[問題] 使用STEP尋找最小BIC

看板R_Language作者killer01 (killer01)時間6年前 (2018/04/22 12:40)推噓2(2推 0噓 20→)

留言22則, 2人參與討論串1/1

我是R新手, 第一次使用這個語言. 使用STEP function 尋找最小的BIC, 在建立模型時我新增了一些 interaction term 到模型中, context1$abilsq <- (context1$abil)^2 context1$educsq <- (context1$educ)^2 context1$expersq <- (context1$exper)^2 context1$abileduc <- context1$abil*context1$educ context1$abilexper<- context1$abil*context1$exper context1$educexper<- context1$educ*context1$exper model2A <- lm(log(wage)~abil+educ+exper+abilsq+educsq +expersq+abileduc+abilexper+educexper , data = context1) model2B <- lm(log(wage)~abil+educ+exper+abilsq+educsq +expersq+abil*educ+abil*exper+educ*exper , data = context1) 兩個模型主要的差別在於 interaction term 有無使用* model2minBIC <- step(model2A, direction = "backward", k = log(nrow(context1))) model2minBIC <- step(model2B, direction = "backward", k = log(nrow(context1))) 執行結果似乎 model2B 可以跑出比較小的值, 且兩者跑出來的結果也不一樣. 下面是model2B的結果. Step: AIC=-1545.67 log(wage) ~ exper + abileduc + educexper Df Sum of Sq RSS AIC <none> 342.06 -1545.7 - abileduc 1 13.185 355.25 -1506.3 - exper 1 17.853 359.91 -1490.2 - educexper 1 27.377 369.44 -1458.1 下面是model2A的結果 Step: AIC=-1539.35 log(wage) ~ abil + educ + exper + educ:exper Df Sum of Sq RSS AIC <none> 341.84 -1539.3 - educ:exper 1 2.8843 344.72 -1536.1 - abil 1 10.3991 352.24 -1509.6 想問是什麼原因造成兩個跑出來的結果會不一樣... 謝謝! [關鍵字]: BIC, step -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 97.99.89.205 ※ 文章網址: https://www.ptt.cc/bbs/R_Language/M.1524372044.A.653.html

推