x = Pr ( X (n) βx ) =
|
|
- Lindsay Wilkerson
- 5 years ago
- Views:
Transcription
1 Exercise 93 / page 45 The desity of a variable X i i 1 is fx α α a For α kow let say equal to α α > fx α α x α Pr X i x < x < Usig a Pivotal Quatity: x α 1 < x < α > x α 1 ad We solve i a similar way as i Example 98 page 48 We get a ituitio that the method of pivotal quatities may work for this exercise by observig that is a scale parameter ad X/ is a variable that its distributio does ot deped o The latter ca be easily see by cosiderig the distributio of the variable Y X/ The x y ad < y < 1 ad f Y y α f X y α d y α dy y α 1 α y α 1 for < y < 1 For a kow α X is a suciet statistic for that ca be foud by usig the Factorizatio Theorem So the pivotal quatity ca be based o the sample oly through X The cumulative distributio of X is Pr X x Pr X 1 x X x idepedet α x < x < Pr X i x idetically distributed [Pr X i x] or alteratively oe ca use Theorem 544 page 9 to derive the distributio of X The cumulative distributio of the pivot quatity QX X Pr QX x Pr X x Pr X x x is α x α < x < 1 1 We eed to d a upper codece limit for ie a iterval of the form X UX The lower boud i this specic case is X because the support of the desity is < x < Observe that QX X is decreasig fuctio of Thus for a iterval with codece coeciet 95 we eed to d a value b such that Pr b < QX 95 where Pr QX > b 1 Pr QX b 1 b α Thus ad the 95% codece iterval is 1 b α 95 b α 5 b 5 1/α C : x < & 5 1/α < Qx : x < < x or x 5 1/α 1 : x < & 5 1/α < x x 5 1/α
2 By Pivotig a cotiuous CDF: We use Theorem 91 ad solve i a similar way as i Example 913 page 433 For a kow α X is a suciet statistic for ad its cumulative distributio is F X x x α < x < as we foud above F X x is a decreasig fuctio of for each x We are iterested i oe-sided iterval that gives a upper codece boud Applyig Theorem 91 for α 5 the upper boud U x is deed as follows: F X x U x 5 x U x Thus the 95% codece iterval for is x By Ivertig a Test Statistic: α 5 U x x 5 1/α x 5 1/α We solve i a similar way as i Example 93 page 43 Sice we eed a 95% upper codece iterval we eed to ivert the acceptace regio of a 5% test of the hypothesis H : H 1 : < where the parameter space of is B X As a test statistic we will use the LRT which is of the form: where λx sup L x α sup B L x α L x α fx i α α α 1 α 1 xi α x α i x < From exercise 71 we kow that the MLE of is X Thus λx sup L x α sup B L x α L x α L x x α α α α x α x i α 1 x i α 1 x α Note that the LRT depeds o the sample oly through x which is a suciet statistic for The rejectio regio is R x : λx c x : x α c
3 For a 5% test we have 5 sup Pr X R Pr X R Pr X Pr c 1/α c [1/α ]α c X α c use eq1 Therefore the rejectio regio of the 5% test is R x : x α 5 the acceptace regio is A x : x α > 5 x : x > 5 1/α ad the 95% codece iterval is Cx : x < < x 5 1/α or x x 5 1/α Note that i this exercise the codece itervals derived by all three methods are the same However this is ot always the case b Based o the data give i exercise ˆα MLE 159 ad x 14 5 Thus the 95% codece iterval is 5 C : 5 < < : 5 < < 543 or / Exercise 94 / page 45 Note that i this exercise λ is used to deote the LRT while λ is the value of the ratio λ σ Y /σ X uder H a Let θ be the vector parameter θ σx σ Y The parameter space of θ is the two-dimesioal space Θ σx σ Y : σ X σ Y > Uder H the parameter space is Θ σx σ Y : σ Y λ σx σ X σ Y > The LRT statistic for the hypothesis is: λx y sup L σx σ Y x y θ Θ L σx σ Y x y sup θ Θ For the deomiator supl σx σ Y x y we have to d the MLE of σ X ad σ Y The likelihood θ Θ fuctio is 3
4 m LσX σy x y fx σxfy σ Y fx i σx fy i σy πσx 1/ e 1 σ X x m i πσy 1/ e 1 σ Y π +m/ σ X y i / σ Y m/ e 1 σ X x i e 1 σ Y y i The log-likelihood is: log Lσ X σ Y x y + m logπ log σ X m log σ Y 1 σ X x i 1 σ Y m yi ad its rst partial derivatives are: σ X log Lσ X σ Y x y σ X 1 + σx x i [ ] σ σx 4 X x i σ Y log Lσ X σ Y x y m σ Y 1 + σy m y i m σ 4 Y [ ] σy y i m Solvig the system σx σy log LσX σ Y x y log LσX σ Y x y i terms of σx ad σ Y we get: ˆσ X 1 x i ad ˆσ Y 1 m m yi Thus the poit ˆσ X ˆσ Y is a critical poit ad usig the secod derivatives we check whether it is maximum The secod partial derivatives are σ X log Lσ X σ Y x y σ Y log Lσ X σ Y x y σ 1 X σx 3 m σ 1 Y σy 3 x i m yi σ 6 X m σ 6 Y [ ] σx x i [ ] σy y i m σ X σ Y l Lσ X σ Y x y The values of log Lσ σx X σ Y x y ad log Lσ σy X σ Y x y at the critical poit are: 4
5 σ X log Lσ X σ Y x y σ X σ Y ˆσ X ˆσ Y σ 6 X < σ Y log Lσ X σ Y x y σ Y ˆσ Y m σ 6 Y < The Hessia matrix of log LσX σ Y x y evaluated at the poit ˆσ X ˆσ Y is H ˆσ X ˆσ Y σx 6 m σy 6 ad its determiat is positive Thus the poit ˆσ X ˆσ Y is the maximum poit of log Lσ X σ Y x y The the deomiator of λx y is supl σx σy x y L ˆσ X ˆσ Y x y π ˆσ +m/ X / m/ ˆσ Y e +m θ Θ Regardig the umerator of λx y uder H σy λ σx with σ X σ Y fuctio is modied as follows: > ad the likelihood L σ X σ Y x y Lσ X λ σ X x y π +m/ λ m/ σ X [ +m/ 1 e σ X x i + 1 ] m λ y i set LσX x y So sup L σx σy x y sup LσX x y θ Θ σx :σ > X The log-likelihood is: log Lσ X x y + m logπ m log λ + m [ log σx 1 x σx i + 1 λ yi ] ad its rst derivative is: d dσ X log Lσ X x y + m σ X + m σ 4 X [ 1 + x σx i + 1 λ σ X [ yi x i + 1 λ y i + m ] ] For d log Lσ dσx X x y we get: [ ˆσ 1 x i m λ yi ] 5
