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I'm working through a pair of papers on Simultaneous Localization and Mapping and I'm having trouble with some of the notation as I lack some formal math education.

The papers can be found
here: http://citeseerx.ist.psu.edu/viewdoc/summary?doi=
and here http://robots.stanford.edu/papers/thrun.seif.pdf

$P_{k|k}$ and $\Sigma_{t}^{-1}$ are both covariance matrices.

What does the $_{k|k}$ mean? I recall from probability, that P(y|x) means the probability of y given x, but that doesn't seem to make sense here.

With $\Sigma_{t}^{-1}$ I thought that $\sum$ is usually used for summation (and I initially confused it with summation!) Is there any significance to $\sigma$ being used to represent the covariance matrix or is it just historical accident?

Some other questions raised while formulating this question (answered in the comments below):

Later, on page 6, there is a formula $P(x_{2} | x_1^{(i)})$. Here I don't know what the superscript $^{(i)}$ represents.

Answer: $x_1^{(i)} \sim P(x_1)$ is the ith sample from the distribution $P(x_1)$

The information vector $b_{t} = \mu_{t}^{T}H_{t}$ looks like the mean times the information matrix, but I thought the mean was a scalar value, so I don't understand the Transpose symbol.

Answer: $\mu_{t}$ is a vector because it is the mean of the $\zeta$

Thanks for everyone who's helped me refine these questions!

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+1 for remembering to mention the paper. –  J. M. Dec 6 '11 at 1:34
@t.b. I was sent here from MO as it's a more appropriate forum. The MO post is now deleted. –  munkhd Dec 6 '11 at 2:59
There were a few answers in the comments of the MO question. You could have incorporated them in your question above. –  Joel Reyes Noche Dec 6 '11 at 3:00
Okay, I removed the crosspost notification. Here's the still relevant part of BR's comment on MO: "if you take another look at the papers, you will see that the notations are clearly defined ($P_{k∣k}$ and $\Sigma_t$ are covariance matrices and $x^{(i)}_1$ is the $i$-th sample from a distribution)." –  t.b. Dec 6 '11 at 3:08
In the paper of Thrun et al., $\mu_t$ is a vector (because $\xi_t$ is a vector). –  Joel Reyes Noche Dec 6 '11 at 3:12

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