kubostat2018d p.2 :? bod size x and fertilization f change seed number? : a statistical model for this example? i response variable seed number : { i
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- かずゆき ながおか
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1 kubostat2018d p.1 I 2018 (d) model selection and kubo@ees.hokudai.ac.jp : : :? AIC : deviance model selection misunderstanding kubostat2018d ( (d) / 44 kubostat2018d ( (d) / 44 number of parameters? 4 GLM 5 GLM : : seed number (A) k = 1 (B) k = 7 Too few parametes? bod size x Too man parameters? bod size x How man parameters do ou need for the best prediction? kubostat2018d ( (d) / 44 kubostat2018d ( (d) / 44 :? :? k? 1. :? seed number (A) k = 1 (B) k = 7 Too few parametes? Too man parameters? bod size x bod size x? number of parameters k? kubostat2018d ( (d) / 44 kubostat2018d ( (d) / 44
2 kubostat2018d p.2 :? bod size x and fertilization f change seed number? : a statistical model for this example? i response variable seed number : { i } explanator variable : bod size {x i } fertilization {f i } sample size f i C: T: control (f i = C): 50 sample (i {1, 2, 50}) fertilization (f i = T): 50 sample (i {51, 52, 100}) i x i probabilit distribution Poisson distribution : linear predictor : β 1 + β 2 x i + β 3 f i link function : log link function kubostat2018d ( (d) / 44 kubostat2018d ( (d) / 44 :? 4 candidate models 4 : (A) constant λ :? 4 candidate models 4 : (B) f model λ i = exp(β 1 ) λ i = exp(β 1 + β 3 f i ) (log likelihood) (log likelihood) > loglik(glm( ~ 1, data = d, famil = poisson)) log Lik (df=1) > loglik(glm( ~ f, data = d, famil = poisson)) log Lik (df=2) kubostat2018d ( (d) / 44 kubostat2018d ( (d) / 44 :? 4 candidate models 4 : (C) x model :? 4 candidate models 4 : (D) x + f model λ i = exp(β 1 + β 2 x i ) λ i = exp(β 1 + β 2 x i + β 3 f i ) (log likelihood) (log likelihood) > loglik(glm( ~ x, data = d, famil = poisson)) log Lik (df=2) > loglik(glm( ~ x + f, data = d, famil = poisson)) log Lik (df=3) kubostat2018d ( (d) / 44 kubostat2018d ( (d) / 44
3 kubostat2018d p.3 :? k increases log L increases AIC : deviance (A) constant λ k = (C) x model k = (B) f model k = fertilization Control (D) x + f model k = Control fertilization 2. AIC : deviance badness of prediction : AIC kubostat2018d ( (d) / 44 kubostat2018d ( (d) / 44 AIC output R glm() deviance : deviance AIC deviance D = 2 log L : deviance > glm( ~ x + f, data = d, famil = poisson) Call: glm(formula = ~ x + f, famil = poisson, data = d) Maximum log likelihood log L : goodness of fit Deviance D = 2 log L : Coefficients: (Intercept) x ft Degrees of Freedom: 99 Total (i.e. Null); Null Deviance: 89.5 Residual Deviance: 84.8 AIC: Residual model k log L Deviance Residual 2 log L deviance constant λ f x x + f saturation Residual Deviance? Null Deviance? AIC? kubostat2018d ( (d) / 44 kubostat2018d ( (d) / 44 AIC : deviance Null deviance, Residual deviance,... AIC : deviance badness of prediction : AIC = 2 log L + 2k Max deviance constant λ x model Look for a model of the smallest AIC AIC Deviance 2 log L () 89.5 (Null Deviance) 85.0 (Residual Deviance) model k log L Deviance Residual 2 log L deviance AIC constant λ f x x + f saturation Min deviance saturation model AIC: A (or Akaike) information criterion kubostat2018d ( (d) / 44 kubostat2018d ( (d) / 44
4 AIC : deviance (estimation)? constant λ ˆβ 1 = 2.04 β 1 = 2.08 parameter estimation kubostat2018d ( (d) / 44 AIC : deviance Is it OK? Goodness of fit is evaluated b using the SAME data set...? constant λ ˆβ 1 = 2.04 log L () biased goodness of fit! kubostat2018d ( (d) / 44 AIC : deviance : constant λ ˆβ 1 = 2.04 E(log L) β 1 = 2.08 ( ) (200 ) kubostat2018d ( (d) / 44 AIC : deviance log likelihood β 1 β 1 β 1 (200 ) ( ) (A) (B) (A) (C) log L = E(log L) = ˆβ 1 = 2.04 β 1 = 2.08 kubostat2018d ( (d) / 44 AIC : deviance kubostat2018d ( (d) / likelihood ratio test kubostat2018d ( (d) / 44 kubostat2018d p.4
