Master s Written Examination

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1 Master s Written Examination Option: Statistics and Probability Spring 05 Full points may be obtained for correct answers to eight questions Each numbered question (which may have several parts) is worth the same number of points All answers will be graded, but the score for the examination will be the sum of the scores of your best eight solutions Use separate answer sheets for each question DO NOT PUT YOUR NAME ON YOUR ANSWER SHEETS When you have finished, insert all your answer sheets into the envelope provided, then seal it

2 Problem Stat 40 X and Y are independent random variables with X exponential(λ) and Y exponential(µ) It is impossible to obtain direct observations of X and Y Instead, we observe the random variables Z and W, where {, Z = X; Z = min{x, Y } and W = 0, Z = Y (a) Find the joint distribution of Z and W (b) Prove that Z and W are independent Solution to Problem (a) We first compute P (Z > z, W = ) P (Z > z, W = ) = P (Z > z, Z = X) Clearly, P (W = ) = P (Z > 0, W = ) = Similarly, we have = P (X > z, X Y ) = = P (Z z, W = ) = z z x λµe λx e µy dydx λe λx e µx dx = λ λ + µ e (λ+µ)z λ Thus λ+µ λ ( ) e (λ+µ)z λ + µ P (Z z, W = 0) = µ ( ) e (λ+µ)z λ + µ (b) It is sufficient to show that P (Z z W = i) = P (Z z) Clearly, we have On the other hand, P (Z z) = e (λ+µ)z P (Z z W = i) = P (Z z, W = i) P (W = i) Utilizing the results in (a), we can directly verify that P (Z z W = i) = P (Z z) Problem Stat 40 Suppose you have a wooden stick with length You first break it into two pieces at a randomly selected point on the stick Then you again break the longer piece at random into two pieces Let X and X be the places where you break the stick the first and second time, respectively Find the joint probability density function of X and X

3 Find the probability that you can make a triangle using the three pieces Solution to Problem Let X and X be the places where you break the stick the first/second time X Unif (0,) Let, x > /, and x (x, ); x f (x x ) =, x < /, and x (0, x ), x Since X Unif (0, ), we have, x > /, and x (x, ); x f(x, x ) = f(x )f (x x ) =, x < /, and x (0, x ), x Now we have two cases x > / In this case, the three pieces have length x, x x and x In order to make a triangle, we need x > x, x + x > x x, and x > x In other words, we need x / < x < / x < / Similarly, the three pieces have length x, x x and x Solving these inequalities gives / < x < x + / Therefore the probability that a triangle can be made using the three pieces is = P(X > /, X / < X < / OR X < /, / < X < X + /) x =/ / x / x dx dx + / /+x x =0 x =/ = log(/) / / log(/) = log 4 x dx dx Problem 3 Stat 4 Let X,, X n be a random sample from X Normal (θ, σ ), θ R is unknown, and σ > 0 is known () Find the maximum likelihood estimator ˆθ mle for mean θ () Is ˆθ mle an efficient estimator for mean θ? (3) Let parameter η = θ, determine its mle ˆη mle and the asymptotic distribution of ˆη mle? Solution to Problem 3 Log likelihood function of the sample is () The likelihood equation is l (θ) = n log ( πσ ) σ l θ = σ (x i θ) (x i θ) = 0 3

