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Theory Of Point Estimation Homework Solution

The homework and project are graded on a 5 point scale. The Midterm and Final will each be normalized to a mean of 5 and a standard deviation of 1. Based on the above formula, the final grades will be roughly on a 15 point scale. The log-sum-exp function returns a value greater than both arguments. You will benefit slightly (up to .693 points) by having high scores in both items rather than one.

theory of point estimation homework solution

There will be approximately 7 weekly homework assignments. These are not handed in. Instead, each student should make a 15 minute appointment during office hours to review the completed assignment in person. Please use the appointment scheduler to find a time that works for you. A student who chooses not to do the assignments may wish to use the time to discuss a project. The lowest homework scores will be dropped. Beyond that, exceptions will not be made for missed or late homework.

In more formal terms, the estimate occurs as a result of point estimation applied to a set of sample data. Points are single values, in comparison to interval estimates, which are a range of values. For example, a confidence interval is one example of an interval estimate.

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Regardless of which option you choose, the course staff will supportyou: put together relevant homeworks for practice, along withsolutions, have sections, with section worksheets, and solutions,homework parties, office hours with additional support beyond theteaching assistants, and some tutoring support.

[iv] Data for other years are available on the NAEP Data Explorer. For Table 1, the starting point of 1984 was chosen because it is the first year all three ages were asked the homework question. The two most recent dates (2012 and 2008) were chosen to show recent changes, and the two years in the 1990s to show developments during that decade.

Solve the ODE using the ode45 function on the time interval [0 20] with initial values [2 0]. The resulting output is a column vector of time points t and a solution array y. Each row in y corresponds to a time returned in the corresponding row of t. The first column of y corresponds to y1, and the second column corresponds to y2.

Solve the van der Pol equation with μ=1 using ode45. The function vdp1.m ships with MATLAB and encodes the equations. Specify a single output to return a structure containing information about the solution, such as the solver and evaluation points.

If tspan has more than two elements [t0,t1,t2,...,tf], then the solver returns the solution evaluated at the given points. However, the solver does not step precisely to each point specified in tspan. Instead, the solver uses its own internal steps to compute the solution, and then evaluates the solution at the requested points in tspan. The solutions produced at the specified points are of the same order of accuracy as the solutions computed at each internal step.

If tspan contains several intermediate points [t0,t1,t2,...,tf], then the specified points give an indication of the scale for the problem, which can affect the value of InitialStep used by the solver. Therefore, the solution obtained by the solver might be different depending on whether you specify tspan as a two-element vector or as a vector with intermediate points.

Structure for evaluation, returned as a structure array. Use this structure with the deval function to evaluate the solution at any point in the interval [t0 tf]. The sol structure array always includes these fields:

The Basis for this course is to develop an understanding of statistical inference. This class will focus on a wide variety of topics; however the general principle is that of making informed decisions from data. This course will focus on various methods of parameter estimation, hypothesis testing, and interval estimation. Topics in decision theory will also be discussed. Both theoretical and computational methods will be examined in this class.

7 Solving it for λ, we obtain ˆλ = X, the method of moments estimator of λ. Maximum likelihood. The p.m.f. of Poisson distribution is P (x) = e λ λx x!, and its logarithm is ln P (x) = λ + x ln λ ln(x!). Thus, we need to maximize ln P (X 1,..., X n ) = ( λ + X i ln λ) + C = nλ + ln λ X i, where C = ln(x!) is a constant that does not contain the unknown parameter λ. Find the critical point(s) of this log-likelihood. Differentiating it and equating its derivative to 0, we get This equation has only one solution λ ln P (X 1,..., X n ) = n + 1 λ ˆλ = 1 n X i = X. X i = 0. Since this is the only critical point, and since the likelihood vanishes (converges to 0) as λ 0 or λ, we conclude that ˆλ is the maximizer. Therefore, it is the maximum likelihood estimator of λ. For the Poisson distribution, the method of moments and the method of maximum likelihood returned the same estimator, ˆλ = X. # 1d (Uniform) 7

Math/Stats 342: Solutions to Homework Steven Miller ( November 17, 2011 Abstract Below are solutions / sketches of solutions to the homework problems from Math/Stats 342: Probability

In this course, you will learn to recognize and solve convex optimization problems that arise in applications across engineering, statistics, operations research, and finance. Examples will be chosen to illustrate the breadth and power of convex optimization, ranging from systems and control theory, to estimation, data fitting, information theory, and machine learning. A tentative list, subject to change, of what we will cover includes: convex sets, functions, and optimization problems; the basics of convex analysis; least-squares, linear and quadratic programs, semidefinite programs, minimax, extremal volume, and other problems; optimality conditions, duality theory, theorems of alternatives, and applications; interior-point algorithms for solving convex optimization problems, and their complexity analysis; applications to signal processing, statistics and machine learning, control, digital and analog circuit design, and finance.

