search for: lambda_2

...d" 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. theta <- c(...

...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 phi...

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 =

...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*} #############...

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 lists, but having tried several versions unsuccessfully I thought I would defer to the collective expertise of R-help. Thanks in advance. url: www.econ.uiuc.edu/~roger/my.html Roger Koenker email r...

...ed in one of the packages in R? This kind of functionality is available in the R package `spatstat' (available on CRAN or from www.spatstat.org) According to the cited website, the O-ring statistic is a rescaled version of the pair correlation function between two types of points: O_12(r) = lambda_2 g_12(r) In spatstat, pair correlation functions are computed by the function 'pcf'. To estimate the cross-type pair correlation function, you do something like pcf(Kcross(X, 1, 2)) where X is a marked point pattern containing points of types 1 and 2. To estimate the intensity of ty...