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1 AMS Practice Midterm Exam Sprig, 018 Name: ID: Sigature: Istructio: This is a close book exam. You are allowed oe-page 8x11 formula sheet (-sided). No cellphoe or calculator or computer is allowed. Cheatig shall result i a course grade of F. The exam will last the etire lecture. Please provide complete solutios for full credit ad also tur i this cover page. Good luck! **** Please ote that this practice exam has more questios tha what will appear i the real midterm. The real midterm will have oly 3 problems each with o more tha 3 questios. 1. Let,, i. ~ i. d. (, ) 1 X X N, be a radom sample from the ormal populatio where is assumed kow. Please derive (a) The maximum likelihood estimator for. (b) The method of momet estimators for. (c) Are the above estimator(s) for ubiased? (d) Are these two estimators idetical? (e) Is the MLE a efficiet estimator for? (f) Is the MOME a efficiet estimator for? (g) Is the MLE a best estimator (UMVUE) for? (h) Is the MOME a best estimator (UMVUE) for? (i) Is the sample variace a efficiet estimator for Hit: Cramér-Rao Iequality: Let ˆ h X1, X,, X radom sample from a populatio with pdf f ;? Is it a best estimator (UMVUE) for be ubiased for, where X, i 1,,, is a X x satisfyig all regularity coditios. The 1 1 ˆ l f X x; l f X x; Var E E Solutio: (a) The likelihood fuctio is L = f (x i ; σ ) = 1 πσ exp [ (x i μ) σ ] The log likelihood fuctio is Solvig l = ll = costat l(σ ) (x i μ) i = (π) / [σ ] / exp [ (x i μ) σ ] σ l σ = σ + (x i μ) σ 4 = 0 We obtai the MLE for σ : σ = (X i μ) (b) We ow derive the MOME estimator for σ. Sice the first populatio momet is E[X] = μ that does ot ivolve σ, we oly eed to use the secod populatio ad sample momet. Settig them equal we have:? 1

2 X i1 E( X ) Oe step further we have the MOME estimator of σ : σ = X i μ i (c) Sice E[σ ] = E(X i μ) = σ It is straight-forward to verify that the MLE is a ubiased estimator for σ. = (X i μ) σ Sice E[σ ] = E[X i ] μ = (σ + μ ) μ = σ We have show that the MOME σ = X i μ is a ubiased estimator for σ. (d) The MLE for σ ca be rewritte as: σ = (X i μ) = (X i μx i + μ ) = X i μx + μ Comparig to the MOME, we foud them NOT idetical, although both are ubiased estimators for σ uless μ = 0. (e) Now we calculate the Cramer-Lower boud for the variace of a ubiased estimator for σ : l[f (x; σ 1 (x μ) )] = l [ exp [ πσ σ ]] = costatt 1 (x μ) lσ σ l[f (x; σ )] σ = 1 (x μ) + σ σ 4 l[f (x; σ )] = 1 (x μ) σ σ4 σ 6 E [ l[f (x; σ )] ] = E [ 1 (X μ) σ σ4 σ 6 ] = [ 1 σ 4 σ σ 6] = Thus the Cramer-Rao lower boud is: σ 4

3 σ 4 Therefore we claim that the MLE is a efficiet estimator for σ. We ow show that the MLE for σ : is a efficiet estimator for σ. Let σ = (X i μ) W = (X i μ) σ We kow that W~ χ ad therefore, Var(W) = Thus Var(σ ) = Var ( σ W ) = σ4 σ4 = Therefore, the MLE is a efficiet estimator for σ, ad subsequetly we ca immediately claim it is also the best estimator (UMVUE) for σ our aswer for part (g). (f) Usig the momet geeratig fuctio, we ca show that the variace of the MOME is: Var[σ ] = Var[X i ] = Var[X ] = σ4 + 4μ σ Comparig to the C-R lower boud, we foud the MOME is NOT is a efficiet estimator for σ uless μ = 0. Sice the MOME does ot have lower variace tha the MLE whe µ is ot zero, we claim for (h) that the MOME is ot a best estimator uless μ = 0. ***Below we show the detailed derivatios of [X ], ad thus, Var[σ ] *** Var[X ] = E[X 4 ] (E[X ]) = E[X 4 ] (σ + μ ) Now we derive E[X 4 ] usig the momet geeratig fuctio for X~N(μ, σ ): M(t) = exp(μt + σ t /) The first through the fourth derivatives for M(t) with respect to t is as follows: dm(t) dt = (μ + σ t)exp(μt + σ t /) d M(t) dt = σ exp(μt + σ t /) + (μ + σ t) exp(μt + σ t /) d 3 M(t) dt 3 = 3σ (μ + σ t)exp(μt + σ t /) + (μ + σ t) 3 exp(μt + σ t /) 3

