similar to: Any help understand the FKF package? Thanks.

Dear R users, When running the program below I receive the following error message: fit <- optim(parm, objective, yt = tyield, hessian = TRUE) Error in as.vector(data) : no method for coercing this S4 class to a vector I can't figure out what the problem is exactly. I imagine that it has something to do with "tyield" being a matrix. Any help on explaining what's going on

Hi, A quick question: I have to vectors, say ar1 and ar2 > ar1 [1] "a" "c" "c" "a" attr(,"levels") [1] "a" "b" "c" > ar2 [1] TRUE TRUE FALSE TRUE > table(ar1, ar2) ar2 ar1 FALSE TRUE a 0 2 c 1 1 I would like to obtain: T F a 2 0 b 0 0 c 1 1

Dear all, I have a problem that may be someone of you can help. I am a newbie and do not find how to do it in manuals. I have two arrays, for example: ar1 <- array(data=c(1:16),dim=c(4,4)) ar2 <- array(data=c(1:16),dim=c(4,4)) colnames(ar1)<-c("A","B","D","E") colnames(ar2)<-c("C","A","E","B") > ar1

Dear R users, Please let me know how to plot the forecast in such a model: First I do it simple with ARIMA model that works ok with the codes provided to me at the lecture: arima<-arima(HCPIlong, order=c(1,1,0)) arima.predict<-predict(arima, n.ahead= 5 ) ts.plot(HCPIlong,arima.predict$pred,lty=1:2, main="Forecast of HCPI") But I need to include the additional variable in my

Dear R users, I am trying to carry out MLE of the time-varying CAPM using the FKF package. My approach so far has been to try and adapt the example given in the help file found using ?fkf which demonstrates the MLE of an ARMA(2,1) model. When I attempt to run my R code (given below) I get the following error: Error in fkf(a0 = sp$a0, P0 = sp$P0, dt = sp$dt, ct = sp$ct, Tt = sp$Tt, : Some of

Hello I am trying to use predict from an arma-Garch model (arma(2, 2) + garch(1, 1)) and I am getting the following error: Error en arima(x = object@data, order = c(max(u, 1), 0, max(v, 1)), init = c(ar, : non-stationary AR part from CSS Does anybody know what can be the reason of this error? The model I have estimated is the following: Title: GARCH Modelling Call: garchFit(formula =

Dear R People: I have the output from an arima model fit in an object xxx. I want to verify that the ma1 coefficient is there, so I did the following: > xxx$coef ar1 ar2 ma1 intercept 1.3841297 -0.4985667 -0.9999996 -0.1091657 > str(xxx$coef) Named num [1:4] 1.384 -0.499 -1 -0.109 - attr(*, "names")= chr [1:4] "ar1" "ar2"

Estimated people, I have two matrices: ar1 <- array(data=c(1:16),dim=c(4,4)) ar2 <- array(data=c(1,2,3,3,5:16),dim=c(4,4)) They only differ in the fourth row. I would like to compare them in order to know which columns are equal. The following works, but I would like to have a better solution, and not to use what someone called "prehistorical loops": for(i in

Dear users, I tried to fit an AR(2) model to data. This the result: > arima(vw,c(3,0,0)) Call: arima(x = vw, order = c(3, 0, 0)) Coefficients: ar1 ar2 ar3 intercept 0.1052 -0.0102 -0.1203 0.0099 s.e. 0.0337 0.0339 0.0338 0.0018 sigma^2 estimated as 0.002934: log likelihood = 1293.16, aic = -2576.33 Now, ar2 is not significantly different from

Hello , I have estimated the following model, a sarima: p=9 d=1 q=2 P=0 D=1 Q=1 S=12 In R 2.12.2 Call: arima(x = xdata, order = c(p, d, q), seasonal = list(order = c(P, D, Q), period = S), optim.control = list(reltol = tol)) Coefficients: ar1 ar2 ar3 ar4 ar5 ar6 ar7 ar8 ar9 0.3152 0.8762 -0.4413 0.0152 0.1500 0.0001 -0.0413 -0.1811

