Kalman filter equation

Started by Frank Neuhaus July 29, 2008
Hi

I've got a question on one equation of the Kalman filter.
First off, a quick sync of the terminology (just like on Wikipedia)

x: System state estimate
P: System state estimation error covariance

F: Process Model
H: Measurement Model

z: Measurement
y: Innovation
S: Innovation covariance

K: Kalman Gain

Consider the first equations of the Kalman filter first (leaving out 
subscripts etc):
x=Fx + w (leaving out control input u, w is the zero mean noise)
P=F * P * F^T + Q

The second Formula can be derived from the first using knowledge about 
covariance matrices.
since P is just cov(x) we can compute P as:
P=cov(x) = cov(Fx + w) = cov(Fx) + cov(w) = F * cov(x) * F^T + Q = F * P * 
F^T + Q

The fact that cov(Fx)=F * cov(x) * F^T is just some identity that is true 
for all covariance matrices.
So far so good... Now i tried to do the same thing on the last equation of 
the Kalman Filter.

The Kalman gain is computed as K = P * H^T * S^-1

The update step looks like this:

x=x + K*y

Again we want P=cov(x)

So we write:

P = cov(x) = cov(x + K*y) = cov(x) + cov(K*y) = P + K * cov(y) * K^T
  = P + K * S * K^T = P + (P * H^T * S^-1) * S * (P * H^T * S^-1)^T
  = P + P * H^T * S^-1 * S * S^-T * H * P^T

Now since P is symmetric, P^T=P, S is also symmetric, thus S*S^-T = Identity

  = P + P * H^T * S^-1 * H * P
  = P + K * H * P
  = (Identity + K*H)*P

So as you can see my result is P = (Identity + K*H)*P. When looking up the 
formula in other sources, the result is (Identity !!-!! K*H)*P
I dont see where that - sign comes from though. Can anyone explain that?

Thanks alot :-)
   Frank










On Tue, 29 Jul 2008 19:21:12 +0200, Frank Neuhaus wrote:

> Hi > > I've got a question on one equation of the Kalman filter. First off, a > quick sync of the terminology (just like on Wikipedia) > > x: System state estimate > P: System state estimation error covariance > > F: Process Model > H: Measurement Model > > z: Measurement > y: Innovation > S: Innovation covariance > > K: Kalman Gain > > Consider the first equations of the Kalman filter first (leaving out > subscripts etc): > x=Fx + w (leaving out control input u, w is the zero mean noise) P=F * P > * F^T + Q > > The second Formula can be derived from the first using knowledge about > covariance matrices. > since P is just cov(x) we can compute P as: P=cov(x) = cov(Fx + w) = > cov(Fx) + cov(w) = F * cov(x) * F^T + Q = F * P * F^T + Q > > The fact that cov(Fx)=F * cov(x) * F^T is just some identity that is > true for all covariance matrices. > So far so good... Now i tried to do the same thing on the last equation > of the Kalman Filter. > > The Kalman gain is computed as K = P * H^T * S^-1 > > The update step looks like this: > > x=x + K*y > > Again we want P=cov(x) > > So we write: > > P = cov(x) = cov(x + K*y) = cov(x) + cov(K*y) = P + K * cov(y) * K^T > = P + K * S * K^T = P + (P * H^T * S^-1) * S * (P * H^T * S^-1)^T = P > + P * H^T * S^-1 * S * S^-T * H * P^T > > Now since P is symmetric, P^T=P, S is also symmetric, thus S*S^-T = > Identity > > = P + P * H^T * S^-1 * H * P > = P + K * H * P > = (Identity + K*H)*P > > So as you can see my result is P = (Identity + K*H)*P. When looking up > the formula in other sources, the result is (Identity !!-!! K*H)*P I > dont see where that - sign comes from though. Can anyone explain that? > > Thanks alot :-) > Frank
I think your notation may be a little confusing since x and P in the first half of your post are not the same as the x and P in the second half. The Kalman filter equations are usually split into two steps; prior and posterior. The first half of your post is the prior step and the second half is the posterior step. I would derive the equation you are looking for as follows: x true state x^ posterior state estimate x^^ prior state estimate P prior state error covariance PP posterior state error covariance I identity matrix z measurement H measurement matrix v measurement noise R measurement noise covariance x~ = x - x^ x~~ = x - x^^ P = E{x~ * x~'} = E{(x - x^)(x - x^)' = E{(x~~ - K(z - H'*x^^))(x~~ - K(z - H'*x^^))'} = E{(x~~ - K(H'*x + v - H'*x^^))(x~~ - K(H'*x + v - H'*x^^))'} = E{((I - K*H')*x~~ + K*v)((I - K*H')*x~~ + K*v)'} = (I - K*H')*PP*(I - K*H')' + K*R*K' = (I - K*H')*PP - (I - K*H')*PP*H*K' + K*R*K' = (I - K*H')*PP - PP*H*K' + K*(H'*PP*H + R)*K' = (I - K*H')*PP - PP*H*K' + PP*H*(H'*PP*H + R)^-1*(H'*PP*H + R)*K' = (I - K*H')*PP - PP*H*K' + PP*H*K' = (I - K*H')*PP Best regards, Thomas Arildsen -- All email to sender address is lost. My real adress is at es dot aau dot dk for user tha.
Hi

