4.5 Generalized likelihood ratio test
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1 4.5 Geeralized likelihood ratio test A assumptio that is used i the Athlete Biological Passport is that haemoglobi varies equally i all athletes. We wish to test this assumptio o a sample of k athletes. Let the values of i-th athlete be ormally distributed (i,..., k, i.e. X ij N(µ i, σ 2 i, where j,..., i deote the idividual s measuremets. Assume that all measuremets are idepedet. Write the ull ad alterative hypothesis Null hypothesis: H 0 : σ 2 σ σ 2 k Alterative hypothesis: H : σ 2 i are ot all equal First say that we have oly oe athlete with measuremets. How would we estimate his parameters µ ad σ 2 with the method of maximum likelihood? The likelihood fuctio equals L(x, µ, σ 2πσ exp (x j µ 2 2σ 2, the part of its logarithm that cotais the parameters to be estimated equals log L(x, µ, σ log σ 2σ 2 (x j µ 2.
2 We fid the maximum with respect to µ: 2 σ 2 log L(x, µ, σ µ 0 (x j µ( 2 0 (x j µ 0 µ x j The variace: ṋ σ 2 log L(x, µ, σ σ 0 (x j µ 2 2 σ 3 0 σ 2 + (x j µ 2 0 σ 2 (x j µ 2 Show that uder the alterative hypothesis (for k athletes, the parameter estimates equal ˆµ i xij i ˆσ i 2 i (x ij ˆµ i 2 i The likelihood fuctio uder the alterative hypothesis equals L(x, µ, σ i 2πσi exp (x ij µ i 2 2σ 2 i 2
3 After takig the logarithm of the fuctio, we are left with a sum of terms, each of them cotaiig parameters of oe idividual oly. If iterested i idividual i, we thus do ot eed ay iformatio about the other idividuals ad the result is equal to cosiderig each idividual separately. How do we estimate meas uder the ull hypothesis? Uder the ull hypothesis, σ i equals for all i, ad thus does ot affect our estimate of the idividual averages. The estimates of the averages are thus equal to those uder the alterative hypothesis. What is the variace estimate uder the ull hypothesis? The part of the logarithm of the likelihood fuctio that we are iterested i equals log L(x, µ, σ i log σ 2σ 2 i (x ij µ i 2. After equallig the derivative with respect to σ to 0, we get ˆσ 2 0 i i (x ij ˆµ i 2 How would you test the ull hypothesis with the geeralized likelihood ratio test? Wilks Λ equals (the maximum of the likelihood fuctio uder the alterative hypothesis i the umerator, the maximum uder the ull 3
4 i the deomiator: Λ i ( ( ( 2π σi exp{ (x ij µ i 2 2 σ 2 i i ( 2π σ0 exp{ (x ij µ 0i 2 i i 2π σi 2 σ 2 0 exp{ 2π σ0 exp{ k i (x ij µ i 2 2 σ i 2 i (x ij µ 0i 2 2 σ 0 2 We isert the estimates of the variace i the expoet ad get exp{ 2 i the umerator as well as the deomiator. The two terms cacel out, leavig the logarithm of Λ equal to ( ( i i log Λ log( σ i + log( σ 0 ( ( i log( σ 0 i log( σ i i [log( σ 0 log( σ i ] The 2 log Λ is distributed as χ 2 k, sice we estimate k more parameters uder the alterative hypothesis. i 5 Liear regressio 5. Liear regressio We are iterested i the associatio betwee the umber of hours of learig per week ad the score at the statistics exam. Say we kow that the exam result is distributed coditioally ormal: Y X N(β 0 + β X, σ 2. 4
