Register to reply

MMSE estimators

by prhzn
Tags: estimators, mmse
Share this thread:
prhzn
#1
Apr25-10, 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[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]


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.
Phys.Org News Partner Science news on Phys.org
Climate change increases risk of crop slowdown in next 20 years
Researcher part of team studying ways to better predict intensity of hurricanes
New molecule puts scientists a step closer to understanding hydrogen storage
prhzn
#2
Apr28-10, 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}[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.


Register to reply

Related Discussions
Conditional & uncoditional MSE (in MMSE estimation) Set Theory, Logic, Probability, Statistics 14
Help with Unbiased Estimators Calculus & Beyond Homework 1
LogNormal Mean Estimators Set Theory, Logic, Probability, Statistics 3
Estimators Calculus & Beyond Homework 12
Distribution of the statistic? Set Theory, Logic, Probability, Statistics 2