Kalman filter in which observables depend on lagged observables

Hi,

I am puzzled over how to do the Kalman filter of a system with observables that depend on lagged observables. 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 guess the state variables are z1 and z2. I think x and y are not state variables.

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)+e(x,t)
yt=u+z(2,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 x(t-1),y(t-1) part don't exist in observation, since they are known at time t, they have no bearing whatsoever on the Kalman filter equations, which is concerned with estimating z1 and z2.

2.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?

3.If I am wrong can you please post a solution in matrix language what the Kalman filter equations should be?


 

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John Wayne
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On Sat, 25 Aug 2012 14:43:12 -0700, swh5 wrote:

> Hi,
> 
> I am puzzled over how to do the Kalman filter of a system with
> observables that depend on lagged observables. 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)      (1)
>
> 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)                           (2)
> The observations are xt and yt
> 
> I guess the state variables are z1 and z2. I think x and y are not state
> variables.

x and y certainly look like state variables, as do your z vector.

You never specify what e(n, t) means, and I don't see any tie-in between 
(1) and (2) (my designations), except through this mysterious e(whatever, 
t) function.

> 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.

Not if you just think it.  Do you _know_ it?  How can you test?  Are the 
measurements perfectly (or practically) noise-free?  If you're sure that 
you're measuring x and y with practically no noise, then yes, you could 
take them as given.  But since (1) and (2) don't seem to be affecting one 
another, I'm not sure how you can use x and y to shine any light on z.

> 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)+e(x,t)
> yt=u+z(2,t-1)+e(y,t)                               (3)
> 
> 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)                           (4)
> 
> 1.Am I right that I can pretend that the x(t-1),y(t-1) part don't exist
> in observation, since they are known at time t, they have no bearing
> whatsoever on the Kalman filter equations, which is concerned with
> estimating z1 and z2.
> 
> 2.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?
> 
> 3.If I am wrong can you please post a solution in matrix language what
> the Kalman filter equations should be?

The way that you are presenting this is confused.  At least, I'm confused 
after trying to unwind what you're doing.

I think you need to address my comments about the first two systems you 
describe; if doing so doesn't make the solution pop out at you, then post 
what you know and I'll try to make headway with it.

-- 
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