Econ 413 Exam 13 H ANSWERS Settet er nndelt 9 deloppgaver, A,B,C, som alle anbefales å telle lkt for å gøre det ltt lettere å stå. Svar er gtt <<.. >>. Unfortunately, there s a prntng error n the hnt of problem / secton B(). Ths secton should, therefore, weght less n the evaluaton. Problem 1 A. Suppose the random varable (rv) U s posson dstrbuted wth expected value (wrtten n short U ~ pos( ) ). Then the probablty mass functon (pmf) for U s gven by x f ( x) P( U x) e for x,1,, x! and the moment generatng functon (mgf) by 1 M () t e defned for any real t. e t Use ths to show that E( U) var( U). << Answer: We fnd the dervatves t e 1 M '( t) e e t t t t e 1 e 1 e 1 t t t t t M ''( t) e e e e e e e 1 e Hence, E( U) M '() and E U M ''(), gvng var( U) E U E( U) >>
B. ) Suppose Z ~ pos( ) where s a parameter. Put Z 1 and show that the pmf of s gven by x 1 f ( x) P( x) e for x 1,,3, ( x 1)! ) In addton to n secton B we ntroduce the rv Y such that the condtonal dstrbuton of Y gven a fxed value x of s posson wth expectaton x where s a new parameter (n short, ( Y x) ~ pos( x) ). Explan why the ont pmf of ( Y, ) can be wrtten (1) y x f ( x, y ;, ) ( x 1)! y! x1 y e x otherwse for x1,, and y,1,, << Answer: ) x 1 Z 1 P( x) P( Z x 1) e ( x 1)! for x 1,, ) follows drectly from the factorzaton, y ( x) x f ( x, y) f ( x) f ( y x) f ( x) e...etc. >> y! C. Usng (e.g.) the law of total expectaton, show that ) EY ( ) ( 1) ) ) var( ) ( 1) Y Y E [Hnt. Remember that f g(, Y ) s any transformaton Y, then Eg(, Y) x Eg( x, Y) x of (, ) where x can be consdered as a constant.] << Answer: E( Y) E E( Y ) E E( Z 1) ( 1) )
3 ) var( Y) E var( Y ) var E( Y ) E ( ) var ( ) E( Z 1) var( Z 1) ( 1) Y Y 1 1 E x E x E Y x x x x x, mplyng Y Y E E E E ) >> D. Now let denote the total number of brths by mothers n age group 3-39 observed n a country durng one year. Let Y be the number of chldren wth Downs syndrome (whch we wll call DS n short) among these. Assume that the ont dstrbuton of ( Y, ) s as descrbed n secton B wth ont pmf n (1) (.e., consderng the probablty of the absurd event that Y should be larger than as neglgble even f t s, n fact, postve accordng to the model). We are nterested n the yearly rate,, of DS brths defned as the expected number of DS per 1 brths. ) Explan why 1. ) Let ( 1, Y1 ),(, Y ),,( n, Y n) be d random pars representng observatons of ( Y, ) sampled over n 1 years. The common pmf s gven n (1). Show that the maxmum lkelhood estmators (mle) of,, are gven by ˆ 1, ˆ Y 1 n, ˆ 1 ˆ n 1 where denotes the sample mean, n. ) Calculate the mle estmates of,, usng the data n table 1. Table 1 Brths wth Down s syndrome (DS), over 1 years. Mothers age 3-39. (Data smulated from nformaton n problem ) Year 1 3 4 5 6 7 8 9 1 Total No. of brths, x 73 1 14 5 1 7 935 9 37 6 98 1 774 4 53 39 Brths wth DS, y 9 1 13 11 1 17 11 15 1 7 16
4 << Answer: ) Y gves the number DS per brth, and 1 Y the number of DS per 1 ) brths for an observaton of (, ). The ont pmf s n Y Hence the rate, E Y x y 1 1. n y x 1 ( 1) y x y x n x e f ( x1, y1,, xn, yn) e 1 ( x 1)! y! 1 ( x 1)! y! gvng log lkelhood, l(, ) c (ln ) ( x 1) (ln ) y n x where c does not depend on the parameters. y l x gves mle, ˆ Y ( x 1) l 1 n gves mle, ˆ ( 1) 1 n Becausee of the substtuton nvarance property of mle s, we get the mle of smply as ˆ 1 ˆ. ) From table 1 we get ˆ obs 39 1 1 389, ˆ obs 16 39.53, and ˆ 1 ˆ obs obs.53 x E. Show that the moment method estmators (mme) for and, based on the two frst order moments, E( ) and E( Y ) are the same as the maxmum lkelhood estmators n secton D. <<< Answer:
