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Derive Yourself a Kalman Filter

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Re: Derive Yourself a Kalman Filter

#31

Post by your-nanny »

gdy wrote: post #28
Waterluvian wrote: post #2

"It's really quite simple. Here, let me show you all these formulas..." This reminds me of last week. I bought a house and want to get into woodworking so I looked up intro videos on YouTube. "It's easy. Just follow me over to this table saw and router and planer and all these other tools you don't own." I know I'm not being fair. But a recurring frustration I have is when experts claim it's easy or simple or for beg…

"it can be insulting" Insulting, really? Somebody cares to try to explain you something and the word you can find is "insulting"? It's like everyone owes it to you to chew knowledge in small enough bits that you can swallow them without any effort. What a kindergarden.

Well, I dont agree with OP but i think you miss the mark. it's not deigning to provide an explanation that would be insulting, but the (apparent) promise of it being an accessible and intuitive explanation, and quite obviously not being one that OP considered insulting.

Re: Derive Yourself a Kalman Filter

#32

Post by your-nanny »

SiempreViernes wrote: post #18
Waterluvian wrote: post #2

"It's really quite simple. Here, let me show you all these formulas..." This reminds me of last week. I bought a house and want to get into woodworking so I looked up intro videos on YouTube. "It's easy. Just follow me over to this table saw and router and planer and all these other tools you don't own." I know I'm not being fair. But a recurring frustration I have is when experts claim it's easy or simple or for beg…

I don't know man, the link you clicked on says " Derive Yourself a Kalman Filter", so complaining about a formal treatment seems misguided rather than simply unfair.

you are assuming that the reader is familiar with this specific understanding of "derive". someone not familiar with formal methods wouldn't.

Re: Derive Yourself a Kalman Filter

#33

Post by richajak »

A few months ago I try to use Kalman Filter as one of the filters to stabilize the GPS from mobile phones due to various factors: loss of signal, jumpy stationary coordinate, fake GPS, etc. I found that Kalman Filter was not giving me satisfactory result, adjusting its parameters many times without further improvement. Finally, I just need to write my own simple filter, without fancy math filters (kalman/etc) as I can differentiate the speed of car/motorbike/pedestrian, distance of jumpy signal (less than 20 meters in four directions), accuracy of gps signal in the city/suburb, etc.

Re: Derive Yourself a Kalman Filter

#34

Post by mlevental »

your-nanny wrote: post #32
SiempreViernes wrote: post #18

Earlier quoted context omitted.

I don't know man, the link you clicked on says " Derive Yourself a Kalman Filter", so complaining about a formal treatment seems misguided rather than simply unfair.

you are assuming that the reader is familiar with this specific understanding of "derive". someone not familiar with formal methods wouldn't.

what other understanding of derive is there???

Re: Derive Yourself a Kalman Filter

#35

Post by gdy »

your-nanny wrote: post #31
gdy wrote: post #28

Earlier quoted context omitted.

"it can be insulting" Insulting, really? Somebody cares to try to explain you something and the word you can find is "insulting"? It's like everyone owes it to you to chew knowledge in small enough bits that you can swallow them without any effort. What a kindergarden.

Well, I dont agree with OP but i think you miss the mark. it's not deigning to provide an explanation that would be insulting, but the (apparent) promise of it being an accessible and intuitive explanation, and quite obviously not being one that OP considered insulting.

I understood OP exactly as you said. This doesn't justify OP's whining, not one bit.

Re: Derive Yourself a Kalman Filter

#36

Post by srean »

bonoboTP wrote: post #24
kgwgk wrote: post #22

Earlier quoted context omitted.

> it may be good to know that it is not fully rigorous What is the problem with A|B=b being a random variable? (Apart from you unfamiliarity with the concept, I mean.) Edit: I don’t say there are no problems, I ask what do you think the problem is? There is no problem in the discrete case. In the continuous setting things are indeed more complicated (but if the limiting process is well defined there are no issues). N…

A random variable is different concept from a distribution. For me personally it is helpful to keep them separate, but I can see that others may not care about the complete conceptual picture. In the PDF file linked above I can see conditional probabilities, conditional distributions and conditional expectation etc, which are all valid and rigorous. I can see that the author thinks it's a good idea to merge these int…

I think it will help if you think in terms of conditioning on (for example, a coarser sigma algebra). You would get another random variable that is measurable on the sigma algebra you conditioned on. If that is coarser so would be the new function you obtained by conditioning.

Re: Derive Yourself a Kalman Filter

#37

Post by bonoboTP »

srean wrote: post #36
bonoboTP wrote: post #24

Earlier quoted context omitted.

A random variable is different concept from a distribution. For me personally it is helpful to keep them separate, but I can see that others may not care about the complete conceptual picture. In the PDF file linked above I can see conditional probabilities, conditional distributions and conditional expectation etc, which are all valid and rigorous. I can see that the author thinks it's a good idea to merge these int…

I think it will help if you think in terms of conditioning on (for example, a coarser sigma algebra). You would get another random variable that is measurable on the sigma algebra you conditioned on. If that is coarser so would be the new function you obtained by conditioning.