Hello , I have estimated the following model, a sarima: p=9 d=1 q=2 P=0 D=1 Q=1 S=12 In R 2.12.2 Call: arima(x = xdata, order = c(p, d, q), seasonal = list(order = c(P, D, Q), period = S), optim.control = list(reltol = tol)) Coefficients: ar1 ar2 ar3 ar4 ar5 ar6 ar7 ar8 ar9 0.3152 0.8762 -0.4413 0.0152 0.1500 0.0001 -0.0413 -0.1811

I'm using the fixed argument in arima. Shouldn't ar4, ar5, and ar6 display as zero in the output? Call: arima(x = window(log(hhprice), start = c(1990, 1), end = c(2003, 3)), order = c(7, 1, 0), xreg = window(ts.union(exa1 = lag(exa, -1), exa12 = lag(exa, -12), exb1 = lag(exb, -1), exc1 = lag(exc, -1), exc12 = lag(exc, -12)), start = c(1990, 1), end = c(2003, 3)),

Relative newbie so would welcome comments re structure of a self-referential relationship I''ve got. It''s started off similar to the Person HABTM friends in the Rails recipes books (working fine), but I needed to turn the join table into a full-blown model as I wanted to add attributes to it. Few tricky bits dealing with the new :through structure and understanding how :scope

Hi dear sirs, I am wondering why the fit of the time serie x with an arima and the fit of diff(x) with an arma (same coeff p & d) differ one from another here are the output of R: %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% > modelarma<-arma(diff(x),c(7,5)) > modelarma Call: arma(x = diff(x), order = c(7, 5)) Coefficient(s): ar1 ar2 ar3 ar4 ar5 ar6 ar7 ma1 ma2 0.06078

Andrew Ruef wrote: > Evening, > > I was writing some code that tried to insert calls to the > llvm.annotation intrinsic function, which has a signature of (i32, > i8*, i8*, i32). The code is below. > > void addAnnotation( BasicBlock *block, Function *F) > { > string foo = "foo"; > string bar = "barr"; > > Type

Dear all, The standard call to ARIMA in the base package such as arima(y,c(5,0,0),include.mean=FALSE) gives a full 5th order lag polynomial model with for example coeffs Coefficients: ar1 ar2 ar3 ar4 ar5 0.4715 0.067 -0.1772 0.0256 -0.2550 s.e. 0.1421 0.158 0.1569 0.1602 0.1469 Is it possible (I doubt it but am

Evening, I was writing some code that tried to insert calls to the llvm.annotation intrinsic function, which has a signature of (i32, i8*, i8*, i32). The code is below. void addAnnotation( BasicBlock *block, Function *F) { string foo = "foo"; string bar = "barr"; Type *charTy = Type::getInt8Ty(block->getContext()); ArrayType *s1Ty =

Hi, I'm trying to estimate the parameters of a state space model of the following form measurement eq: z_t = a + b*y_t + eps_t transition eq y_t+h = (I -exp(-hL))theta + exp(-hL)y_t+ eta_{t+h}. The problem is that the distribution of the innovations of the transition equation depend on the previous value of the state variable. To be exact: y_t|y_{t-1} ~N(mu, Q_t) where Q is a diagonal

I've posted this before but have not been able to locate what I'm doing wrong. I cannot determine how the forecast is made using the estimated coefficients from a simple AR(2) model when there is an exogenous variable. Does anyone know what the problem is? The help file for arima doesn't show the model with any exogenous variables. I haven't been able to locate any documents

Hello I want to fit an AR model were two of the coefficients are fixed to zero (the second and third ar-coefficients). I used the "arima" function with the "fixed" argument but the ar3 coefficient is not set to zero: ============================================== > arima(Y, order=c(4,0,0), xreg=1:23, fixed=c(NA,0,0,NA,NA,NA)) Call: arima(x = Y, order = c(4, 0, 0), xreg =