4 d 4 M(t) dt 4 = 3σ 4 exp(μt + σ t /) + 6σ (μ + σ t) exp(μt + σ t /) + (μ + σ t) 4 exp(μt + σ t /) Therefore we have: E[X 4 ] = d4 M(t) dt 4 t=0 = 3σ 4 + 6μ σ + μ 4 Thus: Var[X ] = E[X 4 ] (E[X ]) = E[X 4 ] (σ + μ ) = σ 4 + 4μ σ Therefore we have: Var[σ ] = Var[X i ] = Var[X ] = σ4 + 4μ σ (g) Alteratively, we ca derive that the MLE is the best estimator directly (rather tha usig the efficiet estimator is also a best estimator argumet) as follows: The populatio pdf is: f(x; σ ) = 1 (x μ) πσ e σ = 1 1 πσ e σ (x μ) So it is a regular expoetial family, where the red part is w(σ ) ad the gree part is t(x). Thus T(X) = (X i μ) i is a complete & sufficiet statistic (CSS) for σ. It is easy to see that the MLE for σ is a fuctio of the complete & sufficiet statistic as follows: σ = (X i μ) = T(X) I additio, we kow that the MLE is a ubiased estimator for σ as we have doe i part (c): E(σ ) = E i (X i μ) = i E(X i μ) = i Var(X i) σ Sice the MLE σ is a ubiased estimator ad a fuctio of the CSS, σ is the best estimator (UMVUE) for σ by the Lehma-Scheffe Theorem. (j) The sample variace is also a ubiased estimator for σ : Let S = (X i X ) 1 V = (X i X ) σ We kow that V~ χ 1 ad therefore, Var(V) = ( 1) Thus 4

5 Var(S ) = Var ( σ V 1 ) = σ 4 σ4 ( 1) = ( 1) 1 > σ4 Therefore, the sample variace is either a efficiet estimator or a best estimator for σ whe μ is kow.. Suppose that the radom variables Y 1, Y,, Y satisfy Y i = βx i + ε i, i = 1,,, where x 1, x,, x are fixed costats, ad ε 1, ε,, ε are idepedet Gaussia radom variables with mea 0 ad variace σ (ukow). This is called a regressio through the origi. Please derive: (1) The (ordiary) least squares estimator (LSE) for β. () The maximum likelihood estimator (MLE) for β. Is the MLE the same as the LSE? (3) The distributio of the MLE. (4) Ca you fid a way to derive the method of momet estimator (MOME) for β? If so, please derive. Solutio: (1) To miimize SS β = 0 x i(y i βx i ) SS = (y i βx i ) : = 0 β = x iy i x i y = β x is the least squares regressio lie through the origi. () Y i ~N(βx i, σ ), i = 1,, Sice Y 1, Y,, Y are idepedet to each other, the likelihood is: L = 1 πσ e The log likelihood is: (y i βx i ) σ = (πσ ) / exp [ (y i βx i ) σ ] ll = l(πσ ) (y i βx i ) σ Take derivative with respect to β, ad set it to zero, we obtai the MLE: x iy i (3) Let β = x i θ i =, i = 1,, x i The θ i Y i ~N(θ i βx i, θ i σ ), i = 1,, Furthermore, they are idepedet to each other. We have the momet geeratig fuctio for β : x i 5

6 Therefore we foud: M β (t) = E[exp(tβ )] = E [exp (t θ i Y i )] = E[exp(tθ i Y i )] = exp [tθ i βx i + 1 t θ i σ ] = exp [t θ i βx i = exp [tβ + 1 t σ x i ] β ~N (β, σ x i ) i + 1 t θ i σ ] i (4) Yes, for example, By settig up: We obtai a MOME: W i = Y i βx i ~N(0, σ ), i = 1,, W = Y βx = EW = 0 β = Y x 3. Let X1, X,, X be a radom sample from the trucated expoetial distributio with pdf: f ( x) exp x, if x ; ad f ( x) 0, if x. Please (a) Derive the method of momet estimator of θ; (b) Derive the MLE of θ. (c) Are the MOME ad the MLE ubiased estimators for θ? (d) Compare the MSE of the MOME ad the MLE for θ. Which oe is a better estimator for θ? Solutio: (a) First we derive the populatio mea as follows. x x x x x x 1 E X xf ( x) dx xe dx e xe dx e e dx xe e e xe e e e Settig the populatio mea equal to the sample mea, we have 1 X ; thus the MOME for θ is W = ˆ X 1 (b) The Likelihood is L exp x exp x ; for x, i 1,, for mi X i1 i i1 i i i ˆ mi i Thus the likelihood is maximized whe θ achieves its largest value; thus the MLE for θ is X (c) The MOME is obviously ubiased as E(W) = (1 + θ) 1 = θ. The pdf of the MLE Y = X (1) is: f(y) = e θ y, y θ Thus the MLE is ot ubiased. E(Y) = yf(y) dy = θ + 1 (d) It is easy to show that X θ ~ exp (1), ad Y θ ~ exp (1/) θ 6