4 and its solution is ˆθ = x = n x i, and l θ = n σ < 0, so ˆθ mle = x ) () E (ˆθmle = E ( X) = θ, an unbiased estimator for θ, ) V ar (ˆθmle = V ar ( V ar (X) X) = n = σ n = ni (θ) as Fisher information of the sample is ( ) log f (X, θ) ni (θ) = ne = n θ σ The mle reaches the R-C lower bound, so it is an efficient estimator for θ ) (3) Based on the functional invariant property of the mle, ˆη mle = (ˆθmle = X Based on the asymptotic normal distribution of the mles, we have ( ) n (ˆηmle η) N 0, (η (θ)), I (θ) ie n (ˆη mle η) N (0, 4θ σ ) Problem 4 Stat 4 Let X,, X n and Y,, Y n be independent random samples from two normal distributions N(µ, σ ) and N(µ, σ ), respectively, where σ is the common but unknown variance (a) Find the likelihood ratio for testing H 0 : µ = µ = 0 against all alternatives (b) Rewrite so that it is a function of a statistic Z which has a well-known distribution under null hypotheses (Hint: X i = (X i X + X) ) (c) Give the distribution of Z under null hypotheses Solution to Problem 4 The likelihood function is ( (πσ ) n exp σ ( (X i µ ) + (a) Under the null hypotheses, the MLE of σ is ) (Y j µ ) ) j= ˆσ = X i + j= Y j n 4

5 and ( L(ˆω) = (π) n X i + j= Y ) j n exp ( n) n Under Ω, we have ˆµ = X, ˆµ = Ȳ, ˆσ = (X i X) + j= (Y j Ȳ ), n and Thus, we have (b) Notice that Thus where ( L(ˆΩ) = (π) n (X i X) + j= (Y j Ȳ ) n ) exp ( n) n = L(ˆω) L(ˆΩ) = Xi + j= Y j = Z = ( X i + j= Y j (X i X) + j= (Y j Ȳ ) (X i X) + ) n (Y j Ȳ ) + n X + nȳ j= = ( + Z) n, n X + nȳ (X i X) + j= (Y j Ȳ ) (c) Under null hypotheses, n X /σ χ (), nȳ /σ χ (), (X i X) /σ χ (n ), and j= (Y j Ȳ ) /σ χ (n ) Notice that they are independent with each other Thus (n X + nȳ )/σ χ () and ( (X i X) + j= (Y j Ȳ ) )/σ χ (n ) Consequently, Z F distribution with first df= and second df=n Problem 5 Stat 4 Consider a bivariate observation X = (X, X ) with density f θ (x, x ) = exp{ θx θ x }, x, x, θ > 0 Note: Under this model, X and X are independent (but not iid) exponential random variables with means /θ and θ, respectively Use the factorization theorem to find the minimal sufficient statistic; you don t have to prove the minimal part Find the maximum likelihood estimator of θ Is it sufficient? 3 Show that the minimal sufficient statistic identified in Part is not complete 5

6 Solution to Problem 5 The likelihood for θ is just the joint density of the observations, and it is clear that the shape of the likelihood depends on all of the data, (X, X ) So, by the factorization theorem, there is no non-trivial sufficient statistic The derivative of the log-density is X + X /θ Setting this equal to zero and solving for θ gives the MLE, ie, ˆθ = (X /X ) / This is, of course, a function of the sufficient statistic, but since it s not one-to-one, it cannot itself be a sufficient statistic 3 To show that T = (X, X ) is not complete, we need to find a non-zero function f(t) such that E θ {f(t )} = 0 for all θ Since the exponential distribution is a scale parameter family, we expect that some multiplication or division might cancel out the dependence on θ Indeed, if we take then we get f(t) = f(t, t ) = t t, E θ {f(t )} = E θ (X X ) = E θ (X )E θ (X ) = θ θ = 0, θ Since f is not identically zero, we conclude T is not complete Problem 6 Stat 46 in six consecutive years: The table below shows yearly sales (in $0, 000) of a bookstore Suppose we choose significant level 005 Time Yr- Yr- Yr-3 Yr-4 Yr-5 Yr-6 Sales (a) Test if there exists a trend using total number of runs above and below sample median (b) Use the rank von Neumann (RVN) test to see if there exists a trend (c) Are your conclusions in (a) and (b) consistent? What s your final conclusion? Solution to Problem 6 (a) Arrange the yearly sales in increasing order:,,4,6,7,0 Then the sample median is (4 + 6)/ = 5 Using + or to indicate the yearly sales above and below the sample median, we get,,, +, +, + Then the total number of runs R = According to Table D, the p-value is P (R ) = 0 > 005 We conclude that there is no significant trend at 5% level 6