first - thanks for your reply. Sorry for the notation.

The derivation you posted is pretty much what can also be found on Wikipedia 
and I obviously believe its correct :). I was just curious why _my_ 
computation, based on cov(x) (which works well for the computation of the 
first P as i showed, and works just as well for the innovation covariance S, 
only NEARLY works for this last equation. I believe there is just some small 
thing that I am missing that makes it slightly wrong in this last step :(

Thanks again
   Frank

"Thomas Arildsen" <tha.es-aau-dk@spamgourmet.com> schrieb im Newsbeitrag 
news:48902ee7$0$90263$14726298@news.sunsite.dk...
> On Tue, 29 Jul 2008 19:21:12 +0200, Frank Neuhaus wrote: > >> Hi >> >> I've got a question on one equation of the Kalman filter. First off, a >> quick sync of the terminology (just like on Wikipedia) >> >> x: System state estimate >> P: System state estimation error covariance >> >> F: Process Model >> H: Measurement Model >> >> z: Measurement >> y: Innovation >> S: Innovation covariance >> >> K: Kalman Gain >> >> Consider the first equations of the Kalman filter first (leaving out >> subscripts etc): >> x=Fx + w (leaving out control input u, w is the zero mean noise) P=F * P >> * F^T + Q >> >> The second Formula can be derived from the first using knowledge about >> covariance matrices. >> since P is just cov(x) we can compute P as: P=cov(x) = cov(Fx + w) = >> cov(Fx) + cov(w) = F * cov(x) * F^T + Q = F * P * F^T + Q >> >> The fact that cov(Fx)=F * cov(x) * F^T is just some identity that is >> true for all covariance matrices. >> So far so good... Now i tried to do the same thing on the last equation >> of the Kalman Filter. >> >> The Kalman gain is computed as K = P * H^T * S^-1 >> >> The update step looks like this: >> >> x=x + K*y >> >> Again we want P=cov(x) >> >> So we write: >> >> P = cov(x) = cov(x + K*y) = cov(x) + cov(K*y) = P + K * cov(y) * K^T >> = P + K * S * K^T = P + (P * H^T * S^-1) * S * (P * H^T * S^-1)^T = P >> + P * H^T * S^-1 * S * S^-T * H * P^T >> >> Now since P is symmetric, P^T=P, S is also symmetric, thus S*S^-T = >> Identity >> >> = P + P * H^T * S^-1 * H * P >> = P + K * H * P >> = (Identity + K*H)*P >> >> So as you can see my result is P = (Identity + K*H)*P. When looking up >> the formula in other sources, the result is (Identity !!-!! K*H)*P I >> dont see where that - sign comes from though. Can anyone explain that? >> >> Thanks alot :-) >> Frank > > I think your notation may be a little confusing since x and P in the > first half of your post are not the same as the x and P in the second > half. The Kalman filter equations are usually split into two steps; prior > and posterior. The first half of your post is the prior step and the > second half is the posterior step. > I would derive the equation you are looking for as follows: > > x true state > x^ posterior state estimate > x^^ prior state estimate > P prior state error covariance > PP posterior state error covariance > I identity matrix > z measurement > H measurement matrix > v measurement noise > R measurement noise covariance > > x~ = x - x^ > x~~ = x - x^^ > > P = E{x~ * x~'} = E{(x - x^)(x - x^)' > = E{(x~~ - K(z - H'*x^^))(x~~ - K(z - H'*x^^))'} > = E{(x~~ - K(H'*x + v - H'*x^^))(x~~ - K(H'*x + v - H'*x^^))'} > = E{((I - K*H')*x~~ + K*v)((I - K*H')*x~~ + K*v)'} > = (I - K*H')*PP*(I - K*H')' + K*R*K' > = (I - K*H')*PP - (I - K*H')*PP*H*K' + K*R*K' > = (I - K*H')*PP - PP*H*K' + K*(H'*PP*H + R)*K' > = (I - K*H')*PP - PP*H*K' + PP*H*(H'*PP*H + R)^-1*(H'*PP*H + R)*K' > = (I - K*H')*PP - PP*H*K' + PP*H*K' > = (I - K*H')*PP > > Best regards, > > Thomas Arildsen > -- > All email to sender address is lost. > My real adress is at es dot aau dot dk for user tha.
On Wed, 30 Jul 2008 11:13:34 +0200, Frank Neuhaus wrote:

> Hi > > first - thanks for your reply. Sorry for the notation. > > The derivation you posted is pretty much what can also be found on > Wikipedia and I obviously believe its correct :). I was just curious why > _my_ computation, based on cov(x) (which works well for the computation > of the first P as i showed, and works just as well for the innovation > covariance S, only NEARLY works for this last equation. I believe there > is just some small thing that I am missing that makes it slightly wrong > in this last step :( > > Thanks again > Frank > > "Thomas Arildsen" <tha.es-aau-dk@spamgourmet.com> schrieb im Newsbeitrag > news:48902ee7$0$90263$14726298@news.sunsite.dk... >> On Tue, 29 Jul 2008 19:21:12 +0200, Frank Neuhaus wrote: >> >>> Hi >>> >>> I've got a question on one equation of the Kalman filter. First off, a >>> quick sync of the terminology (just like on Wikipedia) >>> >>> x: System state estimate >>> P: System state estimation error covariance >>> >>> F: Process Model >>> H: Measurement Model >>> >>> z: Measurement >>> y: Innovation >>> S: Innovation covariance >>> >>> K: Kalman Gain >>> >>> Consider the first equations of the Kalman filter first (leaving out >>> subscripts etc): >>> x=Fx + w (leaving out control input u, w is the zero mean noise) P=F * >>> P * F^T + Q >>> >>> The second Formula can be derived from the first using knowledge about >>> covariance matrices. >>> since P is just cov(x) we can compute P as: P=cov(x) = cov(Fx + w) = >>> cov(Fx) + cov(w) = F * cov(x) * F^T + Q = F * P * F^T + Q >>> >>> The fact that cov(Fx)=F * cov(x) * F^T is just some identity that is >>> true for all covariance matrices. >>> So far so good... Now i tried to do the same thing on the last >>> equation of the Kalman Filter. >>> >>> The Kalman gain is computed as K = P * H^T * S^-1 >>> >>> The update step looks like this: >>> >>> x=x + K*y >>> >>> Again we want P=cov(x) >>> >>> So we write: >>> >>> P = cov(x) = cov(x + K*y) = cov(x) + cov(K*y) = P + K * cov(y) * K^T >>> = P + K * S * K^T = P + (P * H^T * S^-1) * S * (P * H^T * S^-1)^T = >>> P + P * H^T * S^-1 * S * S^-T * H * P^T >>> >>> Now since P is symmetric, P^T=P, S is also symmetric, thus S*S^-T = >>> Identity >>> >>> = P + P * H^T * S^-1 * H * P >>> = P + K * H * P >>> = (Identity + K*H)*P >>> >>> So as you can see my result is P = (Identity + K*H)*P. When looking up >>> the formula in other sources, the result is (Identity !!-!! K*H)*P I >>> dont see where that - sign comes from though. Can anyone explain that? >>> >>> Thanks alot :-) >>> Frank
[cut...] OK, I see what you mean. Looking at it again now, I think you may be wrong in your assumption that cov(x + K*y) = cov(x) + cov(K*y). Since y = z - H^T * x (in your notation), you are forgetting cross-covariance between x and y when you split the term up as you do. See if that will correct the problem. Best regards, Thomas Arildsen -- All email to sender address is lost. My real adress is at es dot aau dot dk for user tha.
"Thomas Arildsen" <tha.es-aau-dk@spamgourmet.com> schrieb im Newsbeitrag 
news:48903eb1$0$90263$14726298@news.sunsite.dk...