5 E( ) 1 and E( Y) ( 1) gve the mme equatons 1 and Y ( 1), gvng the mme s: Y Y 1, whch are equal to the mle s, ˆ and ˆ n D. F. ) Show 1 that the Fsher nformaton matrx for one observaton (par) s gven by 1 I(, ) 1 ) Develop an approxmate 95% confdence nterval (CI) for the rate,, of DS per 1 brths, based on the fact that, () n D ˆ N, n 1 whch follows from general mle theory. You do not need to ustfy () here, but explan how Slutsky s lemma s relevant for the constructon of the CI. ) Calculate the observed value of the CI n ) based on the data. <<< Answer: ): Usng the ont pmf n (1), we get ln f const. ( x 1)ln yln x, gvng ln f x 1 ln f y 1, x, and 1 Remember that mle theory for d vectors (e.g., pars) works n the same way as for one-dmensonal observatons, gvng consstent and asymptotcally normally dstrbuted estmators. In partcular, f (,, ) are the parameters, the (, ) -th element of the Fsher nformaton matrx for one 1 ln f observaton (vector) s, as before, m ( ) E, where f s the common pmf (or pdf) of the vectors, wth correspondng random varables substtuted for the arguments n f.
6 ln f ln f x 1 ln f y,,, and the elements n the Fsher matrx, 1 11 1 Y 1 m11 E, m 1 m1, m E n ): () mples that ( 1) D ˆ N(, 1). Then, ˆ, ˆ beng consstent, Slutsky s lemma mples n( ˆ 1) ( ˆ 1) ( 1) ˆ n D N(, 1) ˆ ˆ 1 x1 snce, the functon g( x, y) beng contnuous, the expresson n the square y bracket must converge to 1 n probablty. Usng ths we obtan an (for large n) 95% CI for ˆ ˆ 1.96 n( ˆ 1) Snce the transformaton, 1, s ncreasng, the nterval for, gven by ˆ 1 ˆ 1.96 n( ˆ 1) wll have the same degree of confdence as the nterval for. ): The observed nterval for becomes 1.57 1.96.47.57.9.435,.619 Problem
7 Introducton. Table shows the number of mothers of chldren wth Down s syndrome (DS) for brths n Australa from 194 195, grouped accordng to the age of the mother at the brth. Table (,) Age group (,) Total number of brths ( x ) Number of mothers wth DS chld ( y ) Rate of DS per 1 brths (1,1) < 35 555 15.4 (,1) -4 7 931 18.6 (3,1) 5-9 53 45 8.8 (1,) 3-34 17 97 194 1.13 (,) 35-39 86 46 97 3.45 (1,3) 4-44 4 498 4 9.8 (,3) >44 1 77 37 1.68 Total 78 157 1 119 We combne the 7 age groups nto three larger groups ndcated by the second ndex,, for age group (, ). Addng up we get the data for the larger age groups n table 3. Table 3 Data for the larger age groups ( k ndcates how many age groups from table age group conssts of.) k Larger age group Total number of brths ( x ) Number of mothers wth DS chld ( y ) Rates of DS per 1 brths 1 3 < 3 496 936 351.71 3-39 57 16 491 1.91 3 4 6 5 77 1.57 Total 78 157 1 119 The problem s to compare the expected rate of DS for dfferent age groups. As part of our model we assume that the expected rate of DS per brth, denoted by the parameter s constant wthn the larger age group, for 1,,3. On the other hand we assume, that the expected number of brths, denoted by 1 where s a parameter, may vary between the 7 smaller age groups n table.