7 (b) The ranks of yearly sale, rank(x i ) s, are,, 3, 5, 4, 6 Thus the test statistic NM = 5 [rank(x i ) rank(x i+ )] = = 4 According to Table S, the p-value is P (NM 4) = 0047 We conclude that there is a significant trend at 5% level (c) The conclusions in (a) and (b) are not consistent The reason is that the RVN test is more powerful than the test based on total number of runs We follow the RVN test and conclude that there is a significant trend Problem 7 Stat 43 Consider a without-replacement sample of size from a population of size 5, with joint inclusion probabilities π = π 45 = 05, π 3 = π 34 = 0, π 3 = π 4 = 005, π 4 = π 35 = 0, and π 5 = π 5 = 008 (a) Calculate the inclusion probability π i for this design (b) The variance Horvite-Thompson estimator can be written as V (ˆt HT ) = N N ( π i ) t i + π i N k i π ik π i π k π ik t i t k Show that it is equivalent to V (ˆt HT ) = N N k i π i π k π ik π ik ( ti t ) k π i π k (Hint: N k i π ik = (n )π i ) (c) Suppose that t i = i, i =,, 5 Find ˆV HT (ˆt HT ) and ˆV SY G (ˆt HT ) for sample {, } Which one is more stable in general? Solution to Problem 7 (a) π = π + π 3 + π 4 + π 5 = 04, π = π + π 3 + π 4 + π 5 = 038, π 3 = π 3 + π 3 + π 34 + π 35 = 037, π 4 = π 4 + π 4 + π 34 + π 45 = 04, π 5 = π 5 + π 5 + π 35 + π 45 = 043 (b) Utilizing that fact that N k i π ik = (n )π i, we can draw the conclusion after some ( ) t simple algebra through the expansion of the term i π i t k π k 7

8 (c) For S = {, }, we have ˆV HT (ˆt HT ) = i S ( π i ) t i π i + i S k Sk i π ik π i π k π ik t i π i t k π k For S = {, }, we have ˆV HT (ˆt HT ) = ( π ) t π = 057 ˆV HT (ˆt HT ) = i S k Sk i + ( π ) t π π i π k π ik π ik ˆV SY G (ˆt HT ) = π π π π = π π π π t π t π ( ti t ) k π i π k ( t t ) π π SYG one is generally more stable among the two estimators Problem 8 Stat 45 Suppose that X = (X,, X n ) are iid samples from a twocomponent normal mixture model with density f(x) = αn(x µ, ) + ( α)n(x ν, ) The parameter of interest is θ = (α, µ, ν), where α [0, ] Derive the EM algorithm for computing the maximum likelihood estimator of θ Solution to Problem 8 Introduce missing data Z = (Z,, Z n ), where Z i is a label for the mixture component that X i was sampled from That is, { if X i N(µ, ) Z i = i =,, n 0 if X i N(ν, ), The complete data is Y = (X, Z), and the complete-data log-likelihood is log L Y (θ) = Z i {log α + (X i µ) } + ( Z i ){log( α) + (X i ν) } Now we need the conditional distribution of Z, given X This is effectively an application of Bayes formula, and we can write Z i (X i, θ (t) ) Ber(ω (t) i ), where ω (t) i = α (t) N(X i µ (t), ), i =,, n α (t) N(X i µ (t), ) + ( α (t) )N(X i ν (t), ) Then we have the following E- and M-steps 8