> On Wed, 30 Jul 2008 11:13:34 +0200, Frank Neuhaus wrote: > >> Hi >> >> first - thanks for your reply. Sorry for the notation. >> >> The derivation you posted is pretty much what can also be found on >> Wikipedia and I obviously believe its correct :). I was just curious why >> _my_ computation, based on cov(x) (which works well for the computation >> of the first P as i showed, and works just as well for the innovation >> covariance S, only NEARLY works for this last equation. I believe there >> is just some small thing that I am missing that makes it slightly wrong >> in this last step :( >> >> Thanks again >> Frank >> >> "Thomas Arildsen" <tha.es-aau-dk@spamgourmet.com> schrieb im Newsbeitrag >> news:48902ee7$0$90263$14726298@news.sunsite.dk... >>> On Tue, 29 Jul 2008 19:21:12 +0200, Frank Neuhaus wrote: >>> >>>> Hi >>>> >>>> I've got a question on one equation of the Kalman filter. First off, a >>>> quick sync of the terminology (just like on Wikipedia) >>>> >>>> x: System state estimate >>>> P: System state estimation error covariance >>>> >>>> F: Process Model >>>> H: Measurement Model >>>> >>>> z: Measurement >>>> y: Innovation >>>> S: Innovation covariance >>>> >>>> K: Kalman Gain >>>> >>>> Consider the first equations of the Kalman filter first (leaving out >>>> subscripts etc): >>>> x=Fx + w (leaving out control input u, w is the zero mean noise) P=F * >>>> P * F^T + Q >>>> >>>> The second Formula can be derived from the first using knowledge about >>>> covariance matrices. >>>> since P is just cov(x) we can compute P as: P=cov(x) = cov(Fx + w) = >>>> cov(Fx) + cov(w) = F * cov(x) * F^T + Q = F * P * F^T + Q >>>> >>>> The fact that cov(Fx)=F * cov(x) * F^T is just some identity that is >>>> true for all covariance matrices. >>>> So far so good... Now i tried to do the same thing on the last >>>> equation of the Kalman Filter. >>>> >>>> The Kalman gain is computed as K = P * H^T * S^-1 >>>> >>>> The update step looks like this: >>>> >>>> x=x + K*y >>>> >>>> Again we want P=cov(x) >>>> >>>> So we write: >>>> >>>> P = cov(x) = cov(x + K*y) = cov(x) + cov(K*y) = P + K * cov(y) * K^T >>>> = P + K * S * K^T = P + (P * H^T * S^-1) * S * (P * H^T * S^-1)^T = >>>> P + P * H^T * S^-1 * S * S^-T * H * P^T >>>> >>>> Now since P is symmetric, P^T=P, S is also symmetric, thus S*S^-T = >>>> Identity >>>> >>>> = P + P * H^T * S^-1 * H * P >>>> = P + K * H * P >>>> = (Identity + K*H)*P >>>> >>>> So as you can see my result is P = (Identity + K*H)*P. When looking up >>>> the formula in other sources, the result is (Identity !!-!! K*H)*P I >>>> dont see where that - sign comes from though. Can anyone explain that? >>>> >>>> Thanks alot :-) >>>> Frank > [cut...] > > OK, I see what you mean. Looking at it again now, I think you may be > wrong in your assumption that cov(x + K*y) = cov(x) + cov(K*y). Since y = > z - H^T * x (in your notation), you are forgetting cross-covariance > between x and y when you split the term up as you do. See if that will > correct the problem.
Hi again I think you are right. I looked up the rules for cov computations again and came up with this: cov(x + K*y) = cov(x + K*y,x + K*y) = cov(x,x) + cov(x,K*y) + cov(K*y,x) + cov(K*y,K*y) = P + cov(x,y)*K^T + K*cov(y,x) + K*cov(y,y)*K^T = P + K*S*K^T + cov(x,y)*K^T + K*cov(y,x) = P + K*S*K^T + cov(x,z-H*x)*K^T + K*cov(z-H*x,x) = P + K*S*K^T + (cov(x,z)-cov(x,H*x))*K^T + K*(cov(z,x)-cov(H*x,x)) = P + K*S*K^T + (cov(x,z)-P*H^T)*K^T + K*(cov(z,x)-H*P) = (assuming cov(x,z)=0, cov(z,x)=0) P + K*S*K^T - P*H^T*K^T - K*H*P = P + K*S*(S^-T*H*P^T) - P*H^T*K^T - K*H*P = (S*S^-T=Identity) P + K*H*P^T - P*H^T*K^T - K*H*P = (P symmetric) P + K*H*P - P*H^T*K^T - K*H*P = P-P*H^T*K^T = (I-H^T*K^T)*P The result I am getting is the transpose of the correct formula. That probably does not matter since the result is symmetric anyway, right? Thanks Frank
