Estimation of unknown vector's amplitude with Gaussian noise
Estimation of unknown vector's amplitude with Gaussian noise
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user1834069 · External communityPost link
External question — Cross Validated Stack Exchange
Author: user1834069
Original post: https://stats.stackexchange.com/questions/91274
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I have the following model:
y
= P
v
+
n
Where y is the vector of observations, v is a unit vector and n is a Gaussian random noise whose covariance matrix is the identity matrix. P is a positive scalar. All the vectors are reals of size N.
I'd like to estimate the amplitude $P^2$ in the mse sense. All I know about v is that it's normalized and that each of its values has an equal chance of being positive or negative.
I can use the maximum likelihood for both P and v, getting the LS solution(in this case it's just setting the estimator to y), and then squaring it to get an estimate for $P^2$. A better estimate would be $y^T y -N$ which is unbiased and has a lower mse. Trading some variance for bias, we can improve that by setting the estimator to zero if $y^T y$ is smaller than N.
Is there a way to do better than this? If not, how can I prove this is the optimal solution ?
Can the ML be used in this case without directly estimating v (which is not needed)?
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Robert Bassett · External communityPost link
External answer — Cross Validated Stack Exchange
Author: Robert Bassett
Original post: https://stats.stackexchange.com/a/567750
License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/
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This is an old question but I'll throw together an answer anyway.
In statistical signal processing, its common to estimate
$P^2$
via the leading eigenvalue of the covariance of
$\mathbf{y}$
. The argument goes as follows.
We assume that
$P$
and
$\mathbf{n}$
are mean zero, that
$\mathbf{v}$
is fixed, and that
$P$
and
$\mathbf{n}$
are uncorrelated. Then some algebra shows that the covariance of
$\textbf{y}$
is
$$E[\mathbf{y} \mathbf{y}^{T}] = E[P^2] \mathbf{v} \mathbf{v}^{T} + \mathbf{I}.$$
The
$N$
eigenvalues of this matrix in descending order are
$E[P^2] + 1, 1, ..., 1$
. The leading eigenvector is
$\pm \mathbf{v}$
and the remaining eigenvectors are an orthogonal basis of the subspace in
$\mathbb{R}^{N}$
perpendicular to
$\mathbf{v}$
. Therefore to get
$E[P^2]$
we take the leading eigenvalue of
$E[\mathbf{y} \mathbf{y}^{T}]$
and subtract 1.
Finally, we use a plug-in estimator for the parameters above by estimating
$E[P^2]$
as one less than the leading eigenvalue of the
sample covariance matrix
of
$\mathbf{y}$
.
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Quoted from Forex.com.bd-Editorial External answer — Cross Validated Stack Exchange Author: Robert Bassett Source score (net votes, not local likes): 2 Original post: https://stats.stackexchange.com/a/567750 License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/ Adaptation: HTML converted to plain text; contact email addresses removed. This is an old question but I'll throw together an answer anyway. In statistical signal processing, its common to estimate $P^2$ via the leading eigenvalue of the covariance of $\mathbf{y}$ . The argument goes as follows. We assume that $P$ and $\mathbf{n}$ are mean zero, that $\mathbf{v}$ is fixed, and that $P$ and $\mathbf{n}$ are uncorrelated. Then some algebra shows that the covariance of $\textbf{y}$ is $$E[\mathbf{y} \mathbf{y}^{T}] = E[P^2] \mathbf{v} \mathbf{v}^{T} + \mathbf{I}.$$ The $N$ eigenvalues of this matrix in descending order are $E[P^2] + 1, 1, ..., 1$ . The leading eigenvector is $\pm \mathbf{v}$ and the remaining eigenvectors are an orthogonal basis of the subspace in $\mathbb{R}^{N}$ perpendicular to $\mathbf{v}$ . Therefore to get $E[P^2]$ we take the leading eigenvalue of $E[\mathbf{y} \mathbf{y}^{T}]$ and subtract 1. Finally, we use a plug-in estimator for the parameters above by estimating $E[P^2]$ as one less than the leading eigenvalue of the sample covariance matrix of $\mathbf{y}$ .
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