search for: kx1

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2009 Feb 15
0
Kalman Filter - dlm package
...currently trying to use the "dlm" package for Kalman filtering. My model is very simple: Y_t = F'_t Theta_t + v_t Theta_t = G_t Theta_t-1 + w_t v_t ~ N(0,V_t) = N(0,V) w_t ~ N(0,W_t) = N(0,W) Y_ t is a univariate time series (1x1) F_t is a vector of factor returns (Kx1) Theta_t is the state vector (Kx1) G_t is the identity matrix My first challenge is to get the Maximum Likelihood estimators of V and W assuming they are time-invariant (homoscedastic) through the dlmMLE function. In the example provided in the user guide, F is the Identity matrix (diag(2))...
2003 Sep 05
4
Basic Dummy Variable Creation
...through the mailing list archive, I did not come across a simple minded example regarding the creation of dummy variables. The Gauss language provides the command "y = dummydn(x,v,p)" for creating dummy variables. Here: x = Nx1 vector of data to be broken up into dummy variables. v = Kx1 vector specifying the K-1 breakpoints p = positive integer in the range [1,K], specifying which column should be dropped in the matrix of dummy variables. y = Nx(K-1) matrix containing the K-1 dummy variables. My recent mailing list archive inquiry has led me to examine R's "model.matri...
2003 Jul 14
2
Subsetting a matrix
...1.44 1.56 [2,] 2.64 2.86 and now it does know how to behave. But it would save some bother (and having to worry about this sort of thing in the midst of complex matrix algebra) if matrix subsetting worked in a consistent way for all possible subsets including cases which should result in 1xk or kx1 matrices (where, by the way, we could have k=1 and get a 1x1 matrix). Is there perhaps some global option which regulates this sort of behaviour? Or is the underclass always with us? With thanks, Ted. -------------------------------------------------------------------- E-Mail: (Ted Harding) &lt...
2006 Mar 06
7
is there a way to let R do smart matrix-vector operation?
Hi all, I want to substract vector B from A's each column... how can R do that smartly without a loop? > A=matrix(c(2:7), 2, 3) > A [,1] [,2] [,3] [1,] 2 4 6 [2,] 3 5 7 > B=matrix(c(1, 2), 2, 1) > B [,1] [1,] 1 [2,] 2 > A-B Error in A - B : non-conformable arrays [[alternative HTML version deleted]]
2004 Jan 19
18
Hi
Test =) soddsjbmshdwqwu -- Test, yep.