9 E-step We evaluate the Q-function as follows: Q(θ θ (t) ) = E[log L Y (θ) X, θ (t) ] = ω (t) i {log α + (X i µ) } + ( ω (t) i ){log( α) + (X i ν) } M-step Differentiation of Q with respect to each element in θ, setting the equations to zero, and solving, we get the following updates: α (t+) = n ω (t) i, µ (t+) = ω(t) i X i ω(t) i, ν (t+) = ( ω(t) i )X i ( ω(t) i ) Problem 9 Stat 46 matrix (a) Find the period of state 0 A Markov chain X 0, X, has the transition probability P = (b) Is this a regular Markov chain? If yes, determine the limiting distribution If no, explain why Solution to Problem 9 (a) Since there is only one communicating class, the periods of all states are the same Hence we only need to compute the period of state 3 But it is easy to see that the period of state 3 is (b) It is not a regular Markov chain because P n 33 will not be strictly positive for all large n (Note that if the chain starts from state 3, the probability it comes back to state 3 in odd number of steps is zero) Problem 0 Stat 46 A Markov chain has a state space S = {0,, } and the following transition probability matrix If the chain starts from state, find the probability that it does not visit state prior its absorption at state 0 9

10 Solution to Problem 0 The question is equivalent to assuming that state is also an absorption state and ask what is the probability that the chain is stopped at state 0, instead of state More precisely, we have a Markov chain with state space S = {0,, } and the following transition probability matrix We want to compute the probability that the chain is eventually stopped at state 0 There are several ways to compute this probability For example, it is computed as = 3 Problem Stat 48 We are interested the effect of nozzle type on the rate of fluid flow Three specific nozzle types are under study Five runs through each of the three nozzle types led to the following results: Nozzle Type Rate of Flow ȳ i = 5 j y ij j (y ij ȳ i ) A B C Construct the ANOVA table and test whether nozzle type has an effect on fluid flow Use significance level α = 005 and F critical value F(005;,) = 389 Solution to Problem The ANOVA table is Source SS df MS F Treatment Error Total where 484 = , ȳ = ( )/3 = 9673, 633 = 5 (( ) + ( ) + ( ) ) Since 34 < F (005;, ) = 389, there is no sufficient evidence to conclude that nozzle type has an effect Problem Stat 48 Y, the model is Consider one explanatory variable x and the response variable Y i = β 0 + β x i + ε i, i =,, n, where the iid errors ε i N (0, σ ) (a) Derive the least square estimates ˆβ 0 and ˆβ as follows such that the fitted line ˆβ 0 + ˆβ x is the best straight line to fit the observed data (x, y ),, (x n, y n ),ie ˆβ = (x i x) y i (x i x), ˆβ 0 = ȳ ˆβ x 0

11 given that ( x i) n (x i ) = (x i x) (b) Prove that the distribution of ˆβ is as follows: ) ˆβ N (β, σ s xx where s xx = (x i x) (c) Based on the sampling distribution of ( ) ˆσ ˆβ β / s xx when σ is unknown, construct a confidence interval for β How to detect the existence of the linear relationship by using this confidence interval? Solution to Problem (a) Least square criterion Q (β 0, β ) = n (Y i β 0 + β x i + ε i ) Take derivative with respect to β 0, β respectively Solve the equation ˆβ = Q (β 0, β ) β 0 = Q (β 0, β ) β = (b) Let c i = (x i x) (x i x), then (Y i β 0 + β x i + ε i ) = 0 (Y i β 0 + β x i + ε i ) x i = 0 (x i x)y i (x i x), ˆβ 0 = ȳ ˆβ x ˆβ = (x i x) y i (x i x) = c i y i, as y i ind N (β 0 + β x i, σ ), then ( ˆβ N c i (β 0 + β x i ), ) c i σ = N (β, σ (c) Take ˆσ = MSE = SSE/ (n ) ( ) ˆσ ˆβ β / t (n ) s xx then the confidence interval for β : ˆβ ± t α (n ) MSE s xx If the confidence interval doesn t cover zero point, then there exists significant linear relationship between the response and the covariate at the given significance level s xx )

Master s Written Examination

Master s Written Examination Option: Statistics and Probability Spring 016 Full points may be obtained for correct answers to eight questions. Each numbered question which may have several parts is worth