5 Updated state of the art
source | link

1. If the rows of $\mathbf{Y}$ are not independent, there is in general nothing one can say about the structure of $\mathbf{y}:=\mathrm{vec}(\mathbf{Y})$ other than $\mathbf{y}\sim N\left((\mathbf{I} \otimes \mathbf{X})\mathrm{vec}(\mathbf{B}), \mathbf{\Psi}\right)$ for some covariance matrix $\mathbf{\Psi}$, which is the classical GLS model.

2. Yes you can, given that you are willing to assume this more restrictive structure and that you also know the matrices. Regarding restrictions, you may notice that this decomposition implies that the autocovariance function for any one pixel is proportional to that of any other.

If you are indeed willing to make this assumption and know the matrices, then just set $\mathbf{\Psi} = \mathbf{\Omega}_n \otimes \mathbf{\Sigma}_d$ and your GLS estimator of $\mathbf{b} = \mathrm{vec}(\mathbf{B})$ is given by $\mathbf{Z}(\mathbf{Z}'\mathbf{\Psi}^{-1}\mathbf{Z})^{-1}\mathbf{Z}'\mathbf{\Psi}^{-1}\mathbf{y}$, where $\mathbf{Z} = \mathbf{I} \otimes \mathbf{X}$.

3. I believe your guess at this point is pretty much correct. The model you suggest for $\mathbf{e}$ is equivalent to assuming a matrix normal distribution for $\mathbf{E}$. This paper considers estimation of the component covariance matrices $\mathbf{\Omega}_n$ and $\mathbf{\Sigma}_d$ using maximum likelihood. Paywall link to published version in Journal of Multivariate Analysis, 2016.

A very short summary ofUpdate:

There is in fact a newer paper (here is an arxiv version) which settles the current state ofuniqueness and existence question for maximum likelihood estimation in the matrix normalthis model is given by their remark on page 6. Namely, suppose

Their Theorem 1 in section 4 gives the following result. Suppose we observe $N$ independent copies of the matrix normal variable $\mathbf{Y}$ and $\mathbf{X} = \mathbf{1}_n$. Then:

  1. If $N < \max(n/d, d/n) + 1$ the MLE of $\mathbf{\Psi}$ does not exist.
  2. If $N > nd$$N > d/n + n/d + 1$ the MLE does existexists and is unique almost surely.
  3. For all $N$ in between it is unknown whether$N \in [\max(n/d, d/n) + 1, d/n + n/d + 1]$ there are examples of when the MLE exists and when it does not, so no general statement can be made.

In particular, note that your case corresponds to having just $N=1$ observation.

1. If the rows of $\mathbf{Y}$ are not independent, there is in general nothing one can say about the structure of $\mathbf{y}:=\mathrm{vec}(\mathbf{Y})$ other than $\mathbf{y}\sim N\left((\mathbf{I} \otimes \mathbf{X})\mathrm{vec}(\mathbf{B}), \mathbf{\Psi}\right)$ for some covariance matrix $\mathbf{\Psi}$, which is the classical GLS model.

2. Yes you can, given that you are willing to assume this more restrictive structure and that you also know the matrices. Regarding restrictions, you may notice that this decomposition implies that the autocovariance function for any one pixel is proportional to that of any other.

If you are indeed willing to make this assumption and know the matrices, then just set $\mathbf{\Psi} = \mathbf{\Omega}_n \otimes \mathbf{\Sigma}_d$ and your GLS estimator of $\mathbf{b} = \mathrm{vec}(\mathbf{B})$ is given by $\mathbf{Z}(\mathbf{Z}'\mathbf{\Psi}^{-1}\mathbf{Z})^{-1}\mathbf{Z}'\mathbf{\Psi}^{-1}\mathbf{y}$, where $\mathbf{Z} = \mathbf{I} \otimes \mathbf{X}$.

3. I believe your guess at this point is pretty much correct. The model you suggest for $\mathbf{e}$ is equivalent to assuming a matrix normal distribution for $\mathbf{E}$. This paper considers estimation of the component covariance matrices $\mathbf{\Omega}_n$ and $\mathbf{\Sigma}_d$ using maximum likelihood. Paywall link to published version in Journal of Multivariate Analysis, 2016.

A very short summary of the current state of maximum likelihood estimation in the matrix normal model is given by their remark on page 6. Namely, suppose we observe $N$ copies of the matrix normal variable $\mathbf{Y}$ and $\mathbf{X} = \mathbf{1}_n$. Then:

  1. If $N < \max(n/d, d/n) + 1$ the MLE of $\mathbf{\Psi}$ does not exist.
  2. If $N > nd$ the MLE does exist.
  3. For all $N$ in between it is unknown whether the MLE exists.

