课程名称︰统计学 课程性质︰数学系选修 课程教师︰陈宏 开课学院：理学院 开课系所︰数学系 考试日期（年月日）︰2015/6/22 考试时限（分钟）：15:30~17:30 试题 : Part 1: 200 points 1. (55 points) Consider a random sample X , X ,..., X from the probability 1 2 n density function 1 + θx f(x;θ) = —————, -1≦x≦1. 2 ^ (a) (20 points) Find θ, the method of moments estimator of θ. ^ (b) (10 points) Is θ an unbiased estimator of θ? Justify your claim. 2 ^ 3 － θ (c) (10 points) Show that Var(θ) = —————. n ^ (d) (15 points) Show that θ is a consistent estimator of θ. 2. (50 points) Let X , X ,..., X be a random sample of size n from a uniform 1 2 n distribution on the interval (0,2θ). (a) (20 points) Find the maximum likelihood estimator of θ. Hint: Use order statistics. (b) (20 points) Find the expected value of the estimator you found in part (a). (c) (10 points) Adjust the estimator you found in part (a) so that it is unbiased. 3. (45 points) Suppose that in one month a stock appreciates $2 with probability 15%, depreciates $2 with probability 10% and remains unchanged with probability 75%. Let X and X denote the return of the stock in 1 2 January and February, respectively. Let S denote the return of the stock in two months. (Note that the initial value of stock is 100 and the return of the second month is calculated based on the value of the stock at the end of the first month.) Assume that the monthly returns are independent(namely, the first month is independent of that of the second month.) Let T be the return of the stock in rwo months. (a) (15 points) Write T in terms of some function of X and X . Give reason 1 2 to justify your claim. (b) (30 points) From now on, assume that X = X X to answer the rest part of 1 2 questions. Find the distribution, theexpected value and standard deviation of X. (Each part is 10 points.) 4. (50 points) In a laboratory on raising mice for scientific study. According to guideline, the population of raised mice should follow a normal distribution with mean 30g and standard deviation 5g. Suppose that 25 rates are sampled such that their weights can be assumed as a random sample from the population of raised mice. Namely, X ,..., X are assumed to be 1 25 independent and identically distributed random variables. Based on the _ sampled weight of rat. X = 27mg and s = 7mg. We now use the framework of 25 25 hypothesis testing to evaluate whether this laboratory really raised the mice according to their claim. 2 2 H : X ~ N(30,5 ) versus H : X ~ N(27,7 ). 0 a Consider the following testing procedure _ | X - 25 | | 25 | Reject H when | √25 (—————) | > c. 0 | 5 | Here c needs to be determined to have probability of type 1 error of no more than 5%. (a) (20 points) Determine c. Please present the reasoning on your choice of c. (b) (10 points) According to the proposed testing procedure, will you reject H ? 0 (c) (20 points) When H holds, what is the probability of committing Type a Ⅱ error? Part 2: 180 points 5. (60 points) Let X , X ,..., X be random variable each one having exponential 1 2 n distribution with parameter 1/λ > 0. The pdf is { 1 { — exp(-x/λ), x > 0 { λ f(x) = { { 0, elsewhere ^ (a) (20 points) Show that θ = n X is an unbiased estimator of λ. (1) ^ ^ ^ 2 (b) (20 points) Find the MSE(θ). Note that MSE(θ) = E(θ-λ) . 1 1 1 ^ (c) (20 points) Find the maximum likelihood estimate λ of λ. Which ^ ^ estimator will you choose between θ and θ and why? Show all your work. 1 2 6. (60 points) Suppose a random sample X ,..., X from a normal distribution 1 n 2 2 N(μ,σ ), with μ given and the variance σ unknown. Calculate the lower bound of variance for any estimator, and compare to that of the sample 2 variance S . 7. (60 points) Consider X ,..., X to be independent with means E(X ) = μ + β 1 n i i 2 and variances Var(X ) = σ . Such a situation could for example occur when i i X are estimators of μ obtained from independent sources and β is the bias i i of the estimator X . We now consider pooling the estimators of μ into a i common estimator by using a linear combination: ^ μ = w X + w X +...+ w X . 1 1 2 2 n n (a) (15 points) If the estimators are unbiased, i.e. if β = 0 for all i, i ^ show that a linear combination μ as above is unbiased if and only if n Σ w = 1. i=1 i (b) (15 points) In the case when β = 0 for all i, show that an unbiased i linear combination has minimum variance when the weights w are inversely i 2 proportional to the variances σ . i ^ (c) (10 points) Show that the variance of μ for optimal weights w is i ^ n -2 Var(μ) = 1/Σ σ . i=1 i (d) (10 points) Next, consider the case where the estimators may be biased so we could have β≠0. Find the mean square error of the optimal linear i combination obtained above, and compare its behavior as n→∞ in the 2 2 biased and unbiased case, when σ = σ , i = 1,..., n. i 2 (e) (10 points) Same as (d), but for a general sequence σ . i --

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