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I have been running the weight decay examples Q2-6, but haven't seen any real improvement in the out-of-sample error compared to no regularization at all. Is that just a feature of this particular problem, or should I recheck my calculations?
Unfortunately (or not), the answers I've been getting do appear as options on the multiple choices. |
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#4
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And yes, I have been using classification error, but that is a good point - I started using the regression residuals and such, but that mistake at least I caught. |
#5
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As I suspected, all my answers on these were wrong. Does anyone have code (R if possible) to show, that I could use for comparison? I'm suspecting my problem was something dumb; even the original linear regression was wrong, and I compared that one with the same answer from the R lm() function.
I'm especially concerned since HW 7 uses all the same data again - so I really need to track this down. |
#6
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#7
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Michael:
Here you go: Code:
#READ IN THE FILES. datos1 <- read.table("in.dta") names(datos1) <- c("X1","X2","Y") datos2 <- read.table("out.dta") names(datos2) <- c("X1","X2","Y") #FOR THE FOLLOWING QUESTIONS, SET UP THE MATRIXES Z <- with(datos1, cbind(rep(1,nrow(datos1)),X1,X2, X1^2,X2^2,X1*X2,abs(X1-X2),abs(X1+X2)) ) Z <- as.matrix(Z) Zout <- with(datos2, cbind(rep(1,nrow(datos2)),X1,X2, X1^2,X2^2,X1*X2,abs(X1-X2),abs(X1+X2)) ) Zout <- as.matrix(Zout) #NOW FIT WITH WEIGHT DECAY USING LAMBDA=10^-3 lambda <- 10^(-3) M <- t(Z)%*%Z + diag(rep(8,1))*lambda w <- solve(M)%*%t(Z)%*%datos1$Y Ym <- as.numeric(sign(Z%*%w)) Ein <- mean(datos1$Y!=Ym) Ym <- as.numeric(sign(Zout%*%w)) Eout <- mean(datos2$Y!=Ym) |
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