In particular, note that your case corresponds to having just $N=1$ observation.

1. If the rows of $\mathbf{Y}$ are not independent, there is in general nothing one can say about the structure of $\mathbf{y}:=\mathrm{vec}(\mathbf{Y})$ other than $\mathbf{y}\sim N\left((\mathbf{I} \otimes \mathbf{X})\mathrm{vec}(\mathbf{B}), \mathbf{\Psi}\right)$ for some covariance matrix $\mathbf{\Psi}$, which is the classical GLS model.

2. Yes you can, given that you are willing to assume this more restrictive structure and that you also know the matrices. Regarding restrictions, you may notice that this decomposition implies that the autocovariance function for any one pixel is proportional to that of any other.

If you are indeed willing to make this assumption and know the matrices, then just set $\mathbf{\Psi} = \mathbf{\Omega}_n \otimes \mathbf{\Sigma}_d$ and your GLS estimator of $\mathbf{b} = \mathrm{vec}(\mathbf{B})$ is given by $\mathbf{Z}(\mathbf{Z}'\mathbf{\Psi}^{-1}\mathbf{Z})^{-1}\mathbf{Z}'\mathbf{\Psi}^{-1}\mathbf{y}$, where $\mathbf{Z} = \mathbf{I} \otimes \mathbf{X}$.

3. I believe your guess at this point is pretty much correct. The model you suggest for $\mathbf{e}$ is equivalent to assuming a matrix normal distribution for $\mathbf{E}$. This paper considers estimation of the component covariance matrices $\mathbf{\Omega}_n$ and $\mathbf{\Sigma}_d$ using maximum likelihood. Paywall link to published version in Journal of Multivariate Analysis, 2016.

Update:

There is in fact a newer paper (here is an arxiv version) which settles the uniqueness and existence question for maximum likelihood in this model.

Their Theorem 1 in section 4 gives the following result. Suppose we observe $N$ independent copies of the matrix normal variable $\mathbf{Y}$ and $\mathbf{X} = \mathbf{1}_n$. Then:

  1. If $N < \max(n/d, d/n) + 1$ the MLE of $\mathbf{\Psi}$ does not exist.
  2. If $N > d/n + n/d + 1$ the MLE exists and is unique almost surely.
  3. For $N \in [\max(n/d, d/n) + 1, d/n + n/d + 1]$ there are examples of when the MLE exists and when it does not, so no general statement can be made.

In particular, note that your case corresponds to having just $N=1$ observation.

4 edited body
source | link

1. If the rows of $\mathbf{Y}$ are not independent, there is in general nothing one can say about the structure of $\mathbf{y}:=\mathrm{vec}(\mathbf{Y})$ other than $\mathbf{y}\sim N\left((\mathbf{I} \otimes \mathbf{X})\mathrm{vec}(\mathbf{B}), \mathbf{\Psi}\right)$ for some covariance matrix $\mathbf{\Psi}$, which is the classical GLS model.

2. Yes you can, given that you are willing to assume this more restrictive structure and that you also know the matrices. Regarding restrictions, you may notice that this decomposition implies that the autocovariance function for any one pixel is proportional to that of any other.

If you are indeed willing to make this assumption and know the matrices, then just set $\mathbf{\Psi} = \mathbf{\Omega}_n \otimes \mathbf{\Sigma}_d$ and your GLS estimator of $\mathbf{b} = \mathrm{vec}(\mathbf{B})$ is given by $\mathbf{Z}(\mathbf{Z}'\mathbf{\Psi}^{-1}\mathbf{Z})^{-1}\mathbf{Z}'\mathbf{\Psi}^{-1}\mathbf{y}$, where $\mathbf{Z} = \mathbf{I} \otimes \mathbf{X}$.

3. I believe your guess at this point is pretty much correct. The model you suggest for $\mathbf{e}$ is equivalent to assuming a matrix normal distribution for $\mathbf{E}$. This paper considers estimation of the component covariance matrices $\mathbf{\Omega}_n$ and $\mathbf{\Sigma}_d$ using maximum likelihood. Paywall link to published version in Journal of Multivariate Analysis, 2016.

A very short summary of the current state of maximum likelihood estimation in the matrix normal model is given by their remark on page 6. Namely, suppose we observe $N$ copies of the matrix normal variable $\mathbf{Y}$ and $\mathbf{X} = \mathbf{1}_n$. Then:

  1. If $N < \max(n/d, d/n) + 1$ the MLE of $\mathbf{\Psi}$ does not exist.
  2. If $N > nd$ the MLE does exist.
  3. For all $N$ in between it is unknown whether the MLE exists.