5 kubostat2018d p.5 Although their procedures are similar... the are total different! AIC ( ) () AIC model selection totall different in their objectives kubostat2018d ( (d) / 44 kubostat2018d ( (d) / 44 Objective? model selection : Look for a model of better prediction : rejection of null hpothesis kubostat2018d ( (d) / 44 統計学的な検定 (NemanPearson framework) statistical test Null Alternative hpothesis hpothesis 帰無仮説対立仮説 glm( ~ 1) どうでもいい 興味ない VS glm( ~ x) 重要! これを主張したい! 非対称性 asmmetricit? kubostat2018d ( (d) / 44 統計学的な検定 (NemanPearson framework) statistical test Null Alternative hpothesis hpothesis 帰無仮説対立仮説 (if...) glm( ~ 1) test! reject 棄却 VS glm( ~ x) support 支持 非対称性 asmmetricit? kubostat2018d ( (d) / 44 統計学的な検定 (NemanPearson framework) statistical test (if...) Null hpothesis 帰無仮説 glm( ~ 1) test! NOT reject VS Alternative hpothesis 対立仮説 glm( ~ x) Sa Nothing!? 非対称性 asmmetricit? kubostat2018d ( (d) / 44
6 kubostat2018d p.6 The same example, again test statistics D 1,2 i i seed number i D: deviance x i bod size x i neglect fertilization treatment (!) x model D 2 = constant λ D 1 = difference in deviance D 1,2 = D 1 D 2 = likelihood ratio? log L 1 L = log L 1 log L 2 2 model k log L Deviance 2 log L constant λ D 1 = x D 2 = null hpothesis alternative hpothesis asmmetricit in test Null hpothesis is junk :... et we are focousing onl on null hpothesis kubostat2018d ( (d) / 44 kubostat2018d ( (d) / 44 How to make null model objective : null hpothesis rejection Null hpothesis is included in Alt hpothesis this is a nested model ( ) observerd D 1,2 = 4.5 () ( ) () () { i } λ i alternative hpothesis : log λ i = β 1 + β 2 x i null hpothesis : log λ i = β 1 ( ) significant not significant is... (Reject ) (Not reject ) TRUE Tpe I error (no problem) NOT true (no problem) Tpe II error asmmetricit in test evaluating onl TpeI error : kubostat2018d ( (d) / 44 kubostat2018d ( (d) / 44 generate D 1,2 distribution D 1,2 : Suppose null hpothesis is TRUE! bootstrap likelihood test How to generate D 1,2 under is TRUE?! constant λ x model D 1,2 ( ˆβ 1 = 2.06 ) D 1,2 D 1,2 D 1,2 > d$.rnd < rpois(100, lambda = mean(d$)) > fit1 < glm(.rnd ~ 1, data = d, famil = poisson) > fit2 < glm(.rnd ~ x, data = d, famil = poisson) > fit1$deviance fit2$deviance generation of random numbers virtual data rpois() ( ) fitting GLM to the virtual data glm() kubostat2018d ( (d) / 44 kubostat2018d ( (d) / 44
7 kubostat2018d p.7 You must define rejection region in advance sa, 5%? 5%? NOT significant significant (5%) A random D 1,2 generator in R get.dd < function(d) # { n.sample < nrow(d) #.mean < mean(d$) # d$.rnd < rpois(n.sample, lambda =.mean) fit1 < glm(.rnd ~ 1, data = d, famil = poisson) fit2 < glm(.rnd ~ x, data = d, famil = poisson) fit1$deviance fit2$deviance # } pb < function(d, n.bootstrap) { replicate(n.bootstrap, get.dd(d)) } kubostat2018d ( (d) / 44 kubostat2018d ( (d) / 44 Generated distribution of D 1,2 = D 1 D 2 Probabilit{ D 1,2 4.5} = = observed D 1,2 D 1,2 = 4.5 constant λ x model D 1,2 (R code is in the next page) > source("pb.r") # reading "pb.r" text file > dd12 < pb(d, n.bootstrap = 1000) > hist(dd12, 100) # to plot histogram > abline(v = 4.5, lt = 2) > sum(dd12 >= 4.5) [1] 38 socalled P value is kubostat2018d ( (d) / 44 kubostat2018d ( (d) / 44 In this case, null hpothesis is rejected In case that P > ? So we can state that alternative hpothesis x model is better than constant λ. i i seed number i can be accepted. D: deviance x i bod size x i x model D 2 = constant λ D 1 = You can conclude NOTHING! You can NOT state that constant λ (Null hpothesis) is better λ asmmetricit in stattest : Null hpothesis is never accepted kubostat2018d ( (d) / 44 kubostat2018d ( (d) / 44
8 kubostat2018d p.8 model selection misunderstanding model selection misunderstanding 4. model selection misunderstanding FAQ kubostat2018d ( (d) / 44 kubostat2018d ( (d) / 44
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