he wrtten Master s Eamnaton Opton Statstcs and Probablty SPRING 9 Full ponts may be obtaned for correct answers to 8 questons. Each numbered queston (whch may have several parts) s worth the same number of ponts. All answers wll be graded, but the score for the eamnaton wll be the sum of the scores of your best 8 solutons. Use separate answer sheets for each queston. DO NO PU YOUR NAME ON YOUR ANSWER SHEES. When you have fnshed, nsert all your answer sheets nto the envelope provded, then seal and prnt your name on t. Any student whose answers need clarfcaton may be requred to submt to an oral eamnaton.
MS Eam, Opton Probablty and Statstcs, SPRING 9. (Stat 4) Let U and V be two ndependent, standard normal random varables. (a) Fnd the jont dstrbuton of U + V and U V. U + V (b) Fnd the dstrbuton of. ( U V). (Stat 4) Let X,, X n be ndependent random varables wth X dstrbuted as N(βa,σ ), for =,,n, where a,..,a n are known (non-random) real numbers at least one of whch s non-zero and σ s known. () Derve the mamum lkelhood estmator (mle) of β. () Is the mle unbased? Justfy your answer. () Fnd the Rao-Cramer lower bound for unbased estmators of β. (v) Is the mle of β effcent (mnmum varance unbased)? Justfy your answer. 3. (Stat 4) Consder one random varable X that has a bnomal dstrbuton wth n = 4 and p = θ. [] Fnd the most powerful test of sze α / 6 based on X only for H : θ =.5 vs. H : θ =.75. [] Calculate the power of your test. [3] Justfy that your test s the most powerful one of sze /6. 4. (Stat 46) A chemst wshes to test the effect of four chemcal agents on the strength of a partcular type of cloth. Because there mght be varablty from one bolt to another, the chemst decdes to use a randomzed block desgn, wth the bolts of cloth consdered as blocks. She selects fve bolts and apples all four chemcals n random order to each bolt. he resultng tensle strengths follow. Bolt 3 4 5 Chemcal A 66 47 6 6 55 Chemcal B 68 48 69 7 55 Chemcal C 66 56 7 68 58 Chemcal D 6 47 64 63 5 Descrbe the basc model of the Fredman test, gve ts test statstc S, and compute S. Eplan also why the Fredman test s here more approprate than the Kruskal-Walls test.
MS Eam, Opton Probablty and Statstcs, SPRING 9 5. (Stat 43) A clent has a fnte populaton of 6 unts and has funds to survey only 3 unts for the purpose of estmatng the mean and ts related standard devaton. here are two samplng plans to be consdered by two survey practtoners. Samplng Plan by Survey Practtoner : Is a smple random samplng plan of sze 3 wthout replacement, SRS (6, 3). Recall that ths s a unform samplng plan wth all samples of sze 3 n ts support. Samplng Plan by Survey Practtoner : Is a controlled samplng plan wth only 4 samples: A- he followng s samples are ecluded from beng surveyed, that s they have zero probablty of selecton. {,,3 }, {,4,5}, {,5,6}, {,3,5}, {,4,6}, {3,4,6}. B- he followng s samples are assgned probablty of / of selecton each: {,,5}, {,3,5}, {,4,6}, {,3,4}, {,3,6}, {4,5,6}. C- he remanng 8 samples are assgned probablty of / of selecton each. Remark: Note that the total probablty over the chosen 4 samples s and the survey s not a unform samplng plan snce s of the samples are assgned twce as much probablty as the remanng 8 samples. Suppose both practtoners mplemented ther survey plans and the sample {,, 5} s selected by both of them and the related survey data are: Y =, Y = 96, and Y 5 = 3. - What are the H estmators and ther values of the populaton mean under both samplng plans? - What are the standard devatons of your estmator under both samplng plans? 3- If you were the consultng statstcan, whch of the above two samplng plans would you recommend to your clent and why? 6. (Stat 46) he number of customers enterng a store on a gven day s Posson dstrbuted wth mean λ =. he amount of money spent by a customer s unformly dstrbuted over (,). Fnd the mean and the varance of the amount of money that the store takes n on a gven day. 3
