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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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    $\begingroup$ It's a typo. Good eye. :-) (Though your title is funny. Every variance matrix is symmetric.) $\endgroup$
    – cardinal
    Dec 12, 2012 at 16:16
  • $\begingroup$ @cardinal Whew! Thank goodness! I thought I was bone-heading something very silly. :-) $\endgroup$
    – Spacey
    Dec 12, 2012 at 16:45
  • $\begingroup$ @cardinal: That's a full answer; you could post it as such so the question doesn't remain unanswered. $\endgroup$
    – joriki
    Dec 12, 2012 at 17:15
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    $\begingroup$ @Mohammad: A symmetric matrix is symmetric about the main diagonal. As cardinal pointed out, every covariance matrix is symmetric. The symmetry about the other diagonal that you're referring to in the title is unrelated to the term "symmetric matrix". If you really want to focus on the symmetry in the title, you could write "Why does this covariance matrix exhibit an additional symmetry?", but I don't think the symmetry is the main point here. Regarding the alignment, I edited the post, so you can look at the source to learn how to do it (right-click and select "Show Math As->TeX Commands"). $\endgroup$
    – joriki
    Dec 12, 2012 at 17:33
  • $\begingroup$ @joriki : Done. $\endgroup$
    – cardinal
    Dec 12, 2012 at 17:40

1 Answer 1

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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 $$ \mathbf C(\mathbf z) = \left(\begin{matrix}1 & a \\ a & a^2 + \sigma_n^2\end{matrix}\right) \> . \tag{4.46} $$

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