Solutions Manual > Georgia Institute Of Technology CS 7641 CSE/ISYE 6740 Mid-term Exam 2 (Fall 2016) Solutions/ Fall 2016 Midterm2

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Georgia Institute Of Technology CS 7641 CSE/ISYE 6740 Mid-term Exam 2 (Fall 2016) Solutions Probability and Bayes’ Rule [14 pts] (a) A probability density function (pdf) is defined by f(x; y) =... (C 0 otherwise (x + 2y) if 0 < y < 1 and 0 < x < 2 (i) Find the value of C [3 pts]. Answer: Z01 Z02 C(x + 2y)dxdy = 4C = 1 Thus, C = 1 4: (ii) Find the marginal distribution of X [2 pts]. Answer: fX(x) = (R 0 otherwise 01 1 4(x + 2y)dy = 14(x + 1) if 0 < x < 2 (iii) Find the joint cumulative density function (cdf) of X and Y [2 pts]. Answer: The complete definition of FXY is: FXY (x; y) = 8>>>>>><>>>>>>: 0 | x ≤ 0 or y ≤ 0 x2y=8 + y2x=4 0 < x < 2 and 0 < y < 1 y=2 + y2=2 x2=8 + x=4 | 2 ≤ x and 0 < y < 1 0 < x < 2 and 1 ≤ y1 2 ≤ x and 1 ≤ y 1 (b) When coded messages are sent, there are sometimes errors in transmission. In particular, Morse code uses \dots" and \dashes", which are known to occur in the proportion of 3:4. This means that for any given symbol, P(dot set) = 3 7 and P(dash sent) = 4 7 : Suppose there is interference on the transmission line, and with probability 18 a dot is mistakenly received as a dash, and vice versa. If we receive a dot, what is the probability that a dot was sent? That is, compute P(dot sentjdot received). [7 pts] Answer: Using Bayes’ Rule, we write P(dot sentjdot received) = P(dot receivedjdot sent)P(dot sent)=P(dot received) = (7=8) × (3=7) (7=8) × (3=7) + (1=8) × (4=7) = 21 25 : 2 2 Maximum Likelihood [12 pts] (a) The independent random variables X1; X2; :::; Xn have the common distribution P(Xi ≤ xjα; β) = 8><>: 0; x < 0 (βx)α; 0 ≤ x ≤ β 1; x > β where the parameters α and β are positive. Find the MLEs of α and β. [7 pts] After differentiating, we can get the P.D.F. as P(Xi = xjα; β) = (0βα; otherwise α x(α-1); 0 ≤ x ≤ β Let x(n) be the maximum value. For any fixed α, the likelihood, L(α; βjx) = 0 if β < x(n), and L(α; βjx) is a decreasing function of β if β ≥ x(n). Thus x(n) is the MLE of β For the MLE of α calculate @ @αhnlogα - nαlogβ + (α - 1)log Qi xii = 0 From this we get, α = n nlogx(n) - log Qi xi 3 (b) Suppose that a particular gene occurs as one of the two alleles (A and a), where allele A has frequency θ in the population. That is a random copy of the gene is A with probability θ and a with probability 1 - θ. Since a diploid genotype consists of two genes, the probability of each genotype is given by: genotype | AA | Aa | aa probability | θ2 | 2θ(1 - θ) | (1 - θ)2Suppose we test a random sample of people and we find that k1 are AA, k2 are Aa, and k3 are aa. Find the MLE of θ [5 pts] likelihood = (θ)2k1(2θ(1 - θ))k2(1 - θ)2k3 log likelihood = constant +2k1ln(θ) + k2ln(θ) + k2ln(1 - θ) + 2k3ln(1 - θ) MLE of θ = | 2k1+k2 2k1+2k2+2k3 [Show More]

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