I need an urgent help in an issue with a state space model for filtering. My state model is like:

$\mathbf{d}_k = \mathbf{d}_{k-1} + \boldsymbol{\eta}_k$ with $\boldsymbol{\eta}_k \sim \mathcal{N}(0, \Sigma_\eta)$

and my observation model is like:

$y_k = \mathbf{d}_k^T \boldsymbol{\Gamma}\mathbf{z}_k + \epsilon_k$, with $\boldsymbol{\epsilon}_k \sim \mathcal{N}(0, \sigma^2_\epsilon)$.

In conventional Kalman Filter, the state vector to be estimated $\mathbf{d}_k$ is accompanied with a constant matrix like $\boldsymbol{\Gamma}$. But in this case it is having an extra term $\mathbf{z}_k$ which is random with $\mathbf{z}_k = [\cos\theta_1~~ \sin\theta_1~~ \cos\theta_2~~ \cos\theta_3~~\sin\theta_3~~\cos\theta_4]^T$ is a vector with $\{\theta_1, \theta_3\}\sim^{\text{iid}} \mathcal{U}(0, 2\pi)$ and $\{\theta_2, \theta_4\}\sim^{\text{iid}} \mathcal{U}(-\pi/2, \pi/2)$, $\mathcal{U}(a,b)$ denoting the uniform distribution with parameters $a$ and $b$.

I want to estimate the $\mathbf{d}_k$ only using Bayesian Principles.

Can anybody help me regarding how to proceed in this case to estimate $\mathbf{d}_k$?

Thanks in advance.

  • $\begingroup$ You need software or you want an explanation regarding a suitable procedure? Also, are you familiar with the Kalman filter? $\endgroup$ – rbm Jun 28 '15 at 14:03
  • $\begingroup$ I am in need of theoretical modeling of this type. In usual Kalman Filtering we dont have the $\mathbf{z}_k$ term. Thats the problem here. $\endgroup$ – Sandipan Karmakar Jun 28 '15 at 14:24
  • $\begingroup$ I don't really see the problem. See Wikipedia: the observation equation for Kalman problems is generally as follows: $\mathbf{z}_k = \mathbf{H}_{k} \mathbf{x}_k + \mathbf{v}_k$. So there you also have an $\mathbf{H}_k$ (depending on $k$). Loosely spoken (ignoring matrix dimensions), if you would define $\mathbf{\Gamma z_k}$ as $\mathbf{H_k}$, why wouldn't this be equivalent to your problem? $\endgroup$ – rbm Jun 28 '15 at 15:15
  • $\begingroup$ Actually problem is if I combine $\boldsymbol{\Gamma}$ and $\mathbf{z}_k$ then $\boldsymbol{\Gamma}\mathbf{z}_k$ would be random as $\mathbf{z}_k$ is random. But in conventional KF, $\mathbf{H}_k$ is assumed constant if known. $\endgroup$ – Sandipan Karmakar Jun 28 '15 at 15:23
  • $\begingroup$ I see. Might be good to make this a bit more explicit in the OP, so that it is easier for people to help you :). You could always use a Gibbs sampling approach btw, but I am not sure if that is what you want (it's not straightforward to give a derivation, but there are packages for that). $\endgroup$ – rbm Jun 28 '15 at 15:27

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