Plan:Lecture 1.(8.01.07) Probability theory.Chapter 1: 1.1, 1.2, 1.3Homework: 1.1, 1.2, 1.4, 1.9, 1.11. Solutions are hereLecture 2.(9.01.07) Probability theory.Chapter 1: 1.1, 1.2, 1.3Homework: 1.6, 1.25. Solutions are hereLecture 3.(11.01.07) Independence. Random variables. Distributions.Chapter 1: 1.3, 1.4, 1.5Homework: 1.13, 1.33, 1.38, 1.52. Solutions are hereLecture 4.(15.01.07) Distributions. Expectations. Moments.Chapter 1: 1.6; Chapter 2: 2.2, 2.3Homework: 1.39, 1.49, 2.20. Solutions are hereLecture 5.(18.01.07) Moment generating function. Distribution of transformations.Chapter 2: 2.1, 2.3Homework: 2.17, 2.18, 2.19. Solutions are hereLecture 6.(22.01.07) Distribution of transformations.Families of distributions.Chapter 2: 2.1, 2.4; Chapter 3: 3.1, 3.2, 3.3Homework: 2.1, 2.12, 2.31. Solutions are hereLecture 7.(23.01.07) Conditional independence. Mixtures.Products of gamma, beta and normal densities.Chapter 1: 1.3, 1.5Homework: 1.47. Solutions are hereLecture 8.(25.01.07) Families of distributions.Chapter 3: 3.2, 3.3, 3.5Homework: 3.21, 3.23, 3.25. Solutions are hereLecture 9.(29.01.07) Multiple random variables.Conditional distributions. Independence.Chapter 4: 4.1, 4.2, 4.3 (something), 4.4 (something), 4.6 (something)Homework: 2.15, 4.1, 4.2Lecture 10.(30.01.07) Joint, marginal, conditional distributions. Hierarchical models.Chapter 4: 4.1, 4.2, 4.4Homework: 4.4Lecture 11.(1.02.07) Conditional expectations. Covariance and correlation. Inequalities.Chapter 3: 3.6 (Theorem 3.6.1); Chapter 4: 4.4 (Theorems 4.4.3, 4.4.7), 4.5, 4.7 (Theorems 4.7.3, 4.7.7)Homework: 4.6, 4.13, 4.15Lecture 12.(5.02.07) Random sample. Statistic. Convergence.Chapter 5: 5.1, 5.2, 5.4, 5.5Homework: 4.41, 4.42, 4.43Lecture 13.(6.02.07) Bayesian statistics. Prior and posteriordistributions. Conjugate families.[1]: 7.2.3, [2]: 4.2Homework: hereLecture 14.(8.02.07) Convergence. Point estimation: method ofmoments.Chapter 5: 5.5.1, 5.5.2, 5.5.3; Chapter 7: 7.1, 7.2.1Homework: 4.58, 4.63, 5.21Lecture 15.(12.02.07) Point estimation. MME and MLE. Evaluation of estimators.Chapter 7: 7.1, 7.2, 7.3.1Homework: 5.22, 5.23, 5.31Lecture 16.(13.02.07) Bayes estimation.[1]: 7.2.3, [2]: 4.3.1Homework: hereLecture 17.(15.02.07) Evaluation of estimators.Chapter 7: 7.3.1, 7.3.2; Chapter 10: 10.1.1Homework: 7.1, 7.2a, 7.6b,cLecture 18.(19.02.07) UMVUEChapter 7: 7.3.2Homework: 7.40, 7.41, 10.1Lecture 19.(20.02.07) Bayes estimation.[1]: 7.2.3, [2]: 4.3.1Homework: [1]: 7.24, 7.26Lecture 20.(22.02.07) Sufficient statistics.Chapter 6: 6.1, 6.2Homework: 6.1, 6.3, 6.6Lecture 21.(12.03.07) Sufficient statistics and optimality.Chapter 6: 6.2, Chapter 7: 7.3Homework: 6.19, 6.21, 7.38Lecture 22.(15.03.07) Sufficient statistics and UMVUE. Exponential families and UMVUE.Chapter 3: 3.4, Chapter 6: 6.2, Chapter 7: 7.3Homework: 6.22, 10.9bLecture 23.(19.03.07) Exponential families and UMVUE.Chapter 3: 3.4, Chapter 6: 6.2, Chapter 7: 7.3Homework: 7.37, 7.47Lecture 24.(20.03.07) Non-informative priors. Bayes hypotheses testing.[1]: 8.2.2, [2]: 3.3Homework: hereLecture 25.(22.03.07) Hypotheses testing.Chapter 8: 8.1, 8.2, 8.3Homework: 8.1, 8.13Lecture 26.(26.03.07) Hypotheses testing.Chapter 8: 8.1, 8.2, 8.3Homework: 8.3, 8.14Lecture 27.(27.03.07) Decision theory.[1]: 7.3.4, 8.3.5, [2]: 1.1, 1.2, 1.3Homework: hereLecture 28.(29.03.07) Evaluation of tests.Chapter 8: 8.3.1, 8.3.2Homework: 8.16, 8.18, 8.19Lecture 29.(12.04.07) UMP tests.Chapter 8: 8.3.2Homework: 8.15, 8.24Lecture 30.(16.04.07) MLR and UMP tests. UIT and IUT.Chapter 8: 8.3.2, 8.2.3Homework: 8.23, 8.31Lecture 31.(17.04.07) Bayes decision rules.[1]: 8.2.2, 8.3.5, [2]: 1.3, 4.3.3Homework: hereLecture 32.(19.04.07) UIT and IUT. Interval estimation.Chapter 8: 8.2.3; Chapter 9: 9.1, 9.2Homework: 8.27, 8.32Lecture 33.(23.04.07) Interval estimation.Chapter 9: 9.1, 9.2.1, 9.2.2, 9.3.1Homework: 9.1, 9.2, 9.13Lecture 34.(24.04.07) Bayesian set estimation. Bayesian prediction.[1]: 9.2.4, 9.3.3, [2]: 4.3.2, 4.3.4Homework: here 350c69d7ab

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