"Frank Neuhaus" <fneuhaus@uni-koblenz.de> schrieb im Newsbeitrag 
news:g6pl7p$l7k$1@cache.uni-koblenz.de...
> "Thomas Arildsen" <tha.es-aau-dk@spamgourmet.com> schrieb im Newsbeitrag > news:48903eb1$0$90263$14726298@news.sunsite.dk... >> On Wed, 30 Jul 2008 11:13:34 +0200, Frank Neuhaus wrote: >> >>> Hi >>> >>> first - thanks for your reply. Sorry for the notation. >>> >>> The derivation you posted is pretty much what can also be found on >>> Wikipedia and I obviously believe its correct :). I was just curious why >>> _my_ computation, based on cov(x) (which works well for the computation >>> of the first P as i showed, and works just as well for the innovation >>> covariance S, only NEARLY works for this last equation. I believe there >>> is just some small thing that I am missing that makes it slightly wrong >>> in this last step :( >>> >>> Thanks again >>> Frank >>> >>> "Thomas Arildsen" <tha.es-aau-dk@spamgourmet.com> schrieb im Newsbeitrag >>> news:48902ee7$0$90263$14726298@news.sunsite.dk... >>>> On Tue, 29 Jul 2008 19:21:12 +0200, Frank Neuhaus wrote: >>>> >>>>> Hi >>>>> >>>>> I've got a question on one equation of the Kalman filter. First off, a >>>>> quick sync of the terminology (just like on Wikipedia) >>>>> >>>>> x: System state estimate >>>>> P: System state estimation error covariance >>>>> >>>>> F: Process Model >>>>> H: Measurement Model >>>>> >>>>> z: Measurement >>>>> y: Innovation >>>>> S: Innovation covariance >>>>> >>>>> K: Kalman Gain >>>>> >>>>> Consider the first equations of the Kalman filter first (leaving out >>>>> subscripts etc): >>>>> x=Fx + w (leaving out control input u, w is the zero mean noise) P=F * >>>>> P * F^T + Q >>>>> >>>>> The second Formula can be derived from the first using knowledge about >>>>> covariance matrices. >>>>> since P is just cov(x) we can compute P as: P=cov(x) = cov(Fx + w) = >>>>> cov(Fx) + cov(w) = F * cov(x) * F^T + Q = F * P * F^T + Q >>>>> >>>>> The fact that cov(Fx)=F * cov(x) * F^T is just some identity that is >>>>> true for all covariance matrices. >>>>> So far so good... Now i tried to do the same thing on the last >>>>> equation of the Kalman Filter. >>>>> >>>>> The Kalman gain is computed as K = P * H^T * S^-1 >>>>> >>>>> The update step looks like this: >>>>> >>>>> x=x + K*y >>>>> >>>>> Again we want P=cov(x) >>>>> >>>>> So we write: >>>>> >>>>> P = cov(x) = cov(x + K*y) = cov(x) + cov(K*y) = P + K * cov(y) * K^T >>>>> = P + K * S * K^T = P + (P * H^T * S^-1) * S * (P * H^T * S^-1)^T = >>>>> P + P * H^T * S^-1 * S * S^-T * H * P^T >>>>> >>>>> Now since P is symmetric, P^T=P, S is also symmetric, thus S*S^-T = >>>>> Identity >>>>> >>>>> = P + P * H^T * S^-1 * H * P >>>>> = P + K * H * P >>>>> = (Identity + K*H)*P >>>>> >>>>> So as you can see my result is P = (Identity + K*H)*P. When looking up >>>>> the formula in other sources, the result is (Identity !!