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Econ 162a: Introduction to Probability and Statistics Midterm Exam, Fall 2001 Instructor: Hanming Fang October 25, 2001 Instructions: You have 80 minutes. Answer every question. Allocate your time as suggested by the points. The total points of this exam is 100. Be sure to answer the questions directly and precisely, and present your argument clearly. Good luck! 1. [Bayes’ Theorem, 15 points] Three plants, C1 , C2 and C3 , produce respectively, 10, 50 and 40 percent of a company’s output. Although plants C1 is a small plant, its manager believes in high quality and only 1 percent of its products are defective. The other two, C2 and C3 , are worse and produce items that are 3 and 4 percent defective, respectively. All plants are sent to a central warehourse. One item is selected and observed to be defective. What is the conditional probability that it comes from plant C1 ? [Hint: Write Ci as the event that a product is produced at firm i; and write D as the event that a product is defective] Suggested Answer: Write Ci as the event that the product is from plant i. From the description of the problem, we know that P (C1 ) = 0.1, P (C2 ) = 0.5, P (C3 ) = 0.4. Denote D as the event that a product is defective. We know that P (D|C1 ) = 0.01, P (D|C2 ) = 0.03, P (D|C3 ) = 0.04. Hence P (C1 |D) = = P (D|C1 ) P (C1 ) P (D|C1 ) P (C1 ) + P (D|C2 ) P (C2 ) + P (D|C3 ) P (C3 ) 0.1 × 0.01 1 = . 0.1 × 0.01 + 0.5 × 0.03 + 0.4 × 0.04 32 2. [Marginal and Conditional Probabilities, 20 points] A national survey of couples showed that 30% of the wives watched the superbowl, and 50% of the husbands. Also, if the wife watched, the probability that the husband watched increased to 60%. For a couple drawn at random, what is the probability that: a. the couple both watch; b. at least one watches; c. neither watches; d. if the husband watches, the wife watches. 1 Suggested Answer: Let H (respectively, W ) be the random variable which equals to 1 if the husband (respectively, the wife) watches the superbowl, and 0 otherwise. We are told that PW (1) = 0.3, PH (1) = 0.5, and furthermore PH|W (1|1) = 0.6. Now let us denote P (h, w) be the joint probability of W = w, H = h. We have PW (1) = P (1, 1) + P (0, 1) = 0.3 PH (1) = P (1, 1) + P (1, 0) = 0.5 P (1, 1) PH|W (1|1) = = 0.6. PW (1) Hence P (1, 1) = 0.6 × 0.3 = 0.18, P (0, 1) = 0.3 − 0.18 = 0.12, and P (1, 0) = 0.5 − 0.18 = 0.32. Finally P (0, 0) = 1 − P (1, 1) − P (1, 0) − P (0, 1) = 0.38. Therefore, the answers are: (a). P (1, 1) = 0.18; (b). 1 − P (0, 0) = 0.62; (c). P (0, 0) = 0.38; (d). PW |H (1|1) = P (1, 1) /PH (1) = 0.18/0.5 = 0.36. 3. [Discrete Distribution, 15 points] A manufacturer of X-ray machines has fitted all machines with an excess radiation warning device, yet experience shows that 10 percent of the machines tested fail to issue the warning. Assume that 10 machines coming off the assembly line during a given month are selected at random. a. What is the probability mass function for the number of faulty machines, denoted by X, found in the sample? [you only need to write down the formulas] b. Calculate the expected value and standard deviation for the random variable X. Suggested Answer: (a). It has a binomial distribution as follows: fX (x) = µ 10 x ¶ 0.1x 0.910−x if x = 0, 1, ..., 10; and fX (x) = 0 otherwise. (b). EX = 0.1 × 10 = 1, VarX = 10 × 0.1 × 0.9 = 0.9. 