Proof of the MMSE Estimator of $\mathbf{\theta}\left[n\right]$

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SUMMARY

The discussion focuses on proving the Minimum Mean Square Error (MMSE) estimator of the time-varying parameter \(\mathbf{\theta}[n]\), defined by the relation \(\mathbf{\theta}[n] = \mathbf{A}\mathbf{\theta}[n-1]\) for \(n \geq 1\). The proof establishes that the MMSE estimator is given by \(\mathbf{\hat{\theta}}[n] = \mathbf{A}^n\mathbf{\hat{\theta}}[0]\), where \(\mathbf{\hat{\theta}}[0]\) is the MMSE estimator of \(\mathbf{\theta}[0]\). This conclusion is reached by recursively applying the deterministic relation and utilizing the properties of conditional expectation.

PREREQUISITES
  • Understanding of MMSE estimation and its mathematical formulation.
  • Familiarity with linear algebra concepts, particularly matrix multiplication.
  • Knowledge of conditional expectation and probability density functions (PDFs).
  • Basic understanding of stochastic processes and their properties.
NEXT STEPS
  • Study the derivation of MMSE estimators in detail, focusing on examples with known PDFs.
  • Explore the properties of deterministic relations in stochastic processes.
  • Learn about the application of matrix exponentiation in time-varying systems.
  • Investigate the implications of MMSE estimators in real-world signal processing applications.
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Students and professionals in statistics, signal processing, and control systems who are looking to deepen their understanding of MMSE estimation and its applications in time-dependent parameter estimation.

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Homework Statement



Consider a parameter [tex]\mathbf{\theta}[/tex] which changes with time according to the deterministic relation

[tex]\mathbf{\theta}\left[n\right] = \mathbf{A}\mathbf{\theta}\left[n-1\right]\; n\geq 1[/tex],

where [tex]\mathbf{A}[/tex] is a known [tex]p\times p[/tex] matrix and [tex]\mathbf{\theta}\left[0\right][/tex] is an unknown parameter which is modeled as a random ([tex]p\times 1[/tex]) vector. Note that once [tex]\mathbf{\theta}\left[0\right][/tex] is specified, so is [tex]\mathbf{\theta}\left[n\right][/tex] for [tex]n\geq 1[/tex].

Prove that the MMSE estimator of [tex]\mathbf{\theta}\left[n\right][/tex] is

[tex]\mathbf{\hat{\theta}}\left[n\right] =\mathbf{A}^n\mathbf{\hat{\theta}}\left[0\right][/tex],

where [tex]\mathbf{\hat{\theta}}\left[0\right][/tex] is the MMSE estimator of [tex]\mathbf{\theta}\left[0\right][/tex], or equivalently,

[tex]\mathbf{\hat{\theta}}\left[n\right] = \mathbf{A}\mathbf{\hat{\theta}}\left[n-1\right][/tex]

Homework Equations



MMSE: [tex]\hat{\theta} = \mathbb{E}\left[\theta|\mathbf{x}\right] = \int p\left(\theta\right)\ln\left(p\left(\theta|\mathbf{x}\right)\right)\mathrm{d}\theta[/tex]

The Attempt at a Solution



So far I haven't got any good attempt as my main problem is how to start. Until now, all exercises about MMSE that I've done have specified information about the PDF's to some of the variables or some other information that has made it more obvious how to start; however, with this I feel like I'm a bit lost. So mainly I'm just looking for a hint on how to start, such that I can do an fair attempt on my own.
 
Last edited:
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Not sure if this is correct, maybe someone can tell or not?

We know that [tex]\mathbf{\theta}[n] = \mathbf{A}\mathbf{\theta}[n-1][/tex] for [tex]n\geq 1[/tex].

Then [tex]\mathbf{\theta}[n] = \mathbf{A}\left(\mathbf{A}\mathbf{\theta}[n-2]\right)[/tex] and so on, resulting in [tex]\mathbf{\theta}[n] = \mathbf{A}^n\mathbf{\theta}[0][/tex].

The MMSE is then

[tex]\mathbf{\hat{\theta}}[n] = \mathbb{E}\left[\mathbf{\theta}[n]|\mathbf{x}\right][/tex]

Doing the same "trick" as above we get

[tex]\mathbf{\hat{\theta}}[n] = \mathbf{A}^n\mathbb{E}\left[\mathbf{\theta}[0]|\mathbf{x}\right][/tex].

We already know that [tex]\mathbf{\hat{\theta}}[0][/tex] is the MMSE estimator of [tex]\mathbf{\theta}[0][/tex]; hence, the proof is complete.
 
Last edited:

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