-!! K*H)*P I >>>>> dont see where that - sign comes from though. Can anyone explain that? >>>>> >>>>> Thanks alot :-) >>>>> Frank >> [cut...] >> >> OK, I see what you mean. Looking at it again now, I think you may be >> wrong in your assumption that cov(x + K*y) = cov(x) + cov(K*y). Since y = >> z - H^T * x (in your notation), you are forgetting cross-covariance >> between x and y when you split the term up as you do. See if that will >> correct the problem. > > > Hi again > > I think you are right. I looked up the rules for cov computations again > and came up with this: > > cov(x + K*y) > = > cov(x + K*y,x + K*y) > = > cov(x,x) + cov(x,K*y) + cov(K*y,x) + cov(K*y,K*y) > = > P + cov(x,y)*K^T + K*cov(y,x) + K*cov(y,y)*K^T > = > P + K*S*K^T + cov(x,y)*K^T + K*cov(y,x) > = > P + K*S*K^T + cov(x,z-H*x)*K^T + K*cov(z-H*x,x) > = > P + K*S*K^T + (cov(x,z)-cov(x,H*x))*K^T + K*(cov(z,x)-cov(H*x,x)) > = > P + K*S*K^T + (cov(x,z)-P*H^T)*K^T + K*(cov(z,x)-H*P) > = (assuming cov(x,z)=0, cov(z,x)=0) > P + K*S*K^T - P*H^T*K^T - K*H*P > = > P + K*S*(S^-T*H*P^T) - P*H^T*K^T - K*H*P > = (S*S^-T=Identity) > P + K*H*P^T - P*H^T*K^T - K*H*P > = (P symmetric) > P + K*H*P - P*H^T*K^T - K*H*P > = > P-P*H^T*K^T > = > (I-H^T*K^T)*P > > The result I am getting is the transpose of the correct formula. That > probably does not matter since the result is symmetric anyway, right?
Oh cool I just got it without the transposition by not replacing K^T but by replacing K and then canceling out S*S^-1 instead of S*S^-T: P + K*S*K^T - P*H^T*K^T - K*H*P = P + P*H^T*S^-1*S*K^T - P*H^T*K^T - K*H*P = P + P*H^T*K^T - P*H^T*K^T - K*H*P = P - K*H*P = (I-KH)*P Ok seems to be right now :-) Thanks alot again
On Jul 29, 7:21&#2013266080;pm, "Frank Neuhaus" <fneuh...@uni-koblenz.de> wrote:
> Hi > > I've got a question on one equation of the Kalman filter. > First off, a quick sync of the terminology (just like on Wikipedia) > > x: System state estimate > P: System state estimation error covariance > > F: Process Model > H: Measurement Model > > z: Measurement > y: Innovation > S: Innovation covariance > > K: Kalman Gain > > Consider the first equations of the Kalman filter first (leaving out > subscripts etc): > x=Fx + w (leaving out control input u, w is the zero mean noise) > P=F * P * F^T + Q > > The second Formula can be derived from the first using knowledge about > covariance matrices. > since P is just cov(x) we can compute P as: > P=cov(x) = cov(Fx + w) = cov(Fx) + cov(w) = F * cov(x) * F^T + Q = F * P * > F^T + Q > > The fact that cov(Fx)=F * cov(x) * F^T is just some identity that is true > for all covariance matrices. > So far so good... Now i tried to do the same thing on the last equation of > the Kalman Filter. > > The Kalman gain is computed as K = P * H^T * S^-1 > > The update step looks like this: > > x=x + K*y > > Again we want P=cov(x) > > So we write: > > P = cov(x) = cov(x + K*y) = cov(x) + cov(K*y) = P + K * cov(y) * K^T > &#2013266080; = P + K * S * K^T = P + (P * H^T * S^-1) * S * (P * H^T * S^-1)^T > &#2013266080; = P + P * H^T * S^-1 * S * S^-T * H * P^T > > Now since P is symmetric, P^T=P, S is also symmetric, thus S*S^-T = Identity > > &#2013266080; = P + P * H^T * S^-1 * H * P > &#2013266080; = P + K * H * P > &#2013266080; = (Identity + K*H)*P > > So as you can see my result is P = (Identity + K*H)*P. When looking up the > formula in other sources, the result is (Identity !!-!! K*H)*P > I dont see where that - sign comes from though. Can anyone explain that? > > Thanks alot :-) > &#2013266080; &#2013266080;Frank
I would say that the equation K = P * H^T * S^-1 uses the posterior (analysed) covariance. Use K = P * H^T * (H * P * H^T + R)^-1 instead. Take care Pavel