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I came across the following statement

$cov(E(\mathbf{z}|y))$ is degenerate in all direction orthogonal to $Span(\mathbf{x}_1, ...,\mathbf{x}_K)$

Vector $\mathbf{z}$ is a centred random vector of size $p$, $y$ is a scalar random variable and the $\mathbf{x}_k$ are $K$ vectors of size $p$.

Does it mean that any vector orthogonal to $Span(\mathbf{x}_1, ...,\mathbf{x}_K)$ is in the kernel of the covariance matrix of $E(\mathbf{z}|y)$?

How would you interpret it?

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2 Answers 2

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Yes.

Plot

This figure shows the situation for $p=2$ where the span of $\{\mathbf{x}_1, \ldots, \mathbf{x}_K\}$ is one-dimensional, shown as a red line through the origin, and the orthogonal space--the kernel of the covariance matrix--also is one-dimensional, shown as a dashed gray line through the origin. Data are shown as points on the red line.

Evidently, the data exhibit no variation in directions parallel to the orthogonal space.

When $\mathbf{z}|y$ is a random variable, a similar picture and the same interpretation hold. Now, any realization of $\mathbf{z}|y$ must lie on the red line. No two realizations can differ by any nonzero element of the orthogonal space.

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In this case the covariance matrix is singular, i.e. it is not of full rank. The directions associated with the zero eigenvalues are degenerate.

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