Displaying 5 results from an estimated 5 matches for "sigmay".
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2010 Aug 12
0
Error: evaluation nested too deeply
...guys,
I have a code in R and it was work well but when I decrease the epsilon
value (indicated in the code) , then I am getting this error
Error: evaluation nested too deeply: infinite recursion /
options(expressions=)?
any help please
y = 6.8;
w = 7.4;
z = 5.7;
muy = 7;
muw = 7;
muz = 6;
sigmay = 0.8;
sigmaz = 0.76;
sigmaw = 0.3;
betayx = 0.03;
betayz = 0.3;
betayw = 0.67
s = c(3.2,0.8)
em = function(W,s) {
a= 1/2*(1/s[2]^2+betayx^2/sigmay^2);
b=
(y-muy+betayx*s[1]-betayz*(z-muz)-betayw*(w-muw))*betayx/sigmay^2+s[1]/s[2]^2;
c=(1/2)*((y-muy+betayx*s[1]-betayz*(z-muz)-betayw*(w-muw)...
2008 Jan 23
2
from a normal bivariate distribution to the marginal one
...I simulated a bivariate normal distribution using these
simple lines:
rbivnorm <- function(n, # sample size
mux, # expected value of x
muy, # expected value of Y
sigmax, # standard deviation of X
sigmay, # standard deviation of Y
rho){ # correlation coefficient
x <- rnorm(n,mux,sigmax)
y <- rnorm(n,muy+rho*sigmay*(x-mux)/sigmax, sigmay*sqrt(1-rho2))
cbind(x,y)
}
In this way I've sampled from a normal bivariat...
2005 Mar 18
1
Bivariate normal distribution and correlation
Suppose I know the value of cumulative bivariate standard normal distribution. How can I solve correlation between variables?
Pekka
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2007 Jul 02
2
how to use mle with a defined function
Hi all,
I am trying to use mle() to find a self-defined function. Here is my
function:
test <- function(a=0.1, b=0.1, c=0.001, e=0.2){
# omega is the known covariance matrix, Y is the response vector, X is the
explanatory matrix
odet = unlist(determinant(omega))[1]
# do cholesky decomposition
C = chol(omega)
# transform data
U = t(C)%*%Y
WW=t(C)%*%X
beta = lm(U~W)$coef
Z=Y-X%*%beta
2004 Aug 30
3
D'agostino test
Hi, Does anyone know if the D'agostino test is available with R ?
Alex