Displaying 8 results from an estimated 8 matches for "lambda_1".
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lambda1
2011 Sep 22
1
Error in as.vector(data) optim() / fkf()
...h "tyield" being a matrix. Any help on explaining what's
going on and how to solve this is much appreciated.
Thank you,
Kristian
library(FKF) #loading Fast Kalman Filter package
library(Matrix) # matrix exponential package
K_1 = 0.1156
K_2 = 0.17
sigma_1 = 0.1896
sigma_2 = 0.2156
lambda_1 = 0
lambda_2 = -0.5316
theta_1 = 0.1513
theta_2 = 0.2055
#test data
tyield <- matrix(data = rnorm(200), nrow =2, ncol =100)
# defining dimensions
m <- 2 # m is the number of state variables
n <- 100 # is the length of the observed sample
d <- 2 # is the number of observed variables.
t...
2011 Nov 12
1
State space model
...nrow=10, ncol=40)
m <- 2 # m is the number of state variables
n <- ncol(x) # is the length of the observed sample
d <- nrow(x) # is the number of observed variables.
h <- 1/52
## creating state space representation of 2-factor CIR model
CIR2ss <- function(K_1, K_2, sigma_1, sigma_2, lambda_1, lambda_2, theta_1,
theta_2, delta_0, delta_1, delta_2) {
## defining auxilary parameters,
phi_11 <- sqrt((K_1+lambda_1)^2+2*sigma_1^2*delta_1)
phi_21 <- sqrt((K_2+lambda_2)^2+2*sigma_2^2*delta_2)
phi_12 <- K_1+lambda_1+phi_11
phi_22 <- K_2+lambda_2+phi_21...
2011 Dec 30
2
Joint modelling of survival data
Assume that we collect below data : -
subjects = 20 males + 20 females, every single individual is independence,
and difference
events = 1, 2, 3... n
covariates = 4 blood types A, B, AB, O
http://r.789695.n4.nabble.com/file/n4245397/CodeCogsEqn.jpeg
?m = hazards rates for male
?n = hazards rates for female
Wm = Wn x ?, frailty for males, where ? is the edge ratio of male compare to
female
Wn =
2003 Sep 04
1
title expressions
The officially sanctioned way to put the expression "lambda_1 = x" in a title
is something like this:
title(substitute(lambda[1] == lamb, list(lamb = x)))
but suppose I have two lambdas and would like something like
"lambda_1 = x_1 , lambda_2 = x_2"
to appear. What then? Undoubtedly I'm missing something blindingly
obvious with list...
2011 Jun 02
4
generating random covariance matrices (with a uniform distribution of correlations)
List members,
Via searches I've seen similar discussion of this topic but have not seen
resolution of the particular issue I am experiencing. If my search on this
topic failed, I apologize for the redundancy. I am attempting to generate
random covariance matrices but would like the corresponding correlations to
be uniformly distributed between -1 and 1.
The approach I have been using is:
2009 May 16
1
maxLik pakage
...8,3,4,34,
32,4,9,9,8,36,44,0,9,27,28,55,72,12,1,
9,0,32,0,0,2,15,5,6,17,
63,61,9,15,15,0,2)
#########goal is found unique maximum for 5-parameter function with using "maxLik" function
#####the function in latex commands is as following:
\begin{eqnarray*}
&&\ell(\xi,\omega,\nu,\lambda_1,\lambda_2)=n\log2-n\log\omega+n\log\Gamma(\frac{\nu+1}{2})-\frac{n}{2}\log(\nu\pi)\\
&-&n\log\Gamma(\frac{\nu}{2})-\frac{\nu+1}{2}\sum_{i=1}^{n}\log(1+\frac{(x_i-\xi)^2}{\omega^2\nu})+\sum_{i=1}^{n}\log\Phi(\lambda_1\frac{(x_i-\xi)}{\sqrt{\omega^2+\lambda_2(x_i-\xi)^2}})
\end{eqnarray*}
###...
2006 Jun 24
2
smoothing splines and degrees of freedom
Hi,
If I set df=2 in my smooth.spline function, is that equivalent to running
a linear regression through my data? It appears that df=# of data points
gives the interpolating spline and that df = 2 gives the linear
regression, but I just want to confirm this.
Thank you,
Steven
2004 Dec 03
3
Computing the minimal polynomial or, at least, its degree
Hi,
I would like to know whether there exist algorithms to compute the
coefficients or, at least, the degree of the minimal polynomial of a square
matrix A (over the field of complex numbers)? I don't know whether this
would require symbolic computation. If not, has any of the algorithms been
implemented in R?
Thanks very much,
Ravi.
P.S. Just for the sake of completeness, a