6 The secod derivative is: d d σ log X Lσ X x y + m σ 1 X σx 3 [ x i + 1 λ yi ] + m σ 6 X [ σ X x i + 1 λ y i + m ] ad evaluated at the critical poit it gets the value: d d σ X log Lσ X x y σ X ˆσ + m ˆσ < [ ] So the maximum occurs at ˆσ 1 +m x i + 1 m λ y i ad the umerator of λx y is sup L σx σy x y sup LσX x y Lˆσ x y θ Θ σx :σ > X Hece the LRT ca be writte as π +m/ λ m/ ˆσ +m/ e +m λx y π +m/ λ m/ ˆσ +m/ e +m π +m/ ˆσ X / ˆσ Y m/ e +m devide by m y i let W X Y let k +m/ X i Y i + m+m/ λ m/ / m m/ ˆσ X / ˆσ Y m/ λ m/ ˆσ +m/ 1 / 1 m/ x i m y i [ λ m/ 1 +m x i + ] +m/ 1 m λ y i + m +m/ x i y i / λ m/ / m m/ x i m y i + m +m/ w / +m/ λ m/ / m m/ w + 1 λ w / k +m/ w + 1λ set λw + 1 λ +m/ We get the idea to use the statistic W X Y above by the questio b of the exercise Note that λx y depeds o the sample poits x ad y oly through the statistics X i ad m Y i which are the suciet statistics for σx ad σ Y i the case of N σ X ad N σ Y respectively The rejectio regio is R x y : λx y c x y : λw c 6
7 where c is such that sup θ Θ Pr λw c α To see how the rejectio regio ca be expressed i terms of the values of W we will study w the mootoicity of the fuctio λw k / +m/ or equivaletly that of the fuctio w+ 1 λ log λw log k + +m log w log w + 1λ The rst derivative of the latter is d log λw dw w + m w + 1 w λ + m λ w + 1 λ w + λ w λ mw w λ w + 1 λ λ mw + w λ w + 1 For d dw log λw we get w /λ m > ad we have: w /λ m log λw + - log λw λw d dw Therefore the rejectio regio i terms of the values of W is We choose k 1 ad k such that R x y : W x y k 1 or W x y k with < k 1 < k sup Pr W x y k 1 or W x y k α θ Θ b Uder H X 1 X is a radom sample from N σx ad Y 1 Y m is aother radom sample idepedet to the rst oe from N λ σx The X i χ σx Y i λ χ σ m ad X the two statistics are idepedet to each other That meas that uder H the variable F X Y X i σ X Y i mλ σ X X i mλ m Y i W X Ymλ F m Note the the distributio of F X Y is idepedet of the parameters σ X ad σ Y Thus the rejectio regio ca be modied as follows: R x y : W x y k 1 or W x y k with < k 1 < k x y : F x y mλ k 1 or F x y mλ k with < k 1 < k x y : F x y c 1 or F x y c with < c 1 < c For a α level test we have α sup θ Θ Pr F X Y c 1 or F X Y c Pr F m c 1 or F m c 7
8 c 1 F mα1 ad c F m1 α where α 1 + α α c Based o questio b we get that the acceptace regio for the test is Aλ x y : F mα1 < F x y < F m1 α x y : F mα1 < W x y mλ < F m1 α Thus the 1 α codece iterval for the ratio λ is C x y λ : F mα1 mw x y < λ < mw x y F m1 α Exercise 95 / page 455 i Ivertig the acceptace regio of LRT statistic: To obtai a 1 α two-sided codece iterval for µ we ca ivert the acceptace regio of the LRT of level α of the hypothesis H : µ µ H 1 : µ µ where the parameter space of µ is M Y with Y X 1 > The parameter space uder H is M µ Sice Y X 1 is a suciet statistic for µ we ca use Theorem 84 ad the LRT is of the form: sup Lµ y µ M λy Lµ y where the likelihood is of the form sup µ M Lµ y exp y µ µ y The likelihood is a icreasig fuctio of µ Hece it reaches its maximum at the maximum possible value of µ with µ y ie for µ y Thus suplµ y Ly y exp y y µ M For the umerator sup µ M Lµ y we have sup Lµ y µ M Lµ y if µ y if µ > y Thus λy Lµ y Ly y exp y µ exp y µ if µ y if µ > y 8
9 The rejectio regio is R x : λy c x : exp y µ c Observe that λy exp y µ y µ is a decreasig fuctio of y Thus exp y µ c y c with c µ ad the rejectio regio ca be expressed i terms of the values of Y as follows: Sice the test is of level α it should hold R x : y c c µ α sup Pr X R µ sup Pr Y c µ Pr Y c µ µ M µ M ˆ ˆ fy µ dy exp y µ dy [ exp y µ ] c c c exp c µ log α c + µ c µ log α Note that the solutio c µ log α satises the coditio c µ sice a 1 ad log α > Thus the rejectio regio of level α LRT for the above hypothesis is R x : y µ log α The acceptace regio is Aµ ad the 1 α codece iterval is µ log α Cx µ : y < ii Usig a Pivotal Quatity: x : y < µ log α with µ y µ : y + log α < µ y To build a pivotal quatity a reasoable strategy is to d a suciet statistic for the parameter i questio ispect its distributio to see whether it is a locatio ad/or scale parameter distributio ad the cosider a variable which icorporates the suciet statistic ad the parameter i questio appropriately i a similar fashio as these variables suggested o page 47 of the book It is give that Y X 1 is suciet statistic for µ The distributio of Y is a locatio parameter distributio with µ beig the locatio parameter We cosider the quatity QX µ X 1 µ Y µ The distributio of QX µ is f Q q e q q > 9