In particular, note that your case corresponds to having just 1$N=1$ observation of the matrix normal variate.

1. If the rows of $\mathbf{Y}$ are not independent, there is in general nothing one can say about the structure of $\mathbf{y}:=\mathrm{vec}(\mathbf{Y})$ other than $\mathbf{y}\sim N\left((\mathbf{I} \otimes \mathbf{X})\mathrm{vec}(\mathbf{B}), \mathbf{\Psi}\right)$ for some covariance matrix $\mathbf{\Psi}$, which is the classical GLS model.

2. Yes you can, given that you are willing to assume this more restrictive structure and that you also know the matrices. Regarding restrictions, you may notice that this decomposition implies that the autocovariance function for any one pixel is proportional to that of any other.

If you are indeed willing to make this assumption and know the matrices, then just set $\mathbf{\Psi} = \mathbf{\Omega}_n \otimes \mathbf{\Sigma}_d$ and your GLS estimator of $\mathbf{b} = \mathrm{vec}(\mathbf{B})$ is given by $\mathbf{Z}(\mathbf{Z}'\mathbf{\Psi}^{-1}\mathbf{Z})^{-1}\mathbf{Z}'\mathbf{\Psi}^{-1}\mathbf{y}$, where $\mathbf{Z} = \mathbf{I} \otimes \mathbf{X}$.

3. I believe your guess at this point is pretty much correct. The model you suggest for $\mathbf{e}$ is equivalent to assuming a matrix normal distribution for $\mathbf{E}$. This paper considers estimation of the component covariance matrices $\mathbf{\Omega}_n$ and $\mathbf{\Sigma}_d$ using maximum likelihood. Paywall link to published version in Journal of Multivariate Analysis, 2016.

In particular, note that your case corresponds to having just 1 observation of the matrix normal variate.

1. If the rows of $\mathbf{Y}$ are not independent, there is in general nothing one can say about the structure of $\mathbf{y}:=\mathrm{vec}(\mathbf{Y})$ other than $\mathbf{y}\sim N\left((\mathbf{I} \otimes \mathbf{X})\mathrm{vec}(\mathbf{B}), \mathbf{\Psi}\right)$ for some covariance matrix $\mathbf{\Psi}$, which is the classical GLS model.

2. Yes you can, given that you are willing to assume this more restrictive structure and that you also know the matrices. Regarding restrictions, you may notice that this decomposition implies that the autocovariance function for any one pixel is proportional to that of any other.

If you are indeed willing to make this assumption and know the matrices, then just set $\mathbf{\Psi} = \mathbf{\Omega}_n \otimes \mathbf{\Sigma}_d$ and your GLS estimator of $\mathbf{b} = \mathrm{vec}(\mathbf{B})$ is given by $\mathbf{Z}(\mathbf{Z}'\mathbf{\Psi}^{-1}\mathbf{Z})^{-1}\mathbf{Z}'\mathbf{\Psi}^{-1}\mathbf{y}$, where $\mathbf{Z} = \mathbf{I} \otimes \mathbf{X}$.

3. I believe your guess at this point is pretty much correct. The model you suggest for $\mathbf{e}$ is equivalent to assuming a matrix normal distribution for $\mathbf{E}$. This paper considers estimation of the component covariance matrices $\mathbf{\Omega}_n$ and $\mathbf{\Sigma}_d$ using maximum likelihood. Paywall link to published version in Journal of Multivariate Analysis, 2016.

A very short summary of the current state of maximum likelihood estimation in the matrix normal model is given by their remark on page 6. Namely, suppose we observe $N$ copies of the matrix normal variable $\mathbf{Y}$ and $\mathbf{X} = \mathbf{1}_n$. Then:

  1. If $N < \max(n/d, d/n) + 1$ the MLE of $\mathbf{\Psi}$ does not exist.
  2. If $N > nd$ the MLE does exist.
  3. For all $N$ in between it is unknown whether the MLE exists.

In particular, note that your case corresponds to having just $N=1$ observation.

3 edited body
source | link

1. If the rows of $\mathbf{Y}$ are not independent, there is in general nothing one can say about the structure of $\mathbf{y}:=\mathrm{vec}(\mathbf{Y})$ other than $\mathbf{y}\sim N\left((\mathbf{I} \otimes \mathbf{X})\mathrm{vec}(\mathbf{B}), \mathbf{\Psi}\right)$ for some covariance matrix $\mathbf{\Psi}$, which is the classical GLS model.

2. Yes you can, given that you are willing to assume this more restrictive structure and that you also know the matrices. Regarding restrictions, you may notice that this decomposition implies that the autocovariance function for any one pixel is proportional to that of any other.