Let's talk about a fair dice roll to make it concrete, and let the rolled number be X and let the event that we rolled an even number be E. P(X=6|E) = 1/3. P(X|E) is a distribution where 1,3,5 has 0 probability mass and 2,4,6 have 1/3 each.

If we consider X|E as a random variable, what is its value if we roll an odd number? Undefined? What does that mean? Random variables always have some value.

Sure you can build a new event space (sigma algebra) but then you can't use random variables over the original one.

Let's consider two independent rolls, X and Y. You can't compute the joint distribution P(Y, (X|E)), it just doesn't make sense as the two "variables" are defined over different spaces. Note that this is not the same as P(X,Y | E). The latter is simple a conditional probability, without any concept "conditional random variables".

Again, this is totally obvious to people who have experience with probabilities, but could be confusing to students. Such cases are where students who try to understand the details may be left more confused than students who just want to get the main idea.

Re: Derive Yourself a Kalman Filter

#38

Post by kgwgk »

bonoboTP wrote: post #24
kgwgk wrote: post #22

Earlier quoted context omitted.

> it may be good to know that it is not fully rigorous What is the problem with A|B=b being a random variable? (Apart from you unfamiliarity with the concept, I mean.) Edit: I don’t say there are no problems, I ask what do you think the problem is? There is no problem in the discrete case. In the continuous setting things are indeed more complicated (but if the limiting process is well defined there are no issues). N…

A random variable is different concept from a distribution. For me personally it is helpful to keep them separate, but I can see that others may not care about the complete conceptual picture. In the PDF file linked above I can see conditional probabilities, conditional distributions and conditional expectation etc, which are all valid and rigorous. I can see that the author thinks it's a good idea to merge these int…

> Practically, if you have two random variables then you can take their joint distribution.

If they are defined in the same sample space.

> a "conditional random variable" does not correspond to any subset of the event space

I would say it's exactly the other way around, the domain of a "conditional random variable" is a subset of the domain of the "unconditioned" random variable (the subset where the conditioning holds).

Re: Derive Yourself a Kalman Filter

#39

Post by srean »

bonoboTP wrote: post #37
srean wrote: post #36

Earlier quoted context omitted.

I think it will help if you think in terms of conditioning on (for example, a coarser sigma algebra). You would get another random variable that is measurable on the sigma algebra you conditioned on. If that is coarser so would be the new function you obtained by conditioning.

Let's talk about a fair dice roll to make it concrete, and let the rolled number be X and let the event that we rolled an even number be E. P(X=6|E) = 1/3. P(X|E) is a distribution where 1,3,5 has 0 probability mass and 2,4,6 have 1/3 each. If we consider X|E as a random variable, what is its value if we roll an odd number? Undefined? What does that mean? Random variables always have some value. Sure you can build a…

Sure you can. The TLDR would be "piecewise constant projection"

I think picking up a standard graduate probability book will clear this up better than any long comment trail. There are no problems defining a coarser sigma algebra using an original one and then defining a function measurable on the new sigma algebra. Note this continues to be an r.v. in the original space as meaurability is preserved. A consistent definition the values of the conditioned r.v. would be the piecewise constant approximation of the original r.v. over the indivisible elements of the coarser sigma algebra.

Let me try another route.

You seem to be accepting of a conditional expectation. Now what is a conditional expectation if not a function. Now all we need is that function be measurable with respect to the new sigma algebra, thats ensured byconstruction. Hope it helped some

Re: Derive Yourself a Kalman Filter

#40

Post by kgwgk »

bonoboTP wrote: post #37
srean wrote: post #36

Earlier quoted context omitted.

I think it will help if you think in terms of conditioning on (for example, a coarser sigma algebra). You would get another random variable that is measurable on the sigma algebra you conditioned on. If that is coarser so would be the new function you obtained by conditioning.

Let's talk about a fair dice roll to make it concrete, and let the rolled number be X and let the event that we rolled an even number be E. P(X=6|E) = 1/3. P(X|E) is a distribution where 1,3,5 has 0 probability mass and 2,4,6 have 1/3 each. If we consider X|E as a random variable, what is its value if we roll an odd number? Undefined? What does that mean? Random variables always have some value. Sure you can build a…

> If we consider X|E as a random variable, what is its value if we roll an odd number? Undefined? What does that mean? Random variables always have some value.

Random variables have some value on their domain, and for the random variable X | E=1 the sample space is restricted to the elementary events {2,4,6} which conform the composite event E=1. The original sample space is partitioned in the subspaces {1,3,5} and {2,4,6} when we condition on the values of the random variable E (0:odd, 1: even).

> Sure you can build a new event space (sigma algebra) but then you can't use random variables over the original one.

I guess we all agree then.

> Let's consider two independent rolls, X and Y. You can't compute the joint distribution P(Y, (X|E)), it just doesn't make sense as the two "variables" are defined over different spaces.

The variables X and Y describing independent rolls are also defined over different spaces and to have a joint distribution you have to define a "common" sample space of the form {x=1,y=1},{x=2,y=1},..,{x=6,y=6}.

You could do the same for a roll of a dice and the toss of a coin. Or do you think that computing the joint distribution of a coin toss and a dice roll doesn't make sense because they are defined over different spaces?

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