# Marginal Covariance of State Vector in a Linear Gaussian State Space Model

Paper: A Unifying Review of Linear Gaussian Models by Roweis & Ghahramani

The generative model is the typical state space model written as \begin{align} \text{state transition equation: }{\bf x}_t &= {\bf A} {\bf x}_{t-1} + {\bf w}_t && {\bf w}_t \sim \mathcal{N} \left( {\bf 0}, {\bf Q} \right) \\ \text{observation equation: }{\bf y}_t &= {\bf C} {\bf x}_t + {\bf v}_t && {\bf v}_t \sim \mathcal{N} \left( {\bf 0}, {\bf R} \right) \end{align} where $${\bf A}$$ is the $$k \times k$$ state transition matrix and $${\bf C}$$ is the $$p \times k$$ observation matrix.

In the paper on. page 2, the authors write

Notice that there is degeneracy in the model: all of the structure in the matrix $$\bf Q$$ can be moved into the matrices $$\bf A$$ and $$\bf C$$. This means we can without loss generality work with models in which $$\bf Q$$ is the identity matrix.

There is a footnote associated with the passage and it reads

In particular, since is it a covariance matrix, $$\bf Q$$ is symmetric positive semi-definite and thus can be diagonalized to the form $$\bf E \Lambda E^{\top}$$ (where $$\bf E$$ is the rotation matrix of eigenvectors and $${\bf \Lambda}$$ is a diagonal matrix of eigenvalues). Thus for any model in which $$\bf Q$$ is not the identity matrix, we can generate an exactly equivalent model using a new state vector $${\bf x}' = {\bf \Lambda}^{-1/2} {\bf E}^{\top} {\bf x}$$ with $${\bf A}' = ( {\bf \Lambda}^{-1/2} {\bf E}^{\top} ) \, {\bf A} \, ( {\bf E} {\bf \Lambda}^{1/2} )$$ and $${\bf C}' = {\bf C} ( {\bf E} {\bf \Lambda}^{1/2})$$ such that the new covariance of $${\bf x}'$$ is the identity matrix: $${\bf Q}' = {\bf I}$$.

Question: How is the new covariance of $$\bf x'$$ equal to the identity matrix?

\begin{align} \text{Var} ({\bf x}') = {\bf \Lambda}^{-1/2} {\bf E}^{\top} \text{Var} ({\bf x}) {\bf E} {\bf \Lambda}^{-1/2} \end{align} equals the identity matrix if $$\text{Var} ({\bf x})$$ is the identity matrix. What in the model tells use that this is the case?

Update Mar 27: I believe the authors are referring to the conditional covariance in the footnote.

From the generative model we can see that the conditional distribution of $${\bf x}_t \, \vert \, {\bf x}_{t-1}$$ is Gaussian with conditional mean $${\bf A} {\bf x}_{t-1}$$ and conditional covariance $${\bf Q}$$.

Using the transformed state vector, the new generative model is now \begin{align} \text{new state transition equation: }{\bf x}'_t &= {\bf A}' {\bf x}'_{t-1} + {\bf w}'_t && {\bf w}'_t \sim \mathcal{N} \left( {\bf 0}, {\bf I} \right) \\ \text{new observation equation: }{\bf y}_t &= {\bf C}' {\bf x}'_t + {\bf v}_t && {\bf v}_t \sim \mathcal{N} \left( {\bf 0}, {\bf R} \right). \end{align}

Thus, the conditional distribution of $${\bf x}'_t \, \vert \, {\bf x}'_{t-1}$$ is Gaussian with conditional mean $${\bf A}' {\bf x}'_{t-1}$$ and conditional covariance $${\bf I}$$.

• yep +1 for the edit--multiply both sides of the state equation by ${\bf \Lambda}^{-1/2} {\bf E}^{\top}$ and rewrite the $\mathbf{x}$s as $\mathbf{x}'$s – Taylor Mar 28 at 2:01

## 1 Answer

I think you have it backwards. $$\bf Q$$, not $$\bf I$$ is the variance of $$\bf x$$.

The intended result can become clearer from some quick matrix multiplications to rearrange your equation.

\begin{align} \text{Var} ({\bf x}') &= {\bf \Lambda}^{-1/2} {\bf E}^{\top} \text{Var} ({\bf x}) {\bf E} {\bf \Lambda}^{-1/2} \\ {\bf \Lambda}^{1/2} \text{Var} ({\bf x}') &= {\bf E}^{\top} \text{Var} ({\bf x}) {\bf E} {\bf \Lambda}^{-1/2} \\ {\bf E}{\bf \Lambda}^{1/2} \text{Var} ({\bf x}') &= \text{Var} ({\bf x}) {\bf E} {\bf \Lambda}^{-1/2} \\ &\vdots \\ {\bf E}{\bf \Lambda}^{1/2} \text{Var} ({\bf x}') {\bf \Lambda}^{1/2} {\bf E}^{\top} &= \text{Var} ({\bf x}) \\ \end{align}

We know from the given noise process that $$\text{Var} ({\bf x}) = {\bf Q}$$. Now if we wager that $$\text{Var} ({\bf x'}) = {\bf I}$$, we see that the left-hand side of the equation becomes our eigendecomposition of $${\bf Q}$$, so the two sides are equal. That is, the variance of $$\bf x'$$ is $$\bf I$$ when the variance of $$\bf x$$ is $$\bf Q$$.

Alternatively, you could go the direct route.

\begin{align} \text{Var} ({\bf x}') &= {\bf \Lambda}^{-1/2} {\bf E}^{\top} \text{Var} ({\bf x}) {\bf E} {\bf \Lambda}^{-1/2} \\ \text{Var} ({\bf x}') &= {\bf \Lambda}^{-1/2} {\bf E}^{\top} {\bf Q} {\bf E} {\bf \Lambda}^{-1/2} \\ \text{Var} ({\bf x}') &= {\bf \Lambda}^{-1/2} {\bf E}^{\top} {\bf E} {\bf \Lambda} {\bf E}^{\top} {\bf E} {\bf \Lambda}^{-1/2} \\ \text{Var} ({\bf x}') &= {\bf \Lambda}^{-1/2} {\bf \Lambda} {\bf \Lambda}^{-1/2} \\ \text{Var} ({\bf x}') &= {\bf I} \\ \end{align}

• I am having a hard time seeing how $\text{Var} ({\bf x}) = {\bf Q}$. I know that $\text{Var} ({\bf w}) = {\bf Q}$ by the way the generative model is defined. How does it follow that they share the same variance? – SOULed_Outt Mar 27 at 2:26
• The conditional variance $\text{Var} ({\bf x}_t \, \vert \, {\bf x}_{t-1} ) = {\bf Q}$. I am certain that the marginal variance $\text{Var} ({\bf x}_t ) = {\bf Q} + \text{something}$ and is only equal to the conditional variance if you make additional assumptions about the model. – SOULed_Outt Mar 27 at 18:23
• @SOULed_Outt the marginal variance doesn't always exist....unless you have a stationary chain, in which case it usually isn't equal to the conditional covariance matrix – Taylor Mar 28 at 2:00