7 Therefore by the formula MSE = Bias + Variace, we ca easily compute the MSE for W ad Y as: 1/ ad / respectively. Thus for sample size, the MSE for the MLE is smaller. 4. The coutig process {N(t), t 0} is said to be a Poisso process havig rate λ, λ > 0, if (i) N(0) = 0; (ii) The process has idepedet icremets; ad (iii) The umber of evets i ay iterval of legth t is Poisso distributed with mea λt. That is, for all s, t 0 (λt) λt P{N(t + s) N(s) = } = e, = 0,1,! Please aswer the followig questios. (a) For this Poisso process, let us deote the time if the first evet by T 1. Further, for > 1, let T deote the elapsed time betwee the (-1) st ad the th evet. The sequece {T, = 1,, } is called the sequece of iterarrival times. Please show that T i.i.d. expoetia(λ), = 1,, ~ (b) The waitig time for the th evet is defied as S = Please derive the distributio of the waitig time S Solutio: T i, 1 (a) P(T > t) = P(time iterval betwee ( 1)th ad th evet is loger tha t) = P(o evet occurs i the time iterval of legth t) = E[P(N(t + S) N(S) = 0 S)] = E[e λt ] Note that {P(N(t + S) N(S) = 0 S = s) = e λt } = e λt t > 0. Thus, the cumulative distributio fuctio is F T (t) = 1 e λt, t > 0, = 1,,3,. Besides, assumptio (ii) implies the idepedece property. Thus, T i.i.d. expoetia(λ), = 1,, ~ (b) Derive the distributio by idetifyig the mgf of S. By (a),t i.i.d. expoetia(λ), = 1,,, ~ S = T i M S (t) = M, 1 Ti (t) = M Ti (t) = [M T1 (t)] M T1 (t) = e ts 0 λe λs ds = 0 λe (λ t)s ds = λ (λ λ t t)e (λ t)s ds 0 = λ, t < λ. λ t Thus, M S (t) = ( λ λ t ) = (1 t λ ) correspodig to Gamma(, λ) with the followig pdf: f S (x) = λ τ() x 1 e λx, x > 0. 7

8 ********* Topics to be covered i the midterm exam********* Dear Studets, There will be 3 problems (each with o more tha 3 sub-questios), take from our lecture otes ad text books covered (except the Bayesia estimators), icludig especially the followig topics: 1. Poit estimators This part is the same as the midterm where we have thoroughly reviewed the. Please review all cocepts covered i lectures icludig (1) geeral methods of derivig estimators icludig -- MLE, MOME, (Ordiary) Least Squares Estimators, ad () geeral properties of estimators such as bias, MSE, efficiet estimators, best estimators, sufficiet statistics, complete statistics, ad the related theorems.. Order statistics Their defiitios, distributios, ad applicatios especially i derivig the MLEs. 3. Poisso process This simple stochastic process provides a good groud for exercisig the cocepts of probabilities, their related properties (idepedece, mutually exclusive evets, etc.), statistical distributio ad sum of variables etc. 4. Probability This is the foudatio of mathematical statistics. At the begiig of the semester, we had revised may related cocepts. 5. Variables, joit distributio, ad variable trasformatio We have three ways to derive the distributio of trasformed variables: the cdf approach, the pdf approach ad the mgf approach. Oe should be fluet i all three. Please pay special attetio to the bivariate ormal radom variable oe should kow both the joit pdf ad the joit mgf ad how to use them, ad the 1-1 trasformatio of two variables especially how to trasform the joit domais. Our midterm exam will be held Thursday, March, 018, 8:30 AM ~ 9:50 AM, at our usual classroom Humaities

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