MS Eam, Opton Probablty and Statstcs, SPRING 9 7. (Stat 47) Home depot stocks wooden planks of length 9 feet. Home Bulders buy such planks and cut them accordng to ther needs. A home bulder needs planks of length 7 feet, planks of length 5 feet, 3 planks of length 4 feet. Suppose only the followng 3 cuttng patterns are used. 7 Feet 5 Feet wth cuttng patterns 3 4 Feet 3 (a). How many 9 feet planks are needed to meet the requrements usng only these cuttng patterns. What s the total length of the wasted porton? (b). Use revsed smple method to check whether there s a potental cuttng pattern that would reduce the number of planks needed. (c). Determne whch pattern has to be replaced by the entry of the new cuttng pattern? (d). How many 9 feet planks are needed to meet the requrements usng only the new cuttng patterns after dscontnung one of the old patterns? 8. (Stat 47) Show that = 5/6, = 5/, = 7 / 6 s optmal to the lnear programmng problem: ma 9 + 4 + 7 3 such that + + 3 6 3 5 + 4 + 3 3 5,, unrestrcted. 9. (Stat 473) On a lnear forest path of unt length covered wth bushes, Joe chooses secretly a locaton to hde and Bob chooses a locaton y to hde. he payoff to Joe from Bob s y y. Show that the strategy whch chooses a random pont y n [,] wth densty f ( ), y gves an epectaton = /6for all. Fnd the value of the game. 4
MS Eam, Opton Probablty and Statstcs, SPRING 9. (Stat 48) A dary company would lke to nvestgate f the mlk s contamnated. In order to nvestgate a possble shpment (batch) effect, the company selects 5 shpments at random. After processng each batch, 6 cartons of mlk are selected at random and are stored for several days. hen the square root of the bactera counts are recorded and denoted by Y j, where =,...,5 and j=,...,6. (a). If the shpment effect s denoted as τ, wrte down ANOVA model and specfy requred dstrbuton of random components n the model. Please state the hypotheses. (b). Complete the followng table and conclude gven sgnfcance level.. Sources df SS MS F Shpment 83.3 Error otal 36. [Gven: F(., 5, 3) = 3.7, F(., 5, 5) = 3.85, F(., 4, 5) = 4.77 ] (c). Estmate the varance component(s).. (Stat 48) In lnear regresson analyss, leave-one-out cross valdaton (LOOCV) procedure s often used to estmate the predcton accuracy of the ftted regresson model. Let ( X, Y) = (, y ); =,..., n be a dataset wth p = (,,..., p) and y R ( X( ), Y ) ( ) y( ) = ( X( ) X( ) ) X ( ) Y( ) R + the -th row. A set of predcton values estmate s defned as, and let be the modfed dataset wthout n LOOCV = n e( ) wth e( ) = y y( ). ( X X) ( X X) (a) Show that( X( ) X( ) ) ( X X) = + ( X X) =. are obtaned and the LOOCV A zz A Hnt: ( A zz ) = A + assumng that A z z s nvertble. z A z e (b) Show that e( ) = ; =,..., n, where H = X( X X) X, h s the -th dagonal element of H h and e= ( e,..., e ) = ( I H) Y. n 5
Statstcs 4&48 MS Eam (Junhu Wang) Sprng Semester 9. Let U and V be two ndependent, standard normal random varables. (a) Fnd the jont dstrbuton of U + V and U V. (b) Fnd the dstrbuton of U + V ( U V). [Soluton] (a) Snce U N, I V, we have U + V U = N, I N,. U V V + ( ) ( ) (b) From (a), U V N,, U V N,, and thus ( ) (, U + V N ), and ( ) ( U V χ ). From (a), U + V and U V are ndependent, whch mples that ( ) U + V and ( ) U V are ndependent as well. herefore, we have U + V ( U V) based on the defnton of t dstrbuton. = ( U + V ) t() ( U V)
