For the model X=Θ+W, and under the usual independence and normality assumptions for Θ and W, the mean squared error of the LMS estimator is

Question

1(1/σ02)+(1/σ12),
where σ02 and σ12 are the variances of Θ and W, respectively.

Suppose now that we change the observation model to Y=3Θ+W. In some sense the “signal" Θ has a stronger presence, relative to the noise term W, and we should expect to obtain a smaller mean squared error. Suppose σ02=σ12=1. The mean squared error of the original model X=Θ+W is then 1/2. In contrast, the mean squared error of the new model Y=3Θ+W is??

Answer 1

Can anyone help with this problem?

Answer 2

0.1

Answer 3

can you explain the answer?

Answer 4

Since Y′ is just Y scaled by a factor of 1/3 , Y′ carries the same information as Y , so that E[Θ∣Y]=E[Θ∣Y′] . Thus, the alternative observation model Y′=Θ+(W/3) will lead to the same estimates and will have the same mean squared error as the unscaled model Y=3Θ+W . In the equivalent Y′ model, we have a noise variance of 1/9 and therefore the mean squared error is

111+11/9=110.

Answer 5

how do you know that we have a noise variance of 1/9 ?

Answer 6

3/2

Answer 7

Since Y' is just Y scaled by a factor of 1/3, Y' carries the same information as Y, so that E[Θ∣Y]=E[Θ∣Y']. Thus, the alternative observation model Y′=Θ+(W/3) will lead to the same estimates and will have the same mean squared error as the unscaled model Y=3Θ+W. In the equivalent Y' model, we have a noise variance of 1/9 and therefore the mean squared error is:
1/((1/1)+(1/(1/9))) = 1/10