search for: dlmfilter

Hello all, I am working with the dlm package, specifcially doing a dlm multivariate Y linear regression using dlmModReg and dlmFilter and dlmSmooth... I have altereted the inputs into dlmModReg to make them time-varying using JFF, JW etc. How do I track the results of the time varying system matrices? For example what I am really interested in is JW - my system variance matrix for each time period - I cannot get R to give me th...

...While forecasting into the future, I will NOT have data for all three variables. So, I am predicting X1 and X2 using two Kalman filters. The code is below x1.model <- dlmModSeas(52) + dlmModPoly(1, dV=5, dW=10) x2.model <- dlmModSeas(52) + dlmModPoly(1, dV=10, dW=10) x1.filter <- dlmFilter(c(train$x1, rep(NA, noofsteps)), x1.model) x2.filter <- dlmFilter(c(train$x2, rep(NA, noofsteps)), x2.model) Now, I am forecasting Y using the predicted X1 and X2 as below pred <- cbind(c(train$x1, x1.filter$f[260:312]),c(train$x2, x2.filter$f[260:312])) Y.model <- dlmModSeas(52)...

...lt;- function (x) { dlmModPoly(order = 2, dV = exp(x[1]), dW = c(0,exp(x[2]))) } fit <- dlmMLE(y=tsdata, parm=c(0,0), build=buildfun) # Warning: a numerically singular 'V' has been slightly perturbed to make it nonsingular fit$conv dlmTsdata <- buildfun(fit$par) tsdataFilter <- dlmFilter(tsdata, mod=dlmTsdata) tsdataSmooth <- dlmSmooth(tsdata, mod=dlmTsdata) plot(tsdata, lwd=2) for (i in 1:10) lines(lty=6, col="blue", dropFirst(dlmBSample(tsdataFilter))[,1]) # looks ok! tsdataForecast <- dlmForecast(tsdataFilter, nAhead=20) sqrtR <- sapply(tsdataForecast$R,...

Hi, I am studying Kalman Filter and it seems to be difficult for me to apply the filter on a simple ARMA. It is easy to construct the state-space model, for instance: dlmModARMA(ar=c(0.4,-0.2),ma=c(0.2,-0.1, sigma2=1) but applying the dlmFilter on it, it doesn't work... I don't know if my problem is clear but if anyone has already worked on Kalman filter, it could be great to advise me! Thank you in advance! Sandrine

Dear R users, I am trying to use the dlm package, and in particular the dlmBSample function. For some reason that I am not able to understand, this function does not work properly and the plot of the result does not make sense, while dlmFilter works perfectly. I think that my_mod is correct, because the output of my_dlmFilter$mod is fine. Where is my mistake or my misunderstanding? This is the code: function (orig_ts){ library(dlm) dV_T <- 20000 dW_T <- c(100,10) m0_T <- rep(0,2) C0_T <- 10000*diag(nrow=2) my_dlmModPoly &lt...

...2468) library(dlm) buildFun <- function(x) { dlmModPoly(1, dV = exp(x[1]), dW = exp(x[2])) } fit <- dlmMLE(CostUSD, parm = c(0,0), build = buildFun) fit$conv dlmCostUSD <- buildFun(fit$par) V(dlmCostUSD) W(dlmCostUSD) #For comparison StructTS(CostUSD, "level") CostUSDFilt <- dlmFilter(CostUSD, dlmCostUSD) CostUSDFore <- dlmForecast(CostUSDFilt, nAhead = 1) after which i return the error message: Error in mod$m[lastObsIndex, ] : incorrect number of dimensions Can anyone offer any insight to this problem? Thanks in advance Mike [[alternative HTML version deleted]]

...r follow a RW or AR(1) model. This is how I think it would be set up; I will have my time-series Y,X, where Y is the response variable this setup should give me a RW process if I have understood the example correctly mydlmModel = dlmModReg(X) + dlmModPoly(order=1) and then run on the dlm model dlmFilter(Y,mydlmModel ) but setting up a AR(1) process is unclear, should I use dlmModPoly or the dlmModARMA to set up the model. And at last but not the least, how do I set up a proper build function to use with dlmMLE to optimize the starting values. Regards Tom -- View this message in context: http:/...

...llowed in the observations. 2) Extractor and replacement functions for the matrices defining a dlm are now available. 3) The function for Kalman smoothing, "dlmSmooth", can take as arguments a data vector and a dlm object. Previously the argument had to be the output from "dlmFilter". 4) In addition to the "+" method function for objects of class "dlm", used to build complex models from simple components, all having the same dimensionality of the observation vector, there is now an outer sum, "%+%", which creates a joint model from...

