Timeline for Retrodiction / Specific filter to obtain initial state
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 13, 2019 at 10:32 | vote | accept | Daniel | ||
| Sep 12, 2019 at 13:13 | answer | added | Daniel | timeline score: 0 | |
| S Sep 5, 2019 at 23:02 | history | bounty ended | CommunityBot | ||
| S Sep 5, 2019 at 23:02 | history | notice removed | CommunityBot | ||
| S Aug 28, 2019 at 21:03 | history | bounty started | Daniel | ||
| S Aug 28, 2019 at 21:03 | history | notice added | Daniel | Draw attention | |
| Aug 27, 2019 at 10:33 | comment | added | Daniel | a) I've tried to understand the derivation of the Kalman smoother, but in vain. In the end, I need an analytical formula for $x_0$, which is why I was trying to derive it myself. b) I have added a second attempt to solve the problem, which seems to almost get me there, but I must be missing something small. Maybe you can spot it? c) And yes, I do have an observation $y_0$. It doesn't strike me as unusual. The difference is small anyway, waiting a while to make the first measurement just changes the expected initial variance. | |
| Aug 27, 2019 at 10:31 | history | edited | Daniel | CC BY-SA 4.0 |
added a second attempt in solving the problem.
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| Aug 26, 2019 at 23:56 | comment | added | Chris Haug | Yes, but it's easiest to derive recursively rather than all at once, first by computing forward each filtering distribution $p(x_t|y_1,...,y_t)$ from $p(x_{t-1}|y_1,...,y_{t-1})$, and then backwards the state posteriors $p(x_{t-1}|y_1,...,y_T)$ from $p(x_t|y_1,...,y_T)$. To be sure, do you really have an observation $y_0$ that corresponds to the state $x_0$ that you care about (as you've written)? Typically the "initial state" is the time right before the first observation, and inference about it is not quite the same. | |
| Aug 26, 2019 at 11:36 | comment | added | Daniel | Thanks for your comment, Chris. I'm not sure, I will check it out. If it is it should be derivable as I tried, right? The discussion on Wikipedia is not very clear to me, it would be nice to get some insight by deriving it for this example. | |
| Aug 25, 2019 at 18:53 | comment | added | Chris Haug | Am I misunderstanding your notation or is what you've called $P(\vec x|\vec y)$ not simply the result of the Kalman smoother? | |
| Aug 25, 2019 at 13:10 | review | First posts | |||
| Aug 25, 2019 at 14:43 | |||||
| Aug 25, 2019 at 13:07 | history | asked | Daniel | CC BY-SA 4.0 |