8 Let for age group (, ) n table and Y denote the total number of brths and the number of DS brths respectvely. We assume that the seven random pars, (, Y ), 1,, k, 1,,3, are ndependent and such that the ont pmf of (, Y ) s gven by (1) n Problem 1. (3) (, Y ) ~ f ( x, y ;, ) 1,, k, 1,,3 Our frst task s to test the null hypothess H aganst H : at least two 1 are dfferent. (4) : 1 3 The log lkelhood functon under the full model (3) s (you do not need to ustfy ths): (5), 3 k l( ) ln f ( x, y ;, ) c ( x 1)ln y ln x 1 1 where c does not depend on the parameters. Questons A. Assume H to be true and let denote the common value of, 1,,3. Show that the mle of s ˆ Y where Y 3 k Y and smlarly for 1 1 <<< Answer: Puttng n (4), the log lkelhood gets 3 k 3 l( ) c ( x 1) ln y ln x c ( x 1) ln y ln x H 1 1 1 1 k l( H ) y y Hence, x for ˆobs, gvng the mle, ˆ Y. x
9 B. ) The mle of under the full model (3) s ˆ Y where and smlarly for statstc s where L, L Y k Y (you do not need to verfy ths). The lkelhood rato test L W ln ln L ln L L are the log lkelhood functons wth the mle estmators substtuted for the parameters under H and the full model respectvely. Show that 3 ˆ W Y ln ˆ 1 [Hnt: Notce that a large part of the expresson wll be the same n both ln L and ln L, and wll, therefore, dsappear when you take the dfference. Notce also that x ˆ y y x ln ˆ.] (***** Prntng error: The last equalty n the hnt should be y x ˆ ) ) Perform the lkelhood rato test for H n (4) based on the data n table and 3. To help the calculatons, we gve the estmates ˆ, obs n table 4 below. 1 Table 4 Mle estmates under full model. Age group ˆ, obs 1 < 3.71 3-39.191 3 >= 4.157 [Hnt. You may assume that the condtons for W beng approxmately ch-square dstrbuted under H are fulflled.] <<< Answer: ): Introducng 3 k d ( x 1)ln ˆ ˆ that does not depend on the lambdas, 1 1 we get (n terms of the observatons) ln L c d y ln ˆ x ˆ and ln L c d y ln ˆ x 3 ˆ 1
1 Snce x ˆ y and x ˆ y y, ths term wll cancel, and we get 3 3 3 3 ˆ ln ln ln ˆ ln ˆ ln ˆ ln ˆ L L y y y y y ln ˆ 1 1 1 1 whch s substtuted n W after changng all observatons to correspondng rv s. ): 1119 Fndng frst ˆ obs.143, we get 78157 Agegroup Sum brths Sum DS, ˆ obs y ln ˆ ˆ obs, obs 1 < 3 496936 351.71-491.516 3-39 5716 491.191 84.191 3 >= 4 65 77.157 118.191 Total 78157 1119 9.8947 So, Wobs 9.9. The degrees of freedom (df) = no. of free parameter under full model (1) mnus no. of free parameter under H (8) = no. of restrctons under H =. The upper.5 crtcal value n s, accordng to table 3 back n Rce, 1.6. So H s reected wth lots of evdence (P-value far below.5). C. Assumng the full model n (3) to be true, there are dfferences between the s, or, n other words, there are dfferences between the rates per 1 brths (.e., the s where 1). The mle estmates of the s can be derved from the lambda estmates n table 4. We are nterested to compare the ncrease n the rate from larger age group 1 to the larger group (.e., parameter 1 1 ) wth the ncrease n the rate from group to 3 (.e., parameter 3 ). ) Calculate an approxmate 95% CI for the dfference, 1, based on the followng nformaton (that you do not need to ustfy): The parameters of nterest are now, ' ( 1,, 3) Accordng to general asymptotc theory the mle, ˆ ' ( ˆ ˆ ˆ 1,, 3), s approxmately (multvarate) normally dstrbuted, approxmately ˆ ~ N(, C)
11 wth the covarance matrx C, beng a consstent estmate, consdered to be known.14 C.743.4338 ) Use the nformaton n ) to test the hypothess, H : 1 aganst H : (usng level of sgnfcance 5%). Formulate a concluson. 1 1 <<< Answer: ): We have 1 3 1 wth mle estmate ˆ ˆ ˆ ˆ obs ( 3 1) obs 7.46. The varance of ˆ can be calculated as (1,,1) C (1,,1)', or drectly (snce the covarances are zero) as var( ˆ ) var( ˆ ) 4var( ˆ ) var( ˆ ).4345 (.659) 3 1, and standard error se( ˆ ).659 The approxmate 95% CI for becomes, ˆ 1.96.659 6.17, 8.75. ): The CI n ) shows that H : can be reected (at 5%) snce s outsde the nterval. A decson theoretcal prncple establshed n the course allows us to conclude (.e., 1 ) nstead of only 1, snce les to the left of the CI. obs