...llowed in the observations. 2) Extractor and replacement functions for the matrices defining a dlm are now available. 3) The function for Kalman smoothing, "dlmSmooth", can take as arguments a data vector and a dlm object. Previously the argument had to be the output from "dlmFilter". 4) In addition to the "+" method function for objects of class "dlm", used to build complex models from simple components, all having the same dimensionality of the observation vector, there is now an outer sum, "%+%", which creates a joint model from...

...= (1, P_t, u_t) * (a1_t, a2_t, a3_t)' + v_t v_t ~ N(0,V) As far as I understand state space modeling the following restrictions on the Variance-covariance matrix W should imply a1_t=a1 and a2_t=a2 which is time invariant: (9) W=[(0,0,0),(0,0,0),(0,0,w_33)] However, if I apply the filter (dlmFilter) (not smoother) on this specification with estimated values for the unknown paramters (w_33 and matrix V) in order to get the series of the state vector (a1_t, a2_t, a3_t)' then for some reason a1_t and a2_t are not constant!!! a3_t isn't either but this is how it is supposed to be. How is...

...end of the mail)   BuildMod <- function(x){  return(dlm(   m0  = x[1],   C0  = x[2],   FF  = 1,   GG  = 1,   V   = x[3],   W   = x[4],   JFF = 1,   X   = X   )) } ModFit  <- dlmMLE(Y,rep(1,4),BuildMod,debug=T) dlmMod  <- BuildMod(ModFit$par) V <- dlmMod$V W <- dlmMod$W ModFilt <- dlmFilter(Y,dlmMod) v <- tail(dlmSvd2var(ModFilt$U.C,ModFilt$D.C),1) m <- tail(ModFilt$m,1)   The results are: V = 5.945003e-05 W = 0.0003086623 v = 9.850526e-05 (the estimated variance for a_t  after we observe the last pair of observations) m = -0.02965614 (the estimated mean for a_t after we obse...

...F  = matrix(1,1,nFactor),   GG  = diag(nFactor),   V   = tail(x,1)^2,   W   = crossprod(L1),   JFF = matrix(1:4,nr=1),   X   = X   )) } ModFit  <- dlmMLE(Y,rep(0.1,nTotal),BuildMod,debug=T) dlmMod  <- BuildMod(ModFit$par) V  = dlmMod$V W  = dlmMod$W m0 = dlmMod$m0 C0 = dlmMod$C0 ModFilt <- dlmFilter(Y,dlmMod) v <- tail(dlmSvd2var(ModFilt$U.C,ModFilt$D.C),1) m <- tail(ModFilt$m,1)   Here is the value of Y: 0.0125678739370109 -0.00241285475528163 0.00386919876129071 -0.00352839097011217 0.00285344714211614 0.00374266510625097 0.00797807743013259 -0.00543459628953192 -0.0138447399853609 -0...

Hi Group, I'm trying to build a model to predict a product's sale price. I'm researching the dlm package. Looks like I should use dlmModPoly, dlmMLE, dlmFilter, dlmSmooth, and finally dlmForecast. I'm looking at the Nile River example and I have a few questions: 1. If I only want to predict future sale price based on observed sale price, I should use a univariate model, correct? 2. how do I initiate value for dV and dW? In the exampl...

...y trends). The model is saved in the object titled "mod." Following the example in the documentation and using the commands below, I am attempting to use the function "dlmSvd2var" to implement SVD and calculate the 90% confidence errors for each time-varying state. outF <- dlmFilter(y,mod) v <- unlist(dlmSvd2var(outF$U.C, outF$D.C)) pl <- dropFirst(outF$m) + qnorm(0.05, sd=sqrt(v[-1])) pu <- dropFirst(outF$m) + qnorm(0.95, sd=sqrt(v[-1])) Since the dataset has 100 observations, I end up with a vector v that comprises 3636 atomic components: (1 + 100) x (6 x 6). If I...

...y trends). The model is saved in the object titled "mod." Following the example in the documentation and using the commands below, I am attempting to use the function "dlmSvd2var" to implement SVD and calculate the 90% confidence errors for each time-varying state. outF <- dlmFilter(y,mod) v <- unlist(dlmSvd2var(outF$U.C, outF$D.C)) pl <- dropFirst(outF$m) + qnorm(0.05, sd=sqrt(v[-1])) pu <- dropFirst(outF$m) + qnorm(0.95, sd=sqrt(v[-1])) Since the dataset has 100 observations, I end up with a vector v that comprises 3636 atomic components: (1 + 100) x (6 x 6). If I...