If you are indeed willing to make this assumption and know the matrices, then just set $\mathbf{\Psi} = \mathbf{\Omega}_n \otimes \mathbf{\Sigma}_n$$\mathbf{\Psi} = \mathbf{\Omega}_n \otimes \mathbf{\Sigma}_d$ and your GLS estimator of $\mathbf{b} = \mathrm{vec}(\mathbf{B})$ is given by $\mathbf{Z}(\mathbf{Z}'\mathbf{\Psi}^{-1}\mathbf{Z})^{-1}\mathbf{Z}'\mathbf{\Psi}^{-1}\mathbf{y}$, where $\mathbf{Z} = \mathbf{I} \otimes \mathbf{X}$.

3. I believe your guess at this point is pretty much correct. The model you suggest for $\mathbf{e}$ is equivalent to assuming a matrix normal distribution for $\mathbf{E}$. This paper considers estimation of the component covariance matrices $\mathbf{\Omega}_n$ and $\mathbf{\Sigma}_n$$\mathbf{\Sigma}_d$ using maximum likelihood. Paywall link to published version in Journal of Multivariate Analysis, 2016.

In particular, note that your case corresponds to having just 1 observation of the matrix normal variate.

1. If the rows of $\mathbf{Y}$ are not independent, there is in general nothing one can say about the structure of $\mathbf{y}:=\mathrm{vec}(\mathbf{Y})$ other than $\mathbf{y}\sim N\left((\mathbf{I} \otimes \mathbf{X})\mathrm{vec}(\mathbf{B}), \mathbf{\Psi}\right)$ for some covariance matrix $\mathbf{\Psi}$, which is the classical GLS model.

2. Yes you can, given that you are willing to assume this more restrictive structure and that you also know the matrices. Regarding restrictions, you may notice that this decomposition implies that the autocovariance function for any one pixel is proportional to that of any other.

If you are indeed willing to make this assumption and know the matrices, then just set $\mathbf{\Psi} = \mathbf{\Omega}_n \otimes \mathbf{\Sigma}_n$ and your GLS estimator of $\mathbf{b} = \mathrm{vec}(\mathbf{B})$ is given by $\mathbf{Z}(\mathbf{Z}'\mathbf{\Psi}^{-1}\mathbf{Z})^{-1}\mathbf{Z}'\mathbf{\Psi}^{-1}\mathbf{y}$, where $\mathbf{Z} = \mathbf{I} \otimes \mathbf{X}$.

3. I believe your guess at this point is pretty much correct. The model you suggest for $\mathbf{e}$ is equivalent to assuming a matrix normal distribution for $\mathbf{E}$. This paper considers estimation of the component covariance matrices $\mathbf{\Omega}_n$ and $\mathbf{\Sigma}_n$ using maximum likelihood. Paywall link to published version in Journal of Multivariate Analysis, 2016.

In particular, note that your case corresponds to having just 1 observation of the matrix normal variate.

1. If the rows of $\mathbf{Y}$ are not independent, there is in general nothing one can say about the structure of $\mathbf{y}:=\mathrm{vec}(\mathbf{Y})$ other than $\mathbf{y}\sim N\left((\mathbf{I} \otimes \mathbf{X})\mathrm{vec}(\mathbf{B}), \mathbf{\Psi}\right)$ for some covariance matrix $\mathbf{\Psi}$, which is the classical GLS model.

2. Yes you can, given that you are willing to assume this more restrictive structure and that you also know the matrices. Regarding restrictions, you may notice that this decomposition implies that the autocovariance function for any one pixel is proportional to that of any other.

If you are indeed willing to make this assumption and know the matrices, then just set $\mathbf{\Psi} = \mathbf{\Omega}_n \otimes \mathbf{\Sigma}_d$ and your GLS estimator of $\mathbf{b} = \mathrm{vec}(\mathbf{B})$ is given by $\mathbf{Z}(\mathbf{Z}'\mathbf{\Psi}^{-1}\mathbf{Z})^{-1}\mathbf{Z}'\mathbf{\Psi}^{-1}\mathbf{y}$, where $\mathbf{Z} = \mathbf{I} \otimes \mathbf{X}$.

3. I believe your guess at this point is pretty much correct. The model you suggest for $\mathbf{e}$ is equivalent to assuming a matrix normal distribution for $\mathbf{E}$. This paper considers estimation of the component covariance matrices $\mathbf{\Omega}_n$ and $\mathbf{\Sigma}_d$ using maximum likelihood. Paywall link to published version in Journal of Multivariate Analysis, 2016.

In particular, note that your case corresponds to having just 1 observation of the matrix normal variate.

2 added 382 characters in body
source | link
1
source | link