You are given the following data:vocabulary V = (w1, w2, w3) and the bigram probability distribution p on V x V given by:!!p(w1, w1) = p(w3, w3) = 1/5%3Dp(w2, w2) = 0,!3!p(w2, w1) = 1/6,!3!p(w1, w3) = 1/5,%3Dp(w2,w3)=1/15p(w1, ") = 2/5 (that is w1 as the first of a pair).p(*, w2) = 1 /6.%3!Calculate p(w1, w2) and p(w2 | w3) using Markov's rule

Answered: Image /qna-images/answer/5c5fe4d6-fdb3-41f7-8f04-4f1ad4cc5d0f.jpg

You are given the following data: vocabulary V (w1, w2, w3) and the bigram probability distribution p on V x V given b p(w1, w1) = p(w3, w3) = 1/5 p(w2, w2) = 0, p(w2, w1) = 1/6, p(w1, w3) = 1/5, p(w2,w3)=1/15 p(w1. ) 2/5 (that is w1 as the first of a pair). p(", w2) = 1 /6. Calculate p(w1, w2) and p(w2 | w3) using Markov's rule

You are given the following data: vocabulary V (w1, w2, w3) and the bigram probability distribution p on V x V given b p(w1, w1) = p(w3, w3) = 1/5 p(w2, w2) = 0, p(w2, w1) = 1/6, p(w1, w3) = 1/5, p(w2,w3)=1/15 p(w1. ) 2/5 (that is w1 as the first of a pair). p(", w2) = 1 /6. Calculate p(w1, w2) and p(w2 | w3) using Markov's rule

Operations Research : Applications and Algorithms

4th Edition

ISBN:9780534380588

Author:Wayne L. Winston

Publisher:Wayne L. Winston

Chapter11: Nonlinear Programming

Section11.4: Solving Nlps With One Variable

Problem 4P

See similar textbooks

Similar questions

A random variable X with two-sided exponential distribution given by has moment generating function given by M X (t)= e^ t +e^ -t -2 t^ 2 . f x (x)= x+1,&-1\\ 1-x,&0<= x<=1 - 1 <= x <= 0 (a) Using M_{X}(t) or otherwise, find the mean and variance of X. (b) Use Chebychev inequality to estimate the tail probability, P(X > delta) , for delta > 0 and compare your result with the exact tail probability.
Let X and Y be two binary, discrete random variables with the following joint probability mass functions. (a) Compute P(X = 0]Y = 1). (b) Show that X and Y are not statistically independent. P(X = 0, y = 1) = P(X = 1, Y = 0) = 3/8 P(X = 0, y = 0) = P(X = 1, Y = 1) = 1/8
Consider values shown in the table below:i=1 (cold) i=2 (allergy) i=3 (stomach pain) p(Hi)0.60.30.1 p(E1 |Hi)0.30.80.3 p(E2 |Hi)0.60.90.0Those values represent (hypothetically) three mutually exclusive and exhaustive hypotheses for the patient’s condition. For example, H1: the patient has a cold, H2: the patient has an allergy, and H3: the patient has stomach pain with their prior probabilities, p(Hi)’s and two conditionally independent pieces of evidence (E1, patient sneezes and E2, patient coughs) which support these hypotheses to differing degrees. Therefore;a) Compute the posterior probabilities for the hypothesis if the patient sneezes. What is the conclusion that can be derived from this condition?b) Based on the answer from the previous result, as the patient coughs are now observed, compute the posterior probabilities for this condition. Explain the results.
Given a two-category classification problem under the univariate case, where there are two training sets (one for each category) as follows: D₁ = (-3,-1,0,4} D₂ = {-2,1,2,3,6,8} Given the test example x = 5, please answer the following questions: have and a) Assume that the likelihood function of each category has certain paramétric form. Specifically, we p(x | w₁) N, 07) p(x₂)~ N(μ₂, 02). Which category should we decide on when maximum-likelihood estimation is employed to make the prediction?
From 1965 to 1974, in U.S. there were M = 17, 857, 857 male livebirths and F = 16, 974, 194 female livebirths. We model the number of male livebirth as a binomial distribution with parameters size = M+F and prob = p. The following code computes the maximum likelihood estimator for p. M <- 17857857 F <- 16974194 ll <- function(p){ dbinom(M, size=M+F, prob=p, log=TRUE) } ps <- seq(0.01, 0.99, by = 0.001) ll.ps <- ll(ps) plot(ps, ll.ps, type='l') phat <- ps[which.max(ll.ps)] abline(v = phat, col='blue') Question: What can we learn from the plot?
From 1965 to 1974, in U.S. there were M = 17, 857, 857 male livebirths and F = 16, 974, 194 female livebirths. We model the number of male livebirth as a binomial distribution with parameters size = M+F and prob = p. The following code computes the maximum likelihood estimator for p. M <- 17857857 F <- 16974194 ll <- function(p){ dbinom(M, size=M+F, prob=p, log=TRUE) } ps <- seq(0.01, 0.99, by = 0.001) ll.ps <- ll(ps) plot(ps, ll.ps, type='l') phat <- ps[which.max(ll.ps)] abline(v = phat, col='blue') Question: An estimator for p, denoted by pˆ, is obtained by ps[which.max(ll.ps)]. Is this the maximum likelihood estimator? Why (explain the code)?
Give an example of a random variable X : {b, c, d, e} → N (Natural Number) with expectation 2, where each of {b, c, d, e} has equal probability
Given 2 patterns at 0.4 and 0.6, estimate probability density analytically using a rectangular window of width 0.3, using a triangular window of width 0.3 and using 1 nearest-neighbour.
Let events A, B, C, and D be given by A = {at least 3 heads}, B = {at most 2 heads}, C = {headson the third toss}, and D = {1 head and 3 tails}. If the probability set function assigns 1/16 toeach outcome in the sample space, find (i) P(A), (ii) P(A ∩ B), (iii) P(B), (iv) P(A∩C), (v) P(D),(vi) P(A∪C), and (vii) P(B ∩ D).
A certain cat shelter has devised a novel way of making prospective adopters choose their new pet. To remove pet owners’ biases regarding breed, age, or looks, they are led blindfolded into a room containing all the cats up for adoption and must bring home whichever they pick up. Suppose you are trying to adopt two cats, and the shelter contains a total of N cats in one of only two colors: black or orange. is it still possible to pick up two black cats with probability ½, given that there is an even number of orange cats in the room? If so, how many cats should be in the room? How many black, how many orange?
Please solve the following problem. Quiz = Pass Quiz = Fail AI = Fail 0.1 0.2 AI = Pass 0.6 0.1 Mid = Pass Mid = Fail AI = Fail 0.2 0.2 AI = Pass 0.5 0.1 Suppose you have three events AI Grade, Quiz, and Mid. Here each event has two possible outcomes, either pass or fail. Additionally, given that AI Grade is observed, Quiz and Mid become independent of each other. Also, out of every 100 students, 30 students fail the AI course. Now, using the joint probability tables given, calculate P(AI Grade=Pass, Quiz=Fail, Mid=Fail).
2. Given a Sample Space S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}, Event A = {1, 3, 4, 7, 9}, and Event B = {3, 7, 9, 11, 12, 13} Find the probability P(A|B). State your answer as a value with one digit after the decimal point.