Displaying 3 results from an estimated 3 matches for "resid2".
Did you mean:
reside
2003 Feb 04
1
test for two samples
Hi R-users,
My question is more methodological one, rather than technical.
I have to samples representing residuals based on two measurements
techniques (resid1,resid2; n=69). I need to compare two samples, to
reject one technique (the worse one), and to keep the one which gave
lower residuals (better one). What to look for? What should I analyse?
Means, variance, std. deviations?
Based on preliminary EDA:
mean_resid1 < mean_resid2 #I would reject resid2...
2013 Apr 22
0
Copula fitMdvc:
...ngth of my initial values is not the same as
the parameter.
Can somebody guide me where my mistake is.
Thanks,
Elisa.
#################################
#### OLS ESTIMATION
fit1 <- lm(Y1 ~ Xi)
fit2 <- lm(Y2 ~ Xi)
fit3 <- lm(Y3~ Xi)
resid1 <- Y1-as.matrix(cbind( 1,Xi))%*%fit1$coef
resid2 <- Y2-as.matrix(cbind( 1,Xi))%*%fit2$coef
resid3 <- Y3-as.matrix(cbind( 1,Xi))%*%fit3$coef
sigma1<-sum(resid1*resid1)/nrow(Xi)
sigma2<-sum(resid2*resid2)/nrow(Xi)
sigma3<-sum(resid3*resid3)/nrow(Xi)
rho12<- cor(resid1, resid2)
rho13<- cor(resid1, resid3)
rho23<- cor(resid2,...
2011 Mar 15
1
Problem with nls.lm function of minpack.lm package.
...to the R-help mailing list; many people that read this list are
interested in nonlinear regression problems, and will be eager to offer
their insights.
Regarding combination of two residual functions: using finite difference
derivatives, it is straightforward. You define a new residual function,
resid2, that evaluates both old residual functions, and returns their
scaled sums. So if you had
foo1(par1, ...)
and foo2(par2, ...)
you define
resid2 <- function(c(par1,par2), ...) {
sse1 <- foo1(par1, ...)
sse2 <- foo2(par2, ...)
return(sse1 * w1 + sse2 * w2)
}
w1 and w2 are we...