Stat 4, Estmaton problem. Sprng 9 Let X ; : : : ; X n be ndependent random varables wth X dstrbuted as N(a, ), for =,...,n, where a ; ::; a n are known (non-random) real numbers at least one of whch s non-zero and s known. () Derve the mamum lkelhood estmator (mle) of. () Is the mle unbased? Justfy your answer. () Fnd the Rao-Cramer lower bound for unbased estmators of. (v) Is the mle of e cent (mnmum varance unbased)? Justfy your answer. Soluton: () he lkelhood s L() = () n= n e ( a) he mle b mnmzes h () = has the soluton h() = n = ( a ) : b = a a : Also, h ( ) b > : Hence b s the mle of. () E b = for all ; hence b s unbased: = aa a () he nformaton n the sample X ; :::; X n s @ ln L() I () = E @ = a : Hence the Rao-Cramer lower bound for the varance of an unbased estmator of s (v) V b unbased. a : = ; the Rao-Cramer lower bound. Hence b s mnmum varance a
[Stat 4, chap. 8, 9] (Je Yang) Consder one random varable X that has a bnomal dstrbuton wth n = 4 and p = θ. [] Fnd the most powerful test of sze α / 6 for H : θ =.5 vs. H : θ =.75 [] Calculate the power of your test. [3] Justfy that your test s the most powerful one of sze /6. [Soluton] [] Snce X follows Bnomal (4,θ ), the lkelhood 4 4 L( θ ; ) = f ( ; θ ) = θ ( θ ), =,,,3,4 herefore, 4 4 3 L(.5; ) 6 L (.5; ) =, L(.75; ) =, = 4 4 4 L(.75; ) 3 3 4 L (.5; ) / L(.75; ) 6 6/3 6/9 6/7 6/8 Pr( X = H ) /6 4/6 6/6 4/6 /6 Pr( X = H ) /56 /56 54/56 8/56 8/56 By Neyman-Pearson theorem, the test wth rejecton regon C : = : L(.5; ) / L(.75; ) k { } s the most powerful one of sze Pr( X C H ). Let k = 6 / 8, then C = { : = 4} leads to the most powerful test of sze / 6. [] he power s Pr( X = 4 H ) = 8/ 56. [3] he test s guaranteed by the Neyman-Pearson theorem to be the most powerful test. Alternatve way to justfy that: Another test wth rejecton regon{ : = } s of sze /6 too. he correspondng power s only /56.
Soluton Statstcs 43-MS Eam Sprng Semester 9 A clent has a fnte populaton of 6 unts and has funds to survey only 3 unts for the purpose of estmatng the mean and ts related standard devaton. here are two samplng plans to be consdered by two survey practtoners. Samplng Plan by Survey Practtoner : Is a smple random samplng plan of sze 3 wthout replacement, SRS (6, 3). Recall that ths s a unform samplng plan wth all samples of sze 3 n ts support. Samplng Plan by Survey Practtoner : Is a controlled samplng plan wth only 4 samples: A- he followng s samples are ecluded from beng surveyed, that s they have zero probablty of selecton. {,,3 }, {,4,5}, {,5,6}, {,3,5}, {,4,6}, {3,4,6}. B- he followng s samples are assgned probablty of / of selecton each: {,,5}, {,3,5}, {,4,6}, {,3,4}, {,3,6}, {4,5,6}. C- he remanng 8 samples are assgned probablty of / of selecton each. Remark: Note that the total probablty over the chosen 4 samples s and the survey s not a unform samplng plan snce s of the samples are assgned twce as much probablty as the remanng 8 samples. Suppose both practtoners mplemented ther survey plans and the sample {,, 5} s selected by both of them and the related survey data are: Y =, Y = 96, and Y 5 = 3. - What are the H estmators and ther values of the populaton mean under both samplng plans? Answer: We observe that the frst ncluson probabltes for both samplng desgn s constant (3/6) for all 6 unts. hus, the H estmator of the populaton mean under both samplng plans s smply the sample mean: (+96+3)/3 = 3. - What are the standard devatons of your estmator under both samplng plans? Answer: We observe also that the second order ncluson probabltes for both samplng plans are constant 3(3-)/6(6-) = 5 for all 5 dstnct pars. hus, thus desgn unbased estmator of the varance of H s smply [(sample varance)/3] [- 3/6)]. Compute the sample varance of the three observatons and plug t n the epresson and take ts postve square root.
