1.18 Machine learning lecture 19  (Page 14/15)

So concretely – oh, and let’s see. In the predict step it turns out that – so what I’m going to do is actually just outline the main steps of the Kalman filter. I won’t actually derive the algorithm and prove it’s correct. It turns out that, I don’t know, working out the actual proof of what I’m about to derive is probably significantly – it’s probably, I don’t know, about as hard, or maybe slightly easier, than many of the homework’s you’ve done all ready. So and you’ve done some pretty amazingly hard homework, so you can work out the proof for yourself. It’s just write out the main outlines and the conclusion of the algorithm. So for the acceptance of the vest T given Y1 after YT. If that is given by that then where – okay? So given ST, having computed the distribution ST given Y1 through YT – and computed the distribution of ST plus one given Y1 through YT as Gaussian, with this mean and this covariance, where you compute the mean and covariance using these two formulas. And just as a point of notation, right? I’m using ST and YT to denote the true states in observations. So the ST is the unknown true state. Okay? ST is whatever state this one is in and you actually don’t know what ST is because you don’t get to observe this.

And, in contrast, I’m using these things like ST given T, ST plus one given T, sigma T given T, and so on. These things are the results of your computations, right? So these things are actually things you compute. So I hope the notations are okay, but these – ST is the unknown true state, right? Whereas these things, ST equals one given T and so on, these are things that you compute inside your algorithm. Okay? So that was the predict step.

And in the update step, you find that – well, okay? And so that’s the updates of the Kalman filter where you compute this in terms of your ST given Y1 through YT. So after having performed the most recent Kalman filter update you find that, right? Your perceived distribution on the estimate of ST plus one, given all your observations so far, is that it’s Gaussian with mean given by this and variance given by that. So, informally, this thing ST plus one given T plus one is our best estimate for ST plus one, right? Given all the observations we’ve had up to that time. Okay?

And, again, the correctness of these equations – the fact that I’m actually computing this mean and covariance of this conditional Gaussian distribution, you can – I’ll leave you to sort of prove that at home if you want. Okay? I’m actually gonna put this together with LQR control in a second, but, so before I do that let me check if you’ve got questions about this? Actually let me erase the board while you take a look at that. Right. Okay. Any questions for Kalman filters? Yeah?

Student: How is it computationally less intensive than compute some drawing Gaussian distribution and then find the conditional –

Instructor (Andrew Ng) :Very quickly. Yeah, of course. So how is this less computationally intensive than the original method I talked about, right? So in the original method I talked about – wow, this is really back and forth. I said, let’s construct a Z, which was this huge Gaussian thing, right? And figure out what the mean and covariance matrix of Z is. So sigma will be like R – it’ll be – well, it’ll be roughly, right? A T by T matrix, right? This is actually – or the T by T is actually T times number of state variables plus number of observation variables by that. This is a huge matrix and as the number of times it increases sigma will become bigger and bigger. So the conditional and marginalization operations require things like computing the inverse of T or subsets of T. So the naïve way of doing this will cos on the order of TQ computation, if you do things naively, right? If – because inverting like a T by T matrix cos on the order of TQ, roughly.