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HW_1 _AMS_570 1.5 Approximately one-third of all human twins are identical (one-egg) and twothirds are fraternal (two egg) twins. Identical twins are necessarily the same sex, with male and female being equally likely. Among fraternal twins, approximately one-fourth are both female, one-fourth are both male, and half are one male and one female. Finally, among all U.S. births, approximately 1 in 90 is a twin birth. Define the following events: π¨ = {π πΌ. πΊ. πππππ πππππππ ππ ππππ πππππππ} π© = {π πΌ. πΊ. πππππ πππππππ ππ ππ πππππππ πππππ} πͺ = {π πΌ. πΊ. πππππ πππππππ ππ πππππ} (a) State, in words, the eventπ¨ β© π© β© πͺ. (b) Findπ·(π¨ β© π© β© πͺ). a. A β© B β© C = {a U.S. birth results in identical twins that are female} b. P (A β© B β© C) = 1/90 * 1/3 * 1/2 = 1/540 1.33 Suppose that 5% of men and 0.25% of women are color-blind. A person is chosen at random and that person is color-blind. What is the probability that the person is male? (Assume males and females to be in equal numbers.) Using Bayes rule π(πΆπ΅|π)π(π) P (M|CB) = π(πΆπ΅|π)π(π) + π(πΆπ΅|πΉ)π(πΉ) = 1 2 0.05β 1 2 1 2 0.05β +0.0025β = 0.9524. 1.41 As in example 1.3.6, consider telegraph signals βdotβ and βdashβ sent in the proportion 3:4, where erratic transmissions cause a dot to become a dash with π π probability π and a dash to become a dot with probabilityπ. (a) If a dash is received, what is the probability that a dash has been sent? (b) Assuming independence between signals, if the message dot-dot was received, what is the probability distribution of the four possible messages that could have been sent? a. P (dash sent | dash rec) = π ( dash rec | dash sent)π ( dash sent) π ( dash rec | dash sent)π ( dash sent) + π ( dash rec | dot sent)π ( dot sent) 32 . 41 (2/3)(4/7) = (2/3)(4/7) + (1/4)(3/7) = b. By a similar calculation as the one in (a) P (dot sent|dot rec) = 27/43. Then we have P (dash sent|dot rec) =16/43.Given that dot-dot was received, the distribution of the four possibilities of what was sent are: Event dash-dash dash-dot dot-dash dot-dot Probability (16/43)2 (16/43)*(27/43) (27/43)*(16/43) (27/43)2 1.51 An appliance store receives a shipment of 30 microwave ovens, 5 of which are (unknown to the manager) defective. The store manager selects 4 ovens at random, without replacement, and tests to see if they are defective. Let X = number of defectives found. Calculate the pmf and cdf of X and plot the cdf. This kind of random variable is called hypergeometric in Chapter 3. The probabilities are obtained by counting arguments, as follows. The c.d.f is a step function with jumps at x=0, 1, 2, 3, and 4. 2.12 A random right triangle can be constructed in the following manner. Let X be a π random angle whose distribution is uniform on(π, π ). For each X, construct a triangle as pictured below. Here, Y = height of the random triangle. For a fixed constant d, find the distribution of Y and EY. We have tan(x) = y/d, therefore tan-1 (y/d) = x, and ππ (π¦) = 2 β ππ 1 π¦ 2 1+( ) π π¦ π π tanβ1 ( ) ππ¦ = 1 π¦ 2 1+( ) π 1 ππ₯ β π = ππ¦. Thus, , 0 < π¦ < β. This is a Cauchy distribution restricted to(0, β), and the mean is infinite. 2.15 Betteley provides an interesting addition law for expectations. Let X and Y be any two random variables and define πΏ β§ π = πππ(πΏ, π) πππ πΏ β¨ π = πππ(πΏ, π). Analogous to the probability law π·(π¨ βͺ π©) = π·(π¨) + π·(π©) β π·(π¨ β© π©), show that π¬(πΏ β¨ π) = π¬πΏ + π¬π β π¬(πΏ β§ π) (Hint: Establish thatπΏ + π = (πΏ β§ π) + (πΏ β¨ π).) Assume without loss of generality that X β€ Y. Then X β¨ Y = Y and X β§ Y = X. Thus, X + Y = (X β§ Y) + (X β¨ Y). Taking expectations: E[X] +E[Y] =E[X + Y] = E [(X β§ Y) + (X β¨ Y)] = E(X β§ Y) + E(X β¨ Y). Therefore, E(X β¨ Y) = EX + EY β E(X β§ Y). π 2.17 A median of a distribution is a value m such that π·(πΏ β€ π) β₯ π and π π β π π·(πΏ β₯ π) β₯ π. (if X is continuous, m satisfiesβ«ββ π(π)π π = β«π π(π)π π = π.) Find the median of the following distribution. (a) π(π) = πππ , π < π < π (b) π(π) = π π (π+ππ ) , ββ < π < β 1 a. π β«0 3 2 3 1 β π₯ ππ₯ = π = 2 , πππ£ππ π = 1 3 (2) = 0.794. b. The function is symmetric about zero, therefore m=0 as long as the integral is finite. +β We know itβs a Cauchy p.d.f andβ«ββ π(π₯)ππ₯ = 1. 