Timeline for How can the Kalman filter be adapted to handle binary observations?
Current License: CC BY-SA 3.0
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| Dec 9, 2016 at 20:54 | comment | added | 8one6 | I'm not sure I see how there is a difference other than the shape of the error distribution. The true state variable is p. I measure p, but i get back a noisy reading. The mean of my reading is p. But the distribution around p is not normal (it is bimodal). So I see this as a potential issue with distribution shape, rather than a broader philosophical difference. No? | |
| Dec 9, 2016 at 15:13 | comment | added | Roger Labbe | I have no idea if the KF can be extended to your way of thinking - I work with them in Newtonian systems, where what I say very much applies. Particle filters do not require assumptions of linear systems or gaussian distributions, so if this is a fruitful line of thinking perhaps you'd be better served with one of them? | |
| Dec 9, 2016 at 15:11 | comment | added | Roger Labbe | Sorry, I hit return. In the update step you work in measurement space, which means the measurement covariance is the error in the measurement of the flip, not the error in the sampling from the state. | |
| Dec 9, 2016 at 15:07 | comment | added | Roger Labbe | I see what you are saying. In your case p is a hidden variable, and you regard a flip as a probabilistically distributed observation of p. But measurement noise in a KF is the noise in the observed variable. I use a GPS to measure position, with error, then I can compute an unobserved variable, such as velocity. In your case the observed variable is the current flip, which you measure perfectly. | |
| Dec 8, 2016 at 17:39 | comment | added | 8one6 | Put another way, my observations are not error-free...just they are distributed "very-non-normally" around their true value. | |
| Dec 8, 2016 at 14:05 | comment | added | 8one6 |
I don't agree with the statement that I have no measurement noise. I flip a coin. At that instant, it has a true probability p of coming up heads (which I do not know). It comes up heads. That doesn't mean p=1.0. The outcome of the single coin toss should be assumed to be a draw from a probability distribution whose parameters are unknown and which we are trying to infer.
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| Dec 8, 2016 at 13:44 | review | First posts | |||
| Dec 8, 2016 at 14:01 | |||||
| Dec 8, 2016 at 13:40 | history | answered | Roger Labbe | CC BY-SA 3.0 |