10 The above ca be easily see by cosiderig the distributio of the trasformed variable Q Y µ Sice y µ the q ad also y q + µ Thus f Q q f Y q + µ d q + µ dq e q+µ µ 1 e q for q > Thus for a 1 α two-sided codece iterval we should d c 1 ad c such that Pr c 1 QX µ c 1 α with < c 1 < c or Pr QX µ < c 1 α 1 ad Pr QX µ > c α with α 1 + α α It is preferable to have two equatios sice we eed to determie two ukow quatities c 1 ad c It has ot bee set α 1 α α/ i order to give a geeral solutio ad also because the desity f Q q e q is ot symmetric ad α 1 Pr QX µ < c 1 ˆ c1 e c 1 1 α 1 c 1 log1 α 1 α Pr QX µ > c c log α ˆ Therefore the 1 α two-sided codece iterval is µ : log1 α 1 y µ log α µ : y + log α c e q [ e q] c 1 1 e c 1 e q [ e q] c e c µ y + log1 α 1 So the three codece itervals are: ˆ µ : y + log α ˆ < µ y based o ivertig the acceptace regio of LRT based o a pivotal quatity ad µ : y + log α µ y + log1 α 1 ˆ µ : y + 1 log α µ y + 1 log1 α by pivotig the CDF of Y foud i Example 913 Note that the rst codece iterval is a special case of the secod whe α 1 ad α α ad the third codece iterval is a special case of the secod whe α 1 α α/ A criterio of compariso of itervals of the same codece coeciet is the legth of the itervals The oe with the shorter legth is preferable The legths of the three itervals matioed above are: ˆ LRT: y log α y log α 1
11 ˆ pivotal quatity: y + log1 α 1 y + log α 1 log1 α 1 log α ad ˆ pivotig the CDF: y + 1 log1 α y + 1 log α 1 log1 α log α To compare the legths it suces to study the fuctio gα 1 log 1 α 1 log α α 1 α 1 α Observe that g log α ad gα/ log1 α log α d gα dα 1 1 α 1 α α 1 1 α 1 α 1 α α 1 > Thus gα 1 is a icreasig fuctio of α 1 reachig its miimum value for α 1 ad the miimum value is g log α Thus the codece iterval based o ivertig the acceptace regio of LRT is the oe with the shorter legth Exercise 934 / page 457 a σ kow We kow that a 1 α codece iterval for µ is µ : x z a/ σ µ x + z a/ σ see Example 91 page 4 ad the legth of the iterval is z a/ σ For a 95% codece iterval the legth is For a legth o more tha σ/4 we get 196 σ 39 σ 39 σ σ b σ ukow I such a case we kow that a 1 α codece iterval for µ is s s µ : x t a/ 1 µ x + t a/ 1 see Example 91 page 49 ad the legth of the iterval is t a/ 1 S For a 95% codece iterval the legth is t 5 1 S 11
12 fuc Figure 1: Fuctio h 1 64t 5 1 χ 9 1 The requiremet regardig the legth is: S Pr t 5 1 σ 9 4 We eed to express the above probability as a fuctio of so that we ca solve the iequality i terms of S Pr t 5 1 σ 4 Pr t S 5 1 Pr χ 1 σ t 5 1 1S Pr σ 1 64t 5 1 Therefore we get: Pr χ t t 5 1 χ t 5 1 χ 9 1 The above is solved umerically for example by tryig dieret values of For 76 we get t 575 χ > That is the smallest value of for which the iequality holds The graph of fuctio h miimum such that h 1 64t 5 1 χ 9 1 is show i Figure 1 We search for the S Commet 1: If the required probability was dieret Pr t 5 1 σ 9 the the 4 solutio for chages The less the probability the smaller the miimum value of is Figure shows the fuctio h whe the probability is set to ad 4 respectively Commet : We ca also solve the iequality 1 64t 5 1 χ 9 1 approximately usig the approximatios that for large a t 5 1 z ad b χ N Thus 1
13 fuctio Figure : Fuctio h whe the probability is ad 4 χ 1 N 1 1 ad χ z The we solve the iequality which is very close to the exact result 76 Exercise 935 / page 457 a I exercise 934 a we showed that for σ kow the legth of a 1 α codece iterval for σ µ is z a/ σ ad E z a/ σ z a/ b I exercise 934 b we showed that for σ ukow the legth of a 1 α codece iterval for µ is t a 1 S Sice the legth depeds o the variable S it is a variable itself The expected legth is S ES E t a 1 t a 1 where Γ ES cσ where c 1Γ regardig the value of c see for example the iformatio give i exercise 75 page 364 Therefore E t a 1 S t a 1 ES t a 1 cσ Thus the expected dierece i legth is equal to σ S σ cσ E z a/ t a 1 z a/ t a 1 σ za/ t a 1 c 13
14 because c 1 ad t a 1 z a/ whe I other words asymptotically the expected legth is Commet: A approximate result ca be derived by usig the approximatio ES σ ote that sice c 1 whe thus ES σ whe The E z a/ σ t a 1 S z a/ σ t a 1 σ σ za/ t a 1 For small values of ad for α 5 the quatities z a/ t a 1 c ad z 5 t 5 1 take the values preseted i the table below We see that the diereces betwee the values are fairly small ad decrease as the sample size icreases z 5 t 5 1 c z 5 t
Direction: This test is worth 150 points. You are required to complete this test within 55 minutes.
Term Test 3 (Part A) November 1, 004 Name Math 6 Studet Number Directio: This test is worth 10 poits. You are required to complete this test withi miutes. I order to receive full credit, aswer each problem
More informationDirection: This test is worth 250 points. You are required to complete this test within 50 minutes.