3- If you were the consultng statstcan, whch of the above two samplng plans would you recommend to your clent and why? Answer: If there s no need or reason to eclude those samples whch survey desgn has ecluded from the support of the samplng plan then the frst survey plan whch wth s SRS(3, 6) s preferred due to ts statstcal optmalty (see the samplng book by Hedayat and Snha). In addton the statstcan does not have to justfy the ecluson of those samples whch survey desgn has done. However, f those samples whch have receved zero probablty of selecton under survey are to be ecluded from the survey then clearly the second survey desgn should be recommended.
Solutons to OR Problems n Masters Eam Sprng 9.E.S. Raghavan Aprl 7, 9 Soluton to Problem 7: Stat 47 From 9 feet boards we want planks 7 feet long, planks 5 feet long and 3 planks 4 feet long only usng cuttng patterns 7 feet 5 feet usng patterns 3,,. 4 feet 3 Say we need,, 3 planks cut n respectve patterns. Solvng 3 = 3 and roundng to the net hghest nteger, we get = 6, = 5, 3 = 9. Snce the fractonal sum s tself > 7 we want to see whether we can manage wth 8 planks. o check ths we have to fnd strct mprovement and ths must correspond to a new cuttng pattern brought n and one of the above pattern elmnated. By revsed smple method we solve for ub = c B where u s the row vector and c B s the row vector [,, ]. We get u = [u, u, u 3 ] = [ 4, u 7 =, ] 7 7. We now look for a cuttng pattern a a a 3 3 Department of Mathematcs, Statstcs and Computer Scence, 85, South Morgan Street #57, Unversty of Illnos at Chcago, Chcago, IL 667-745 Emal: ter@uc.edu
such that 4 7 a + 7 a + 7 a 3 > 7a + 5a + 4a 3 9 hat s 4a + a + a 3 > 7 7a + 5a + 4a 3 9 he knapsack algorthm gves weghts for a, a, a 3 as 4 7, 5, 4 respectvely. hus choosng a =, a =, a 3 = gves the new cuttng pattern. Usng revsed smple method we determne whch pattern of the current has to be dropped out. o do ths we need to solve Bd = a where we solve 3 d d d 3 =. We get d = 4 7, d = 5 7, d 3 = 7. he new soluton clearly has to scrap the last cuttng pattern as d 3 alone s postve. hus the new cuttng patterns are 7 feet 5 feet 4 feet usng patterns 3, 3,. he new cuts to the nearest nteger not below the fractonal soluton s = 6, = 9, 3 = 3 ths meets the demands and we waste (9)(8) ()(7) ()(5) (3)(4) = 3. Prevously we wasted (9)(9) ()(7) ()(5) (3)(4) = 49 feet
Soluton to Problem 8, stat 47 he LP problem has as ts dual ma 9 + 4 + 7 3 such that + + 3 3 6 5 + 4 + 3 + + 3 5,, 3 unrestrcted mn 6y + y + 5y 3 such that y + 5y + y 3 = 9 y + 4y + y 3 = 4 3y + y + y 3 = 7 y, y, y 3 has feasble soluton y =, y =, y 3 = 4. Snce = 5 6, = 5, 3 = 7 6 s feasble for prmal and gves the value of the objectve functon as 44 and so does the dual soluton the two are optmal for the two problems by the dualty theorem. Soluton to Problem 9, Stat 473 Gven the payoff functon K(, y) = y y, y the epected payoff to player I usng whle player II chooses a y at random usng densty f(y), y s gven by K(, y)dy = y = [ y y y = ] + [ y y ( y)dy + ] [ (y )dy y y + y3 3 ] ( y + y )dy = + + [ + 3 ] = 6 By the symmetry of the payoff functon, the same unform strategy s optmal for player I gvng an epectaton 6 for all y. hus the value of the game s 6. 3
