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I want to write the expression for covariance matrix of following linear model

$$x= As+ v$$ Where $A\in R^{N\times p}, s\in R^{p\times 1}$, $v$ consists of MVN noise only with mean $Bz$ and covariance $R=\sigma^2I$, $v\sim \mathcal{N}[Bz, \sigma^2I ]$: $B\in R^{N\times t}, \phi\in R^{t\times 1}, t<N-p$. Now the matrices $A$ and $B$ are linearly independent, i.e $A^TB \neq 0 $ and each matrix is full rank.

Although, I know how to write the covariance matrix expression for following signal $x$ which obeys linear subspace model

$$x=As+n$$ Where $s$ is independent of noise,then covariance $C=AE[ss^T]A^T+\sigma^2I$ for Gaussian noise of zero mean. What if I write the covariance matrix expression for first model as $C=AE[ss^T]A^T+ BE[zz^T]B^T+\sigma^2I$, is it the right way ?

Please suggest as I can not find a reference.

Appreciate your suggestions!

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  • $\begingroup$ Much is unstated in your question; presumably $s$ and $z$ are zero-mean, and you imply that $s$ and $n$ at least are independent; you should make such things explicit. Note that Var(X+Y) = Var(X) + Var(Y) + Cov(X,Y) + Cov(Y,X) (or see Wikipedia for properties of covariance) -- so it depends on whether $E[sz^T] \neq 0$ (or if $z$ is correlated with $n$); if any of those are the case, you'll have more terms. I think this is a duplicate, so I'm trying to locate one. $\endgroup$ – Glen_b -Reinstate Monica Feb 1 '17 at 3:53
  • $\begingroup$ @Glen_b, I have edited my question and added detail. Can you please have a look again? $\endgroup$ – Vendetta Feb 1 '17 at 4:50
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Your statement for the distribution of $v$ should be $v|z\sim\mathcal{N}[Bz,\sigma^2I]$. That is, you specified conditional distribution of $v$ on $z$. If you condition on $z$, the first model is as easy as the second model to describe. The conditional variance of $x$ given $z$ is $$Var(x|z)=A\mathbb{E}[ss^T|z]A^T + \sigma^2I$$ Since you want the unconditional variance, you can apply the familiar formula there. $$Var(x)=E[Var(x|z)]+Var(E[x|z]) = A\mathbb{E}[ss^T]A^T + \sigma^2I + Var(A\mathbb{E}[s|z]+Bz)$$ Similar to what you got, right? Only if you assume that $s$ and $z$ are independent, or more directly $E[s|z]$ is constant (which is a weaker condition), then your formula is right.

[EDIT] You can't do much about the term $Var(A\mathbb{E}[s|z]+Bz)$. You can try to expand as $$AVar(\mathbb{E}[s|z])A^T + ACov(\mathbb{E}[s|z],z)+ Cov(z,\mathbb{E}[s|z])A^T + B\mathbb{E}[zz^T]B^T$$ which has the term you came up with. This also shows that a sufficient condition for your solution being right is that $\mathbb{E}[s|z]=C$, where $C$ is a constant.

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  • $\begingroup$ What this last is representing $Var(AE[s|z]+Bz)$, is it equivalent to what I wrote $BE[zz^T]B^T$? Can you please elaborate a little more? $\endgroup$ – Vendetta Feb 1 '17 at 10:25
  • $\begingroup$ Please see the edit. $\endgroup$ – Julius Feb 1 '17 at 17:20

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