
#1
Apr2510, 09:20 AM

P: 9

1. The problem statement, all variables and given/known data
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[n1\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[n1\right][/tex] 2. Relevant 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)\mat hrm{d}\theta[/tex] 3. 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. 



#2
Apr2810, 04:48 AM

P: 9

Not sure if this is correct, maybe someone can tell or not?
We know that [tex]\mathbf{\theta}[n] = \mathbf{A}\mathbf{\theta}[n1][/tex] for [tex]n\geq 1[/tex]. Then [tex]\mathbf{\theta}[n] = \mathbf{A}\left(\mathbf{A}\mathbf{\theta}[n2]\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. 


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