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In my self study of a statistics book, I came across a page that has confused me somewhat. I am already familiar with covariance matricies, (or maybe not!), and the author's explanation leaves me a little confused.

Here is the page in question:

enter image description here

My question is simply, why is that bottom right entry into the covariance equal to 1, when it fact it seems to me that it should be $a^2 + \sigma_n^2$.

So, I follow everything he is doing, but the bottom right entry to me, seems wrong. The bottom right entry is $var(z_2)$. Thus, by my estimates:

$ cov(z_2,z_2) = var(z_2) = \mathbb{E}(z_2^2) - (\mathbb{E}(z_2))^2 $

Thus, (assuming noise is zero mean, but not that I think it makes a difference anyway):

$ \begin{align} var(z_2) &= \mathbb{E}( (az_1 + n)^2) - 0 \\ &= \mathbb{E}( a^2z_1^2 + 2az_1 + n^2) \\ &= a^2 + \sigma_n^2 \end{align} $

(Sorry about the alignment I am not sure how to make the allignment of the equations nice).

Anyway, to me that should be the answer to the bottom right entry of the covariance matrix.

Am I missing something?

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    @Michael: Please check the timestamps of the edits and comments. Cheers.2014-02-19

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You've found a typo. (Good eye!)

Note, however, that every covariance matrix is symmetric, including, of course, the corrected version of the one you reference. This follows simply from the fact that the covariance matrix is defined as $ \mathbb E\big((\mathbf X - \mu)(\mathbf X - \mu)^T\big) $ for a random vector $\mathbf X \in \mathbb R^n$ with mean vector $\mu$ which satisfies the necessary moment conditions.

In this particular case, the text should have read $ \mathbf C(\mathbf z) = \left(\begin{matrix}1 & a \\ a & a^2 + \sigma_n^2\end{matrix}\right) \> . \tag{4.46} $