Term Test October 3, 003 Name Math 56 Studet Number Directio: This test is worth 50 poits. You are required to complete this test withi 50 miutes. I order to receive full credit, aswer each problem completely
More informationSummary. Recap ... Last Lecture. Summary. Theorem
Last Lecture Biostatistics 602 - Statistical Iferece Lecture 23 Hyu Mi Kag April 11th, 2013 What is p-value? What is the advatage of p-value compared to hypothesis testig procedure with size α? How ca
More informationTopic 9: Sampling Distributions of Estimators
Topic 9: Samplig Distributios of Estimators Course 003, 2018 Page 0 Samplig distributios of estimators Sice our estimators are statistics (particular fuctios of radom variables), their distributio ca be
More informationTopic 9: Sampling Distributions of Estimators
Topic 9: Samplig Distributios of Estimators Course 003, 2018 Page 0 Samplig distributios of estimators Sice our estimators are statistics (particular fuctios of radom variables), their distributio ca be
More informationTopic 9: Sampling Distributions of Estimators
Topic 9: Samplig Distributios of Estimators Course 003, 2016 Page 0 Samplig distributios of estimators Sice our estimators are statistics (particular fuctios of radom variables), their distributio ca be
More information( θ. sup θ Θ f X (x θ) = L. sup Pr (Λ (X) < c) = α. x : Λ (x) = sup θ H 0. sup θ Θ f X (x θ) = ) < c. NH : θ 1 = θ 2 against AH : θ 1 θ 2
82 CHAPTER 4. MAXIMUM IKEIHOOD ESTIMATION Defiitio: et X be a radom sample with joit p.m/d.f. f X x θ. The geeralised likelihood ratio test g.l.r.t. of the NH : θ H 0 agaist the alterative AH : θ H 1,
More informationFirst Year Quantitative Comp Exam Spring, Part I - 203A. f X (x) = 0 otherwise
First Year Quatitative Comp Exam Sprig, 2012 Istructio: There are three parts. Aswer every questio i every part. Questio I-1 Part I - 203A A radom variable X is distributed with the margial desity: >
More informationLast Lecture. Wald Test
Last Lecture Biostatistics 602 - Statistical Iferece Lecture 22 Hyu Mi Kag April 9th, 2013 Is the exact distributio of LRT statistic typically easy to obtai? How about its asymptotic distributio? For testig
More informationLet us give one more example of MLE. Example 3. The uniform distribution U[0, θ] on the interval [0, θ] has p.d.f.
Lecture 5 Let us give oe more example of MLE. Example 3. The uiform distributio U[0, ] o the iterval [0, ] has p.d.f. { 1 f(x =, 0 x, 0, otherwise The likelihood fuctio ϕ( = f(x i = 1 I(X 1,..., X [0,
More informationEstimation for Complete Data
Estimatio for Complete Data complete data: there is o loss of iformatio durig study. complete idividual complete data= grouped data A complete idividual data is the oe i which the complete iformatio of
More informationECE 8527: Introduction to Machine Learning and Pattern Recognition Midterm # 1. Vaishali Amin Fall, 2015
ECE 8527: Itroductio to Machie Learig ad Patter Recogitio Midterm # 1 Vaishali Ami Fall, 2015 tue39624@temple.edu Problem No. 1: Cosider a two-class discrete distributio problem: ω 1 :{[0,0], [2,0], [2,2],
More informationR. van Zyl 1, A.J. van der Merwe 2. Quintiles International, University of the Free State
Bayesia Cotrol Charts for the Two-parameter Expoetial Distributio if the Locatio Parameter Ca Take o Ay Value Betwee Mius Iity ad Plus Iity R. va Zyl, A.J. va der Merwe 2 Quitiles Iteratioal, ruaavz@gmail.com
More informationReview Questions, Chapters 8, 9. f(y) = 0, elsewhere. F (y) = f Y(1) = n ( e y/θ) n 1 1 θ e y/θ = n θ e yn
Stat 366 Lab 2 Solutios (September 2, 2006) page TA: Yury Petracheko, CAB 484, yuryp@ualberta.ca, http://www.ualberta.ca/ yuryp/ Review Questios, Chapters 8, 9 8.5 Suppose that Y, Y 2,..., Y deote a radom
More informationLecture Notes 15 Hypothesis Testing (Chapter 10)
1 Itroductio Lecture Notes 15 Hypothesis Testig Chapter 10) Let X 1,..., X p θ x). Suppose we we wat to kow if θ = θ 0 or ot, where θ 0 is a specific value of θ. For example, if we are flippig a coi, we
More informationIIT JAM Mathematical Statistics (MS) 2006 SECTION A
IIT JAM Mathematical Statistics (MS) 6 SECTION A. If a > for ad lim a / L >, the which of the followig series is ot coverget? (a) (b) (c) (d) (d) = = a = a = a a + / a lim a a / + = lim a / a / + = lim
More informationChapter 6 Principles of Data Reduction
Chapter 6 for BST 695: Special Topics i Statistical Theory. Kui Zhag, 0 Chapter 6 Priciples of Data Reductio Sectio 6. Itroductio Goal: To summarize or reduce the data X, X,, X to get iformatio about a
More information6.3 Testing Series With Positive Terms
6.3. TESTING SERIES WITH POSITIVE TERMS 307 6.3 Testig Series With Positive Terms 6.3. Review of what is kow up to ow I theory, testig a series a i for covergece amouts to fidig the i= sequece of partial
More informationAsymptotics. Hypothesis Testing UMP. Asymptotic Tests and p-values
of the secod half Biostatistics 6 - Statistical Iferece Lecture 6 Fial Exam & Practice Problems for the Fial Hyu Mi Kag Apil 3rd, 3 Hyu Mi Kag Biostatistics 6 - Lecture 6 Apil 3rd, 3 / 3 Rao-Blackwell
More informationResampling Methods. X (1/2), i.e., Pr (X i m) = 1/2. We order the data: X (1) X (2) X (n). Define the sample median: ( n.
