x = √uv and y = � u v the Jacobian is J(u, v) = ������ √v 2√u √u 2√v 1 2√uv − √u 2v√v ������ = − 1 2v and hence g(u, v) = 1 2u2v uv ≥ 1 , u v ≥ 1 b. find the marginal pdf of U and V 6 - u v ............. v = u v = 1 u 1 R h(u) = � u 1 u 1 2u2vdv = 1 2u2 � ln v �u 1 u = ln u u2 u ≥ 1 k(v) =            � +∞ 1/v 1 2u2v du = 1 2 0 ≤ v ≤ 1 � +∞ v 1 2u2v du = 1 2v2 v ≥ 1 Exercise 5 Let X have a beta distribution with parameters α and β; the pdf of X is given by: f(x) = Γ(α + β) Γ(α)Γ(β) xα−1(1 − x)β−1 0 < x < 1 E(X) = � 1 0 Γ(α + β) Γ(α)Γ(β) xα(1 − x)β−1dx = α α + β � 1 0 Γ(α + β + 1) Γ(α + 1)Γ(β) xα(1 − x)β−1dx = α α + β (the last integral is the pdf of a beta distribution with parameters α + 1 and β) E(X2) = � 1 0 Γ(α + β) Γ(α)Γ(β) xα+1(1−x)β−1dx = α(α + 1) (α + β + 1)(α + β) � 1 0 Γ(α + β + 2) Γ(α + 2)Γ(β) xα(1−x)β−1dx = α(α + 1) (α + β + 1)(α + β) hence, V ar(X) = E(X2) − (E(X))2 = α(α + 1) (α + β + 1)(α + β) − � α α + β �2 = αβ (α + β + 1)(α + β) Exercise 6 Fifty numbers are rounded off to the nearest integer and then summed. If the individual round-off errors are uniformly distributed over the interval (−1/2, 1/2). Find the probability that the resultant sum differs from the exact sum by more than 3. Let X1, X2, . . . , X50 be the errors for the 50 numbers; Xi �→ U(−1/2, 1/2), E(Xi) = 0, and V ar(Xi) = 1/12 X1+X2+...+X50 50 ∼ N(0, 1 600) by the central limit theorem. P(−3 < X1 + X2 + . . . + X50 < 3) = P(−3 √ 600 50 < Z < 3 √ 600 50 ) = P(−1.47 < Z < 1.47) = 0.9292 − (1 − 0.9292) = 0.86 (from the normal table)