search for: theta_1

...rix. 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(theta_1, theta_2)...

...<- 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_13 <-...

...data, I need to estimate the parameter theta. I have been doing this by using a negative binomial glm of the same model (except that all the effects are fixed), and estimating mu as the fitted model like so: model_1 <-glm.nb(y~x1+x2+x3, data = datafilename) mu_1 <- fitted(model_1) theta_1 <- theta.ml(y, mu_1, length(data), limit = 10, eps = .Machine$double.eps^0.25, trace = FALSE) Then, I conduct the lmer, using the estimated theta: model_11 <-lmer(y~x1+x2+(1|x3), family = negative.binomial(theta = theta_1, link = “log”), method = “Laplace”) First, I wondere...

...1] * xx[, 4] - xx[, 2] * xx[, 3] z1 <- (xx[, 4] - xx[, 2]) z2 <- - (xx[, 3] - xx[, 1]) return(rq(y~cbind(z1, z2)-1)$coef) } To understand the strategy, note that the area of the triangle formed by the points x_i = (x_i1,x_i2), x_j = (x_j1,x_j2), and theta = (theta_1,theta_2) is given by the determinant, | 1 1 1 | Delta(x_i, x_j, theta) = .5 |y_i1 yj1 theta_1|. |y_i2 yj2 theta_2| Expanding the determinant in the unknown parameters theta gives the l1 regression formulation. Remarkably,...

...en by maximizing?the utility. ? Let U*(i) = max{ U(i,j) | 1 <= j <= m_i } ? Therefore, each outcome was chosen among m_i items by finding the index j such that U(i,j) = U*(i). ? For the utility function U(i,j), there are some independent variables, V_ij(k), 1 <= k <= N ? i.e. ?U(i,j) = theta_1 * V_ij(1) + theta_2 * V_ij(2) + ... + theta_N V_ij(N) ? Further assume the?probability?of choosing H(i,j) in observations i is: ? P(i,j) = exp(U(i,j)) / sum_{j=1}^{m_i} ( exp(U(i,j)) ? Then, I will estimate the parameters of the model theta_1, theta_2, ..., theta_N by maximizing the the log-likel...

...might clarify your thinking to note that a seasonal ARIMA model > is just an ``ordinary'' ARIMA model with some coefficients > constrained to be 0 in an efficient way. E.g. a seasonal AR(1) s = > 4 model is the same as an ordinary (nonseasonal) AR(4) model with > coefficients theta_1, theta_2, and theta_3 constrained to be 0. You > can get the same answer as from a seasonal model by using the > ``fixed'' argument to arima. E.g.: set.seed(42) x <- arima.sim(list(ar=c(0,0,0,0.5)),300) f1 = arima(x,seasonal=list(order=c(1,0,0),period=4)) f2 = arima(...

...0 Inverted MA Roots .53 -.48 gretl: Model 13: ARMAX estimates using the 517 observations 2-518 Estimated using Kalman filter (exact ML) Dependent variable: (1-L) Spot Standard errors based on Outer Products matrix VARIABLE COEFFICIENT STDERROR T STAT P-VALUE theta_1 -0.0491101 0.0439294 -1.118 0.26360 theta_2 -0.248075 0.0439901 -5.639 <0.00001 *** X_1 3.40437 1.21871 2.793 0.00522 *** Mean of dependent variable = 0.613926 Standard deviation of dep. var. = 12.3617...

Hello R People: When using the arima function with the seasonal option, are the seasonal options only good for monthly and quarterly data, please? Also, I believe that weekly and daily data are not appropriate for seasonal parm estimation via arima. Is that correct, please? Thanks, Sincerely, Laura Holt mailto: lauraholt_983 at hotmail.com download!

Dear list-members I am new to R and a statistics beginner. I really like the ease with which I can extract and manipulate data in R, and would like to use it primarily. I've been learning by checking analyses that have already been run in SAS. In an experiment with Y being a response variable, and group a 2-level between-subject factor, and time a 5-level within-subject factor. 2

...@item --help ====================================================================== [-as-] {+as:+} ====================================================================== --- R-intro.texi~ Mon Aug 23 13:24:18 2004 +++ R-intro.texi Mon Aug 23 13:24:18 2004 @@ -4945,7 +4945,7 @@ $\beta_2=\theta_2/\theta_1$. @end tex Supposing a suitable data frame to be set up we could fit this -non-linear regression as +non-linear regression as: @example > nlfit <- glm(y ~ x1 + x2 - 1, ====================================================================== [-below.-] {+below:+} ========================...