Hi, I have a disagreement with friend over if an observable that depend on lagged observable, should be called a state variable. I have the following system: xt=u+z(1,t-1)+x(t-1)+d·[y(t-1)-x(t-1)]+e(x,t) yt=u+z(2,t-1)+y(t-1)-d·[y(t-1)-x(t-1)]+e(y,t) z1 and z2 are AR(1) processes z(1,t)=p1z(1,t-1)+e(1,t) z(2,t)=p2z(2,t-1)+e(2,t) The observations are xt and yt I say the state variables are z1 and z2, but my friend say I am wrong in that the state variable must also include x and y, in fact x and y are both state variables and observables. Since I think that x(t-1) and y(t-1) are observed and known at time t, can I claim that they are known and thus can be thrown out. And as far as the Kalman filter gain and the variance of estimate P , as well as their steady state values are concerned, I can derive K and P and their steady state values by pretending the system is the following, which will yield the exact same K and P: xt=u+z(1,t-1)+x(t-1)+e(x,t) yt=u+z(2,t-1)+y(t-1)+e(y,t) z1 and z2 are AR(1) processes z(1,t)=p1z(1,t-1)+e(1,t) z(2,t)=p2z(2,t-1)+e(2,t) 1. Am I right that I can pretend that the d·[y(t-1)-x(t-1)] part don't exist, they have no bearing whatsoever on the Kalman filter equations. Am I right that the 2nd system will yield the exact same K and P and their respective steady state value as the original system? 2. Can someone please post the complete translation into matrix language, following wikipedia symbol convention on Kalman filter: X(k)=F(k)X(k-1)+B(k)u(k)+w(k) Z(k)=H(k)X(k)+v(k) What will each of the matrices/vectors be? Me and my friend disagree even on the most basics such as what the state variable X(k) should consist of.

# Kalman filter with observables that depend on lagged observables

Started by ●August 29, 2012

Reply by ●August 29, 20122012-08-29

On Wed, 29 Aug 2012 15:08:27 -0500, onthetopo wrote:> Hi, > > I have a disagreement with friend over if an observable that depend on > lagged observable, should be called a state variable. I have the > following system: > > xt=u+z(1,t-1)+x(t-1)+d·[y(t-1)-x(t-1)]+e(x,t) > yt=u+z(2,t-1)+y(t-1)-d·[y(t-1)-x(t-1)]+e(y,t) > > z1 and z2 are AR(1) processes > z(1,t)=p1z(1,t-1)+e(1,t) > z(2,t)=p2z(2,t-1)+e(2,t) > The observations are xt and yt > > I say the state variables are z1 and z2, but my friend say I am wrong in > that the state variable must also include x and y, in fact x and y are > both state variables and observables. > > Since I think that x(t-1) and y(t-1) are observed and known at time t, > can I claim that they are known and thus can be thrown out. And as far > as the Kalman filter gain and the variance of estimate P , as well as > their steady state values are concerned, I can derive K and P and their > steady state values by pretending the system is the following, which > will yield the exact same K and P: > > xt=u+z(1,t-1)+x(t-1)+e(x,t) > yt=u+z(2,t-1)+y(t-1)+e(y,t) > > z1 and z2 are AR(1) processes > z(1,t)=p1z(1,t-1)+e(1,t) > z(2,t)=p2z(2,t-1)+e(2,t) > > 1. Am I right that I can pretend that the d·[y(t-1)-x(t-1)] part don't > exist, they have no bearing whatsoever on the Kalman filter equations. > Am I right that the 2nd system will yield the exact same K and P and > their respective steady state value as the original system?Either you or your friend have posted nearly this exact question already, under the nym "swh5". I answered. Perhaps you could start by reading, and responding to, that answer. In particular, my answer contains some questions that bear heavily on answering your original question; leave those questions unanswered and we're just gabbling in the dark, not driving things toward a helpful solution for you.> 2. Can someone please post the complete translation into matrix > language, following wikipedia symbol convention on Kalman filter: > X(k)=F(k)X(k-1)+B(k)u(k)+w(k) > Z(k)=H(k)X(k)+v(k) > What will each of the matrices/vectors be? Me and my friend disagree > even on the most basics such as what the state variable X(k) should > consist of.Am I capable? Sure. Will I? Well, you respond to answers by asking nearly the exact same question again in a new thread instead of responding to the answers to your questions. This means that you either don't care to look at the responses or you don't care to ask further questions (even "I'm completely flummoxed" is reasonable). -- My liberal friends think I'm a conservative kook. My conservative friends think I'm a liberal kook. Why am I not happy that they have found common ground? Tim Wescott, Communications, Control, Circuits & Software http://www.wescottdesign.com