Jauary 1, 2019 Resamplig Methods Motivatio We have so may estimators with the property θ θ d N 0, σ 2 We ca also write θ a N θ, σ 2 /, where a meas approximately distributed as Oce we have a cosistet estimator
More informationLast Lecture. Unbiased Test
Last Lecture Biostatistics 6 - Statistical Iferece Lecture Uiformly Most Powerful Test Hyu Mi Kag March 8th, 3 What are the typical steps for costructig a likelihood ratio test? Is LRT statistic based
More informationEXAMINATIONS OF THE ROYAL STATISTICAL SOCIETY
EXAMINATIONS OF THE ROYAL STATISTICAL SOCIETY GRADUATE DIPLOMA, 016 MODULE : Statistical Iferece Time allowed: Three hours Cadidates should aswer FIVE questios. All questios carry equal marks. The umber
More informationNovember 2002 Course 4 solutions
November Course 4 solutios Questio # Aswer: B φ ρ = = 5. φ φ ρ = φ + =. φ Solvig simultaeously gives: φ = 8. φ = 6. Questio # Aswer: C g = [(.45)] = [5.4] = 5; h= 5.4 5 =.4. ˆ π =.6 x +.4 x =.6(36) +.4(4)
More information1.010 Uncertainty in Engineering Fall 2008
MIT OpeCourseWare http://ocw.mit.edu.00 Ucertaity i Egieerig Fall 2008 For iformatio about citig these materials or our Terms of Use, visit: http://ocw.mit.edu.terms. .00 - Brief Notes # 9 Poit ad Iterval
More informationLecture 6 Ecient estimators. Rao-Cramer bound.
Lecture 6 Eciet estimators. Rao-Cramer boud. 1 MSE ad Suciecy Let X (X 1,..., X) be a radom sample from distributio f θ. Let θ ˆ δ(x) be a estimator of θ. Let T (X) be a suciet statistic for θ. As we have
More informationINFINITE SEQUENCES AND SERIES
11 INFINITE SEQUENCES AND SERIES INFINITE SEQUENCES AND SERIES 11.4 The Compariso Tests I this sectio, we will lear: How to fid the value of a series by comparig it with a kow series. COMPARISON TESTS
More informationExercises and Problems
HW Chapter 4: Oe-Dimesioal Quatum Mechaics Coceptual Questios 4.. Five. 4.4.. is idepedet of. a b c mu ( E). a b m( ev 5 ev) c m(6 ev ev) Exercises ad Problems 4.. Model: Model the electro as a particle
More informationFACULTY OF MATHEMATICAL STUDIES MATHEMATICS FOR PART I ENGINEERING. Lectures
FACULTY OF MATHEMATICAL STUDIES MATHEMATICS FOR PART I ENGINEERING Lectures MODULE 5 STATISTICS II. Mea ad stadard error of sample data. Biomial distributio. Normal distributio 4. Samplig 5. Cofidece itervals
More informationSequences and Series of Functions
Chapter 6 Sequeces ad Series of Fuctios 6.1. Covergece of a Sequece of Fuctios Poitwise Covergece. Defiitio 6.1. Let, for each N, fuctio f : A R be defied. If, for each x A, the sequece (f (x)) coverges
More informationQuestions and Answers on Maximum Likelihood
Questios ad Aswers o Maximum Likelihood L. Magee Fall, 2008 1. Give: a observatio-specific log likelihood fuctio l i (θ) = l f(y i x i, θ) the log likelihood fuctio l(θ y, X) = l i(θ) a data set (x i,
More information4. Hypothesis testing (Hotelling s T 2 -statistic)
4. Hypothesis testig (Hotellig s T -statistic) Cosider the test of hypothesis H 0 : = 0 H A = 6= 0 4. The Uio-Itersectio Priciple W accept the hypothesis H 0 as valid if ad oly if H 0 (a) : a T = a T 0
More informationSTAT Homework 7 - Solutions
STAT-36700 Homework 7 - Solutios Fall 208 October 28, 208 This cotais solutios for Homework 7. Please ote that we have icluded several additioal commets ad approaches to the problems to give you better
More informationThe variance of a sum of independent variables is the sum of their variances, since covariances are zero. Therefore. V (xi )= n n 2 σ2 = σ2.
SAMPLE STATISTICS A radom sample x 1,x,,x from a distributio f(x) is a set of idepedetly ad idetically variables with x i f(x) for all i Their joit pdf is f(x 1,x,,x )=f(x 1 )f(x ) f(x )= f(x i ) The sample
More informationPlease do NOT write in this box. Multiple Choice. Total
Istructor: Math 0560, Worksheet Alteratig Series Jauary, 3000 For realistic exam practice solve these problems without lookig at your book ad without usig a calculator. Multiple choice questios should
More information1 Introduction to reducing variance in Monte Carlo simulations
Copyright c 010 by Karl Sigma 1 Itroductio to reducig variace i Mote Carlo simulatios 11 Review of cofidece itervals for estimatig a mea I statistics, we estimate a ukow mea µ = E(X) of a distributio by
More informationStat 421-SP2012 Interval Estimation Section
Stat 41-SP01 Iterval Estimatio Sectio 11.1-11. We ow uderstad (Chapter 10) how to fid poit estimators of a ukow parameter. o However, a poit estimate does ot provide ay iformatio about the ucertaity (possible
More informationStatistical Theory MT 2008 Problems 1: Solution sketches
Statistical Theory MT 008 Problems : Solutio sketches. Which of the followig desities are withi a expoetial family? Explai your reasoig. a) Let 0 < θ < ad put fx, θ) = θ)θ x ; x = 0,,,... b) c) where α
More informationLECTURE 14 NOTES. A sequence of α-level tests {ϕ n (x)} is consistent if
LECTURE 14 NOTES 1. Asymptotic power of tests. Defiitio 1.1. A sequece of -level tests {ϕ x)} is cosistet if β θ) := E θ [ ϕ x) ] 1 as, for ay θ Θ 1. Just like cosistecy of a sequece of estimators, Defiitio
More informationSOME THEORY AND PRACTICE OF STATISTICS by Howard G. Tucker
SOME THEORY AND PRACTICE OF STATISTICS by Howard G. Tucker CHAPTER 9. POINT ESTIMATION 9. Covergece i Probability. The bases of poit estimatio have already bee laid out i previous chapters. I chapter 5
More informationLecture 3. Properties of Summary Statistics: Sampling Distribution
Lecture 3 Properties of Summary Statistics: Samplig Distributio Mai Theme How ca we use math to justify that our umerical summaries from the sample are good summaries of the populatio? Lecture Summary
More informationProblem Set 4 Due Oct, 12
EE226: Radom Processes i Systems Lecturer: Jea C. Walrad Problem Set 4 Due Oct, 12 Fall 06 GSI: Assae Gueye This problem set essetially reviews detectio theory ad hypothesis testig ad some basic otios
More informationStatistical Theory MT 2009 Problems 1: Solution sketches
Statistical Theory MT 009 Problems : Solutio sketches. Which of the followig desities are withi a expoetial family? Explai your reasoig. (a) Let 0 < θ < ad put f(x, θ) = ( θ)θ x ; x = 0,,,... (b) (c) where
More informationHOMEWORK I: PREREQUISITES FROM MATH 727
HOMEWORK I: PREREQUISITES FROM MATH 727 Questio. Let X, X 2,... be idepedet expoetial radom variables with mea µ. (a) Show that for Z +, we have EX µ!. (b) Show that almost surely, X + + X (c) Fid the
More informationMATH 320: Probability and Statistics 9. Estimation and Testing of Parameters. Readings: Pruim, Chapter 4
MATH 30: Probability ad Statistics 9. Estimatio ad Testig of Parameters Estimatio ad Testig of Parameters We have bee dealig situatios i which we have full kowledge of the distributio of a radom variable.