4. [Distribution of Functions of Random Variables, 15 points] Let X1 , X2 , ..., Xn be independent and identically distributed random variables from a Uniform distribution on [0, 1] . Let Y = max {X1 , ..., Xn } and Z = min {X1 , ..., Xn } . (a) Find the PDF of the random variables Y and Z; (b) Use the results of (a) to calculate EY and EZ. Suggested Answer: (a). First find the CDF of Y and Z. The support of both Y and Z are [0, 1] . FY (y) = Pr (Y ≤ y) = Pr (max {X1 , ..., Xn } ≤ y) = Pr (X1 ≤ y, ..., Xn ≤ y) = Pr (X1 ≤ y) × ... × Pr (Xn ≤ y) = yn , Hence fY (y) = ny n−1 , 0 ≤ y ≤ 1. 2 FZ (z) = Pr (Z ≤ z) = Pr (min {X1 , ..., Xn } ≤ z) = 1 − Pr (X1 > z, ..., Xn > z) = 1 − (1 − z)n Hence fZ (z) = n (1 − z)n−1 , 0 ≤ z ≤ 1. (b). ¯1 ny n+1 ¯¯ n yny dy = = . EY = ¯ n+1 0 n+1 0 Z 1 Z 1 Z 1 n−1 n−1 zn (1 − z) dz = n (1 − z) dz − (1 − z) n (1 − z)n−1 dz EZ = 0 0 0 ¯ n+1 ¯1 n (1 − z) ¯ = 1+ ¯ ¯ n+1 Z 1 n−1 0 n 1 = 1− = . n+1 n+1 5. [Conditional Distributions, 15 points] Define a continuous random vector (X, Y ) by the joint p.d.f fXY (x, y) = e−y , 0 < x < y < +∞. (a) Verify that fXY (x, y) defined above indeed is a valid p.d.f. (b) Compute the conditional p.d.f of Y given X = x. Suggested Answer: (a). Notice that f (x, y) ≥ 0 everywhere and Z ∞Z ∞ Z ∞ £ −y ¤¯∞ −e ¯x dx e−y dydx = 0 x 0 Z ∞ £ ¤¯∞ e−x dx = −e−x ¯0 = 1. = 0 (b). To find the conditional p.d.f of Y giveng X = x, we first find the marginal density of X. Z ∞ £ ¤¯∞ fX (x) = e−y dy = −e−y ¯x = e−x , 0 < x < ∞ x Hence fY |X (y|x) = fXY (x, y) = fX (x) ½ e−y e−x 0 if y ≤ x = ex−y if y > x. [It is important to write out the domain of the density] 6. [Moment Generating Function and Expectations, 20 points] The p.d.f of random variable X is given by fX (x) = θxθ−1 , 0 < x < 1 where θ > 0 is a parameter. (caution: notice the support of r.v. X) 3 (a) Find the c.d.f of X; (b) Let Y = −2θ ln X. Find the p.d.f of Y ; (c) Calculate the moment generating function of Y ; (d) Calculate EY and V arY. Suggested Answer: (a). For x ∈ (0, 1) Z x FX (x) = θz θ−1 dz = z θ |x0 = xθ . 0 [This should be clear enough.] (b). The support of Y is (0, +∞) . The CDF of Y is Hence the pdf is FY (y) = Pr (Y ≤ y) = Pr (−2θ ln X ≤ y) ³ y´ = Pr ln X ≥ − ³ ³2θ y ´´ = Pr X ≥ exp − 2θ h ³ y ´iθ = 1 − exp − ´ ³ y2θ = 1 − exp − 2 fY (y) = (c). The m.g.f of Y is ³ y´ 1 exp − , y > 0. 2 2 Z ³ y´ ∞ 1 ety exp − dy MY (t) = Ee = 2 2 0 µ ¶ Z ∞ 1 (1 − 2t) y exp − dy = 2 2 0 · µ ¶¸¯ (1 − 2t) y ¯¯∞ 1 × − exp − = ¯ 1 − 2t 2 0 1 1 = ,t < . 1 − 2t 2 tY [It is important to specify the domain of t on which MY (t) is defined. (d). EY = M 0 (0) = 2 EY 2 = M 00 (0) = 8 Hence V arY = EY 2 − [EY ]2 = 4. [Alternatively one can recognize that MY (t) above is the m.g.f of exponential with parameter λ = 2 and use the formula for exponential: EY = λ, V arY = λ2 ]. 4