SA 48 -Sprng 9 (Jng Wang) A dary company would lke to nvestgate f the mlk s contamnated. In order to nvestgate a possble shpment (batch) effect, the company selects 5 shpments at random. After processng each batch, 6 cartons of mlk are selected at random and are stored for several days. hen the square root of the bactera counts are recorded and denoted by Y j, where =,..., 5 and j =,..., 6. (a). If the shpment effect s denoted as τ, wrte down ANOVA model and specfy dstrbuton of random components n the model. Please state the hypotheses. Soluton: he shpment (batch) effect s a random effect. Hence a random-effect ANOVA model can be wrtten as Y j = µ + τ + ε j, where =,..., 5, j =,..., 6. In addton, τ and ε j are ndependently dstrbuted wth respectvely normal dstrbutons, τ N (, σ τ ) and εj N (, σ ) for any and j. o nvestgate f the batch effect s sgnfcant s equvalent to test H : σ τ = aganst H : σ τ >. (b). Complete the followng table and conclude based on. sgnfcance level. Soluton: Sources df SS M S F Shpment 4 83..8 9. Error 5 557..3 otal 9 36. Snce the observed statstc F o = 9. > F (., 4, 5) = 4.77, the test s sgnfcant at level α =.,.e. there s consderable varaton among the shpments. (c). Estmate the varance component(s). Soluton: he error varance can be estmated by the MSE drectly, ˆσ = MS Error =.3. he varance component for the random effect s estmated as follows (sample sze for each batch s the same) ˆσ τ = n (MS reatment MS Eroor ) = (.8.3) = 9.75. 6
. In lnear regresson analyss, leave-one-out cross valdaton (LOOCV) procedure s often used to estmate the predcton accuracy of the ftted regresson model. Let ( X, Y) = (, y ); =,..., n be a dataset wth R + ( X, Y ) ( ) ( ) p = (,,..., p) and y R, and let be the modfed dataset wthout the -th row. A set of predcton values y( ) = ( X( ) X( ) ) X Y ( ) ( ) are obtaned and the LOOCV estmate s defned as n n ( ) ( ). = = LOOCV = n e = n ( y y ) ( X X) ( X X) (a) Show that( X( ) X( ) ) = ( X X) + ( X X). A zz A Hnt: ( A zz ) = A + assumng that z A z A z z s nvertble. e (b) Show that e( ) = ; =,..., n, where H = X( X X) X, h s the -th dagonal h element of H and e= ( e,..., e ) = ( I H) Y. [Soluton] n (a) he equalty follows mmedately after the fact that X ( ) X( ) = X X formula n the Hnt. and the (b) Note that X ( ) Y( ) = X Y y and h = ( X X ) ( ) ( ), we thus have ( ) = ( ) = ( ( ) ( ) ) ( ) ( ) e y y y X X X Y = y X X X Y y ( ) ( ) ( X X) ( X X) = y ( X X) + ( X Y y) h = y X X X Y X X ( X X) y + ( X X) ( X X) X Y ( ) ( ) h h h = y β hy + β y h h e = y β =. h h h where ( β = X X) X Y. ( X X) h y