More informationLecture 6 Simple alternatives and the Neyman-Pearson lemma
STATS 00: Itroductio to Statistical Iferece Autum 06 Lecture 6 Simple alteratives ad the Neyma-Pearso lemma Last lecture, we discussed a umber of ways to costruct test statistics for testig a simple ull
More informationLecture 2: Monte Carlo Simulation
STAT/Q SCI 43: Itroductio to Resamplig ethods Sprig 27 Istructor: Ye-Chi Che Lecture 2: ote Carlo Simulatio 2 ote Carlo Itegratio Assume we wat to evaluate the followig itegratio: e x3 dx What ca we do?
More informationLecture 6 Chi Square Distribution (χ 2 ) and Least Squares Fitting
Lecture 6 Chi Square Distributio (χ ) ad Least Squares Fittig Chi Square Distributio (χ ) Suppose: We have a set of measuremets {x 1, x, x }. We kow the true value of each x i (x t1, x t, x t ). We would
More informationMath 113, Calculus II Winter 2007 Final Exam Solutions
Math, Calculus II Witer 7 Fial Exam Solutios (5 poits) Use the limit defiitio of the defiite itegral ad the sum formulas to compute x x + dx The check your aswer usig the Evaluatio Theorem Solutio: I this
More informationIntroductory statistics
CM9S: Machie Learig for Bioiformatics Lecture - 03/3/06 Itroductory statistics Lecturer: Sriram Sakararama Scribe: Sriram Sakararama We will provide a overview of statistical iferece focussig o the key
More informationSince X n /n P p, we know that X n (n. Xn (n X n ) Using the asymptotic result above to obtain an approximation for fixed n, we obtain
Assigmet 9 Exercise 5.5 Let X biomial, p, where p 0, 1 is ukow. Obtai cofidece itervals for p i two differet ways: a Sice X / p d N0, p1 p], the variace of the limitig distributio depeds oly o p. Use the
More information( µ /σ)ζ/(ζ+1) µ /σ ( µ /σ)ζ/(ζ 1)
A eective CI for the mea with samples of size 1 ad Melaie Wall James Boe ad Richard Tweedie 1 Abstract It is couterituitive that with a sample of oly oe value from a ormal distributio oe ca costruct a
More information6 Sample Size Calculations
6 Sample Size Calculatios Oe of the major resposibilities of a cliical trial statisticia is to aid the ivestigators i determiig the sample size required to coduct a study The most commo procedure for determiig
More information32 estimating the cumulative distribution function
32 estimatig the cumulative distributio fuctio 4.6 types of cofidece itervals/bads Let F be a class of distributio fuctios F ad let θ be some quatity of iterest, such as the mea of F or the whole fuctio
More informationLecture 6 Chi Square Distribution (χ 2 ) and Least Squares Fitting
Lecture 6 Chi Square Distributio (χ ) ad Least Squares Fittig Chi Square Distributio (χ ) Suppose: We have a set of measuremets {x 1, x, x }. We kow the true value of each x i (x t1, x t, x t ). We would
More informationMath 10A final exam, December 16, 2016
Please put away all books, calculators, cell phoes ad other devices. You may cosult a sigle two-sided sheet of otes. Please write carefully ad clearly, USING WORDS (ot just symbols). Remember that the
More information5. Likelihood Ratio Tests
1 of 5 7/29/2009 3:16 PM Virtual Laboratories > 9. Hy pothesis Testig > 1 2 3 4 5 6 7 5. Likelihood Ratio Tests Prelimiaries As usual, our startig poit is a radom experimet with a uderlyig sample space,
More informationThis exam contains 19 pages (including this cover page) and 10 questions. A Formulae sheet is provided with the exam.
Probability ad Statistics FS 07 Secod Sessio Exam 09.0.08 Time Limit: 80 Miutes Name: Studet ID: This exam cotais 9 pages (icludig this cover page) ad 0 questios. A Formulae sheet is provided with the
More informationThe Poisson Process *
OpeStax-CNX module: m11255 1 The Poisso Process * Do Johso This work is produced by OpeStax-CNX ad licesed uder the Creative Commos Attributio Licese 1.0 Some sigals have o waveform. Cosider the measuremet
More informationProperties and Hypothesis Testing
Chapter 3 Properties ad Hypothesis Testig 3.1 Types of data The regressio techiques developed i previous chapters ca be applied to three differet kids of data. 1. Cross-sectioal data. 2. Time series data.
