Mathematics > QUESTION PAPER (QP) > CS 361 Probability & Statistics for Computer Science-University of Illinois - Homework 8 (All)

CS 361 Probability & Statistics for Computer Science-University of Illinois - Homework 8

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CS 361 Probability & Statistics for Computer Science-University of Illinois - Homework 8 CS361 Probability & Statistics University of Illinois at Urbana-Champaigns Problem 1 Since this is the probability density function, using the log-likehood function logL(θ) = logP(Djθ) = log Y xi2D P(xijθ) Substituting the give formula in question to have: , NX i =1 log(θe−θxi) , NX i =1 log(θ) − θxi Differenting both sides of the equation to have: d dθ logL(θ) = 0 = NX i =1 1 θ − xi = N θ − NX i =1 xi Maniputating the equation to have: N θ = NX i =1 xi , θ = N PN i=1 xi 1CS361 Probability & Statistics University of Illinois at Urbana-Champaigns Problem 2 By the formula in the lecture note, the formula of MLE of λ of poisson distribution would be: θ^ = PN i ki N . a.) Day one: θ^1 = PN i ki N = 3 + 1 + 4 + 2 4 = 2:5 Day two: θ^2 = PN i ki N = 2 + 1 + 2 3 = 1:666 Day three: θ^3 = PN i ki N = 3 + 2 + 2 + 1 + 4 5 = 2:4 b.) Day Four: θ^3 = PN i ki N = 13 6 = 2:166 c.) All day combined: θ^ = PN i ki N = 3 + 1 + 4 + 2 + 2 + 1 + 2 + 3 + 2 + 2 + 1 + 4 + 13 4 + 3 + 5 + 6 = 2:222 2CS361 Probability & Statistics University of Illinois at Urbana-Champaigns Problem 3 According to the lecture note, the MLE derivatoin of geometric example is: L(θ) = (1 − θ)k−1θ; θ^ = 1 K a.) Substitute K with r + 1 in the question to have: θ^ = 1 r + 1 . Assuming there are x zero slot in the roulette, the probability of zero slot can be represented by x x + 36 . Manipulating the equations to have: 1 r + 1 = x x + 36 , x = 36 r As a result, the total number of slots in the roulette would be 36 r + 36. b). This estimate is unreliable, since our data base is not large enough to make the MLE reliable. MLE needs more data to have a precise estimate of the number we want. c). We apply the Log-likehood function to do the estimate. logL(θ) = logP(Djθ) = log Y xi2D P(xijθ) logL(θ) = logP(Djθ) = log Y xi2D P(xijθ) = kX i =1 log((1 − θ)riθ) , kX i =1 (ri)log((1 − θ) + log(θ) Taking the differential of both sides of equation and both of them would be equal to zero: d dθ logL(θ) = 0 = kX i =1 ( −ri 1 − θ + 1 θ ) = Pk i=1 −ri 1 − θ + k θ , θ kX i =1 (−ri) = (θ − 1)(k) , θ^ = k k − Pk i=1(−ri) Assuming there are x zero slot in the roulette, the probability of zero slot can be represented by x x + 36 . Manipulating the equations to have: θ^ = x x + 36

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