2.20 A couple decides to continue to have children until a daughter is born. What is the expected number of children of this couple? From Example 1.5.4, if X = number of children until the first daughter, then π(π = π) = (1 β π)πβ1 π, where p = probability of a daughter. Thus, X is geometric random variable, and β πΈπ = β π β (1 β π) π=1 πβ1 β β π=1 π=1 π π (1 β π)π = βπ β β π = βπ β β [β(1 β π)π ] ππ ππ π 1 1 = βπ β [ β 1] = . ππ π π Therefore, if p=1/2, the expected number of children is two. 2.28 Let ππ denote the nth central moment of a random variable X. Two quantities of interest, in addition to the mean and variance, are πΆπ = ππ ππ πππ πΆ = π (ππ )π/π πππ The value πΆπ is called the skewness and πΆπ is called the kurtosis. The skewness measures the lack of symmetry in the pdf. The kurtosis, although harder to interpret, measures the peakedness or flatness of the pdf. (a) Show that if a pdf is symmetric about a pointπΆ, then πΆπ = π. (b) Calculate πΆπ for π(π) = πβπ , π β₯ π, a pdf that is skewed to the right. (c) Calculate πΆπ for each of the following pdfs and comment on the peakedness of each. π(π) = π βππ π(π) = π(π) = ππ β πβ π , ββ < π < β π , βπ < π < π π π β|π| π , ββ < π < β π a. β π 3 β 3 π3 = β« (π₯ β π) π(π₯)ππ₯ = β« (π₯ β π) π(π₯)ππ₯ + β« (π₯ β π)3 π(π₯)ππ₯ ββ 0 ββ π β = β« π¦ 3 π(π¦ + π)ππ¦ + β« π¦ 3 π(π¦ + π)ππ¦ (πβππππ ππ π£πππππππ) ββ 0 β β = β« βπ¦ 3 π(βπ¦ + π)ππ¦ + β« π¦ 3 π(π¦ + π)ππ¦ 0 (π(βπ¦ + π) 0 = π(π¦ + π), ππ’π π‘π π π¦πππ‘πππ π. π. π) =0 b. For π(π₯) = π βπ₯ , π1 = π2 = 1, therefore, πΌ3 = π3 . β β π3 = β«0 (π₯ β 1)3 π βπ₯ ππ₯ = β«0 (π₯ 3 β 3π₯ 2 + 3π₯ β 1)π βπ₯ ππ₯ = 3! β 3 β 2! + 3 β 1 = 2. c. Each distribution hasπ1 = 0, therefore we must calculate π2 = πΈπ 2 πππ π4 = πΈπ 4 . π πΌ4 = π42 = 3. π2 = 1, π4 = 3, For (i), 1 2 1 π 2 π4 π2 = 2, π4 = 24, For (iii), 9 πΌ4 = π42 = 5. π2 = 3 , π4 = 5 , For (ii), πΌ4 = π2 = 6. 2 Therefore, (iii) is the most peaked one, (i) next, and (ii) is least peaked. 2.33 In each of the following cases verify the expression given for the moment generating function, and in each case use the mgf to calculate EX and VarX. (a) π·(πΏ = π) = πβπ ππ π! π , π΄πΏ (π) = ππ(π βπ) , π = π, π, β¦ ; π > π π (b) π·(πΏ = π) = π(π β π)π , π΄πΏ (π) = πβ(πβπ)ππ , π = π, π, β¦ ; π < π < π (c) ππΏ (π) = π βππ π π β (πβπ)π πππ π‘π₯ a. ππ (π‘) = ββ β π₯=0 π πΈπ = π π π π , π΄πΏ (π) = πππ+ππ π βπ ππ₯ π₯! =π βπ β ββ π₯=0 , ββ < π < β, ββ < π < β, π > π (ππ π‘ ) π₯! π₯ π‘ = π βπ β π ππ = π π(π π‘ β1) . π π2 ππ (π‘)|π‘=0 = π; πΈπ 2 = 2 ππ (π‘)|π‘=0 = π2 + π ππ‘ ππ‘ ππππ = πΈπ 2 β (πΈπ)2 = π π β π‘π₯ π₯ π‘ π₯ b. ππ (π‘) = ββ π₯=0 π β π(1 β π) = π βπ₯=0((1 β π)π ) = 1β(1βπ)π π‘ πΈπ = π 1βπ π2 π(1 β π) + 2(1 β π)2 ππ (π‘)|π‘=0 = ; πΈπ 2 = 2 ππ (π‘)|π‘=0 = ππ‘ π ππ‘ π2 ππππ = πΈπ 2 β (πΈπ)2 = c. ππ (π‘) = β β«ββ π π‘π₯ β 1 1βπ π2 1 β2π βπ π β (π₯βπ)2 2π2 ππ₯ = 2 2 π‘π₯+π2 β βπ₯ β2ππ₯β2π 2 2π β« π β2ππ ββ 1 ππ₯. π₯ 2 β 2ππ₯ β 2π 2 π‘π₯ + π 2 = π₯ 2 β 2(π + π 2 π‘)π₯ ± (π + π 2 π‘)2 + π 2 = (π₯ β 2 (π + π 2 π‘)) β [2ππ 2 π‘ + (π 2 π‘)2 ], Then we haveππ (π‘) = π 1 2 2 π‘ π ππ‘+2π . 1 2 ππ‘+ π2 π‘ 2 β β β β« π β2ππ ββ 1 (π₯β(π+π2 π‘)) 2π2 2 1 2 2 π‘ ππ₯ = π ππ‘+2π β1= π π2 2 πΈπ = ππ (π‘)|π‘=0 = π; πΈπ = 2 ππ (π‘)|π‘=0 = π 2 + π 2 ππ‘ ππ‘ ππππ = πΈπ 2 β (πΈπ)2 = π 2 .