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.265/15.070J Fall 2013 Lecture 3 9/11/2013. Large deviations Theory. Cramér s Theorem
MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.265/5.070J Fall 203 Lecture 3 9//203 Large deviatios Theory. Cramér s Theorem Cotet.. Cramér s Theorem. 2. Rate fuctio ad properties. 3. Chage of measure techique.
More informationLesson 10: Limits and Continuity
www.scimsacademy.com Lesso 10: Limits ad Cotiuity SCIMS Academy 1 Limit of a fuctio The cocept of limit of a fuctio is cetral to all other cocepts i calculus (like cotiuity, derivative, defiite itegrals
More informationx 2 x x x x x + x x +2 x
Math 5440: Notes o particle radom walk Aaro Fogelso September 6, 005 Derivatio of the diusio equatio: Imagie that there is a distributio of particles spread alog the x-axis ad that the particles udergo
More informationMATH 472 / SPRING 2013 ASSIGNMENT 2: DUE FEBRUARY 4 FINALIZED
MATH 47 / SPRING 013 ASSIGNMENT : DUE FEBRUARY 4 FINALIZED Please iclude a cover sheet that provides a complete setece aswer to each the followig three questios: (a) I your opiio, what were the mai ideas
More informationSTAT431 Review. X = n. n )
STAT43 Review I. Results related to ormal distributio Expected value ad variace. (a) E(aXbY) = aex bey, Var(aXbY) = a VarX b VarY provided X ad Y are idepedet. Normal distributios: (a) Z N(, ) (b) X N(µ,
More informationAAEC/ECON 5126 FINAL EXAM: SOLUTIONS
AAEC/ECON 5126 FINAL EXAM: SOLUTIONS SPRING 2015 / INSTRUCTOR: KLAUS MOELTNER This exam is ope-book, ope-otes, but please work strictly o your ow. Please make sure your ame is o every sheet you re hadig
More information1 Approximating Integrals using Taylor Polynomials
Seughee Ye Ma 8: Week 7 Nov Week 7 Summary This week, we will lear how we ca approximate itegrals usig Taylor series ad umerical methods. Topics Page Approximatig Itegrals usig Taylor Polyomials. Defiitios................................................
More informationEECS564 Estimation, Filtering, and Detection Hwk 2 Solns. Winter p θ (z) = (2θz + 1 θ), 0 z 1
EECS564 Estimatio, Filterig, ad Detectio Hwk 2 Sols. Witer 25 4. Let Z be a sigle observatio havig desity fuctio where. p (z) = (2z + ), z (a) Assumig that is a oradom parameter, fid ad plot the maximum
More information4.1 Sigma Notation and Riemann Sums
0 the itegral. Sigma Notatio ad Riema Sums Oe strategy for calculatig the area of a regio is to cut the regio ito simple shapes, calculate the area of each simple shape, ad the add these smaller areas
More informationEcon 325 Notes on Point Estimator and Confidence Interval 1 By Hiro Kasahara
Poit Estimator Eco 325 Notes o Poit Estimator ad Cofidece Iterval 1 By Hiro Kasahara Parameter, Estimator, ad Estimate The ormal probability desity fuctio is fully characterized by two costats: populatio
More informationSTATISTICAL INFERENCE
STATISTICAL INFERENCE POPULATION AND SAMPLE Populatio = all elemets of iterest Characterized by a distributio F with some parameter θ Sample = the data X 1,..., X, selected subset of the populatio = sample
More informationGoodness-of-Fit Tests and Categorical Data Analysis (Devore Chapter Fourteen)
Goodess-of-Fit Tests ad Categorical Data Aalysis (Devore Chapter Fourtee) MATH-252-01: Probability ad Statistics II Sprig 2019 Cotets 1 Chi-Squared Tests with Kow Probabilities 1 1.1 Chi-Squared Testig................
More informationUnbiased Estimation. February 7-12, 2008
Ubiased Estimatio February 7-2, 2008 We begi with a sample X = (X,..., X ) of radom variables chose accordig to oe of a family of probabilities P θ where θ is elemet from the parameter space Θ. For radom
More information62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 +
62. Power series Defiitio 16. (Power series) Give a sequece {c }, the series c x = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + is called a power series i the variable x. The umbers c are called the coefficiets of
More informationStatistical Inference Based on Extremum Estimators
T. Rotheberg Fall, 2007 Statistical Iferece Based o Extremum Estimators Itroductio Suppose 0, the true value of a p-dimesioal parameter, is kow to lie i some subset S R p : Ofte we choose to estimate 0
More informationDistributions of Functions of. Normal Random Variables Version 27 Jan 2004
Distributios of Fuctios of Normal Radom Variables Versio 27 Ja 2004 The Uit or Stadard) Normal The uit or stadard ormal radom variable U is a ormally distributed variable with mea zero ad variace oe, i.
More informationMATH 10550, EXAM 3 SOLUTIONS
MATH 155, EXAM 3 SOLUTIONS 1. I fidig a approximate solutio to the equatio x 3 +x 4 = usig Newto s method with iitial approximatio x 1 = 1, what is x? Solutio. Recall that x +1 = x f(x ) f (x ). Hece,
More informationSTAT Homework 1 - Solutions
STAT-36700 Homework 1 - Solutios Fall 018 September 11, 018 This cotais solutios for Homework 1. Please ote that we have icluded several additioal commets ad approaches to the problems to give you better
More information4x 2. (n+1) x 3 n+1. = lim. 4x 2 n+1 n3 n. n 4x 2 = lim = 3
Exam Problems (x. Give the series (, fid the values of x for which this power series coverges. Also =0 state clearly what the radius of covergece is. We start by settig up the Ratio Test: x ( x x ( x x
More informationStatistical Inference (Chapter 10) Statistical inference = learn about a population based on the information provided by a sample.