5 Read the parameter estimates from the output below. Iterpret the results - which ull hypotheses are tested ad how? Call: lm(formula y ~ x Residuals: Mi Q Media 3Q Max Coefficiets: Estimate Std. Error t value Pr(> t (Itercept * x *** --- Sigif. codes: 0 *** 0.00 ** 0.0 * Residual stadard error:.44 o 8 degrees of freedom Multiple R-squared: 0.845, Adjusted R-squared: F-statistic: 35.3 o ad 8 DF, p-value: The parameter estimates equal β 0 9.2, β 3.7, σ.4. We ve tested two ull hypotheses: H 0it : β 0 0 i H 0 : β 0. We are usually oly iterested i the secod oe sice it cosiders the associatio betwee the variables i the populatio. Sice the method of maximum likelihood was used to estimate the parameters, we kow that the parameter estimates are approximately ormally distributed aroud the true values (for a sufficietly large. The stadard error is estimated based o the data, hece we use the t test: T β 3, 7 SE β 0, 6 5, 9 We kow that the radom variable T is distributed approximately as t with 8 degrees of freedom (we spet 2 df with the estimatio of SE. This is amed the Wald test. How would the ull hypothesis H 0 : β 0 be tested with the geeralized likelihood ratio test? Hit: Wheever possible, use the results of the previous exercise Cosider first the estimates uder the ull hypothesis. Uder the ull hypothesis, the average equals for all idividuals (does ot deped o 5
6 X, so the results of the previous exercise ca be directly used by writig β 0 istead of µ. The maximum uder the ull hypothesis thus equals { L 0 (y, x, β 0, σ ( 2π σ exp i β σ 2 { ( 2π σ exp i β (y i β 0 β x i 2 { ( 2π σ exp 2 Uder the alterative hypothesis, previous results ca be used for the estimatio of σ σ 2 (y i β 0 x i β 2 The likelihood fuctio equals: L(y, x, β 0, β, σ its maximum is L A (y, x, β 0, β, σ ( 2πσ exp{ (y i β 0 β x i 2 2σ 2 { ( 2π σ exp (y i β 0 β x i 2 2 σ 2 { ( 2π σ exp { ( 2π σ exp 2 (y i β 0 β x i 2 2 (y i β 0 β x i 2 6
7 Wilks Λ equals Λ L A L 0 ( exp { 2π σ A 2 ( exp { 2π σ 0 2 σ 0 σ A (y i β 00 2 (y i β 0A x i βa 2 The value of the maximum uder the alterative is calculated by isertig the estimated values of β 0 ad β, to calculate the maximum uder the ull, the β 0 i the ull model must be estimated. The resultig value of Λ is distributed as χ 2 uder the ull hypothesis. Uderstadig the ideas i R: Let X be a uiformly distributed variable (betwee 0 ad 20, rouded dow, β 0 5, β 4, σ 0. Geerate a sample of size 0, plot the data usig a scatterplot ad add the estimated ad the populatio lie. > set.seed( > <- 0 #sample size > beta0 <- 5 > beta <- 4 > sigma <- 0 > x <- floor(ruif(*20 #values of x (rouded dow > x <- sort(x #sort the values of x > y <- rorm(,meabeta0+beta*x,sdsigma #geerirate from ormal distr. > plot(x,y #scatterplot > popul <- beta0 + beta*x #populatio value of the lie > lies(x,popul,col"grey",lwd2 #add to the plot (grey > fit <- lm(y~x #estimated lie > summary(fit #look at the estimates > beta0h <- fit$coef[] #estimated beta0 > betah <- fit$coef[2] #estimated beta > predictio <- beta0h + betah*x > lies(x,predictio,lwd2 #add the estimated lie Fid the result of the geeralized likelihood ratio test for the simulated example with R 7
8 > fit0 <- lm(y~ #estimate the lie uder the ull > res0 <- y - fit0$coef #residuals uder the ull > resa <- y - beta0h - betah*x #residuals uder the alterative > logl0 <- -.5**log(sum(res0^2 #loglik (without cost. terms uder the ull > logla <- -.5**log(sum(resA^2 #loglik (without cost. terms uder the alter. > Lambda <- 2*(loglA-logl0 #Wilks lambda > -pchisq(lambda, #likelihood ratio test [] 4.048e-05 8
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