Statistical Iferece (Chapter 10) Statistical iferece = lear about a populatio based o the iformatio provided by a sample. Populatio: The set of all values of a radom variable X of iterest. Characterized
More informationINFINITE SEQUENCES AND SERIES
INFINITE SEQUENCES AND SERIES INFINITE SEQUENCES AND SERIES I geeral, it is difficult to fid the exact sum of a series. We were able to accomplish this for geometric series ad the series /[(+)]. This is
More informationOutput Analysis and Run-Length Control
IEOR E4703: Mote Carlo Simulatio Columbia Uiversity c 2017 by Marti Haugh Output Aalysis ad Ru-Legth Cotrol I these otes we describe how the Cetral Limit Theorem ca be used to costruct approximate (1 α%
More informationChapter 13: Tests of Hypothesis Section 13.1 Introduction
Chapter 13: Tests of Hypothesis Sectio 13.1 Itroductio RECAP: Chapter 1 discussed the Likelihood Ratio Method as a geeral approach to fid good test procedures. Testig for the Normal Mea Example, discussed
More informationLecture 7: Properties of Random Samples
Lecture 7: Properties of Radom Samples 1 Cotiued From Last Class Theorem 1.1. Let X 1, X,...X be a radom sample from a populatio with mea µ ad variace σ
More informationn 3 ln n n ln n is convergent by p-series for p = 2 > 1. n2 Therefore we can apply Limit Comparison Test to determine lutely convergent.
06 微甲 0-04 06-0 班期中考解答和評分標準. ( poits) Determie whether the series is absolutely coverget, coditioally coverget, or diverget. Please state the tests which you use. (a) ( poits) (b) ( poits) (c) ( poits)
More informationStatistical inference: example 1. Inferential Statistics
Statistical iferece: example 1 Iferetial Statistics POPULATION SAMPLE A clothig store chai regularly buys from a supplier large quatities of a certai piece of clothig. Each item ca be classified either
More informationMath 152. Rumbos Fall Solutions to Review Problems for Exam #2. Number of Heads Frequency
Math 152. Rumbos Fall 2009 1 Solutios to Review Problems for Exam #2 1. I the book Experimetatio ad Measuremet, by W. J. Youde ad published by the by the Natioal Sciece Teachers Associatio i 1962, the
More informationf(x) dx as we do. 2x dx x also diverges. Solution: We compute 2x dx lim
Math 3, Sectio 2. (25 poits) Why we defie f(x) dx as we do. (a) Show that the improper itegral diverges. Hece the improper itegral x 2 + x 2 + b also diverges. Solutio: We compute x 2 + = lim b x 2 + =
More informationSimulation. Two Rule For Inverting A Distribution Function
Simulatio Two Rule For Ivertig A Distributio Fuctio Rule 1. If F(x) = u is costat o a iterval [x 1, x 2 ), the the uiform value u is mapped oto x 2 through the iversio process. Rule 2. If there is a jump
More informationChapter 4. Fourier Series
Chapter 4. Fourier Series At this poit we are ready to ow cosider the caoical equatios. Cosider, for eample the heat equatio u t = u, < (4.) subject to u(, ) = si, u(, t) = u(, t) =. (4.) Here,
More information7.1 Convergence of sequences of random variables
Chapter 7 Limit Theorems Throughout this sectio we will assume a probability space (, F, P), i which is defied a ifiite sequece of radom variables (X ) ad a radom variable X. The fact that for every ifiite
More informationMIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS
MIDTERM 3 CALCULUS MATH 300 FALL 08 Moday, December 3, 08 5:5 PM to 6:45 PM Name PRACTICE EXAM S Please aswer all of the questios, ad show your work. You must explai your aswers to get credit. You will
More informationExpectation and Variance of a random variable
Chapter 11 Expectatio ad Variace of a radom variable The aim of this lecture is to defie ad itroduce mathematical Expectatio ad variace of a fuctio of discrete & cotiuous radom variables ad the distributio
More informationMidterm Exam #2. Please staple this cover and honor pledge atop your solutions.
Math 50B Itegral Calculus April, 07 Midterm Exam # Name: Aswer Key David Arold Istructios. (00 poits) This exam is ope otes, ope book. This icludes ay supplemetary texts or olie documets. You are ot allowed
More informationLecture 3: MLE and Regression
STAT/Q SCI 403: Itroductio to Resamplig Methods Sprig 207 Istructor: Ye-Chi Che Lecture 3: MLE ad Regressio 3. Parameters ad Distributios Some distributios are idexed by their uderlyig parameters. Thus,
More informationData Analysis and Statistical Methods Statistics 651
Data Aalysis ad Statistical Methods Statistics 651 http://www.stat.tamu.edu/~suhasii/teachig.html Suhasii Subba Rao Review of testig: Example The admistrator of a ursig home wats to do a time ad motio
More informationAn Introduction to Randomized Algorithms
A Itroductio to Radomized Algorithms The focus of this lecture is to study a radomized algorithm for quick sort, aalyze it usig probabilistic recurrece relatios, ad also provide more geeral tools for aalysis
More informationKurskod: TAMS11 Provkod: TENB 21 March 2015, 14:00-18:00. English Version (no Swedish Version)
Kurskod: TAMS Provkod: TENB 2 March 205, 4:00-8:00 Examier: Xiagfeg Yag (Tel: 070 2234765). Please aswer i ENGLISH if you ca. a. You are allowed to use: a calculator; formel -och tabellsamlig i matematisk
More informationCHAPTER 10 INFINITE SEQUENCES AND SERIES
CHAPTER 10 INFINITE SEQUENCES AND SERIES 10.1 Sequeces 10.2 Ifiite Series 10.3 The Itegral Tests 10.4 Compariso Tests 10.5 The Ratio ad Root Tests 10.6 Alteratig Series: Absolute ad Coditioal Covergece
More information9. Simple linear regression G2.1) Show that the vector of residuals e = Y Ŷ has the covariance matrix (I X(X T X) 1 X T )σ 2.
LINKÖPINGS UNIVERSITET Matematiska Istitutioe Matematisk Statistik HT1-2015 TAMS24 9. Simple liear regressio G2.1) Show that the vector of residuals e = Y Ŷ has the covariace matrix (I X(X T X) 1 X T )σ
More informationCastiel, Supernatural, Season 6, Episode 18
13 Differetial Equatios the aswer to your questio ca best be epressed as a series of partial differetial equatios... Castiel, Superatural, Seaso 6, Episode 18 A differetial equatio is a mathematical equatio
More information