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> It's just Bayesian inference to update the posterior. I've had this discussion with a number of smart people (math PhDs and so on) over the years, and I've n

> It's just Bayesian inference to update the posterior. I've had this discussion with a number of smart people (math PhDs and so on) over the years, and I've never gotten a good explanation. I understand in the very abstract sense how Kalman filters (and particle filters and so on) can be seen as Bayesian, but when it gets to actually applying Bayes' theorem it really gets complicated very fast. For instance, if you're estimating a 3D position (that will be A) and your measurements are some 2D subspace (these will be B), the least squares approach works nicely (normalized products of Normal PDFs). However, I'm not sure I understand a single component of the expression P(A|B) = P(B|A) P(A) / P(B) in this context. What does A|B or B|A even mean when A is a 3D Gaussian and B is 2D? What are the dimensionalities of these operations? What does it mean to divide a product of pdfs? I don't expect an answer, but I always wince a bit when I see Bayes in this context. For Normal distributions, I prefer thinking of it as least squares.

ssivark 9y ago
It's Bayesian because we are estimating a (posterior) distribution for the state of the system, rather than just a "point" estimate. Restricting to only the Gaussian distribution (while a nice example) is losing the forest for the trees. > However, I'm not sure I understand a single component of the expression P(A|B) = P(B|A) P(A) / P(B) in this context. What does A|B or B|A even mean when A is a 3D Gaussian and B is 2D? What are the dimensionalities of these operations? What does it mean to divide a product of pdfs? I'm not able to understand why any of that is bothersome. Suppose A is the system's state (x,y,z) and B is the sensor reading (measures just the projection along the x-axis, such as a depth measurement). Since the sensor is not perfect, it could generate a range of readings for any given position of the object, specified by p(B[x]|A[x,y,z]) -- suppose you're give one value of B which is sampled from that distribution (say that was a Gaussian with standard deviation sigma). Given that value, you need to estimate the likelihood p(A|B). Alternatively, you can imagine performing variational inference for p(A|B) in the family of Gaussian distributions (which happens to contain the true distribution in this case). Whichever way we go about it, p(A|B) is a Gaussian with the same standard deviation sigma, centered at the measured B. (Intuitively, we can see that a measurement B will constrain the "x" component of A but not the other two coordinates) p(A) is the "prior" you get as an extrapolation from your knowledge of the system's past state. However complicated/nonlinear the system's evolution function, you can find p(A, t) either as a closed form expression, or by doing simulations, etc. (typically from p(A,t-1) but note, for example, that this general formulation can be used even if your system's dynamics does not satisfy the Markov property). If you have both p(A|B) and p(A) manifestly in the form of probability density functions, multiplying them is easy. At the end of this process, it is good enough to have an un-normalized posterior for A, which specifies the relative probability of different states. Dividing by p(B) is equivalent to normalizing the candidate posterior distribution. More generally, if the prior and the likelihood are specified by moments, or by some variational parameters, or by simulations, we can think of other ways to combine them.
- civility 9y ago
  > I'm not able to understand why any of that is bothersome. Ok. It's bothersome to me because Bayes gets trotted out in the context of Kalman filters as though it is shedding light on the topic or providing rigor to the approach, but I haven't found anyone who can really describe what the elements of that expression are without hand waiving. You've switched from capital 'P' to lowercase 'p' in your message (which I take to mean Probability and pdf respectively). If that was intentional, it opens a new batch of questions for what the notation means. Anyways, maybe I'm just a slow learner, too pedantic, or something else. If you're interested, read my other reply where I go into it more.
  - gugagore 9y ago
    I'm not the poster you're responding to, but your observation about the distinction between P and p is crucial. The version of Bayes' rule that is relevant for the Kalman filter is the third one here: https://en.wikipedia.org/wiki/Bayes'_theorem#Random_variables https://en.wikipedia.org/wiki/Bayes'_theorem#Random_variable... ("if both X and Y are continuous").
rsp1984 9y ago
However, I'm not sure I understand a single component of the expression P(A|B) = P(B|A) P(A) / P(B) in this context. You've come to the right place :) In the context of least squares estimation you can usually ignore the denominator P(B) and only look at P(A|B) = P(B|A) P(A). Real mathematicians (!= me) cringe here because the numerator alone isn't a proper PDF but since we're only seeking the value where that PDF has its maximum, that's ok. Good, so now we have P(A|B) ~ P(B|A) P(A). What you want to know is the maximum of P(A|B), that is "where's the most probable position of my 3D point A given all the 2D measurements B"? Let's look at the solution P(B|A) P(A). The first factor is the likelihood function, the second factor is the prior. The prior is easy: It's some prior belief about the distribution of A. So if you know the position of your 3D point somewhat (even without looking at B), say you know it must be in a range of coordinates, or you know that Z must be positive or any other statement about A that you know must hold, that information would go into the prior. P(B|A) on the other hand is your likelihood. That's kind of the "meat" of Bayes. Your likelihood is a distribution over B, given A. Meaning, if you knew your 3D point A, how would the PDFs of your 2D observations of it look like? Note that the likelihood is to be understood as a function of state, meaning the domain of it is A. So to summarize argmax(P(A|B)) = argmax(P(B|A) P(A)), meaning to get the most probable location of your 3D point you need to look in your likelihood domain A (in this case the 3D space) where the 3D points are that would produce the observations B that you have and then also take into account your prior beliefs about A, to narrow down that set of candidate points. In the concrete case of least squares both likelihood and prior are assumed to be gaussian, so if you multiply them you get a new gaussian with a single maximum. Neat!
- civility 9y ago
  I appreciate the kind reply, but I don't think you understood the gist of my complaint. I have a degree (just undergrad) in math, and I've implemented Kalman filters, Kalman smoothers, information filters, particle filters and so on at least a dozen times. I know what operations to perform, and I even have an intuition about why they work. When I complain about the Bayes theorem way of describing or thinking about Kalman filters, I really mean that I don't understand what almost anything in that expression means. The notation is intuitive if read as English, but opaque to what operations are being performed on what mathematical objects. Capital P means probability of an event or predicate, and A is some abstraction of the thing I'm trying to estimate, and B is the abstraction of a new observation or measurement. Think of it this way: I have some knowledge of my prior estimate, and I'm able to characterize it as a Normally distributed random variable. It has a mean and a covariance in 3D space (let's call those x and P). There is a multivariate function for that pdf (lowercase p); it's an exponential of the negative square of the Mahalanobis distance normalized to have a unit volume. The same thing applies for my measurement, except it's only a 2D mean and covariance (let's call those z and R). I already know that I can use my H matrix, some matrix inverses and multiplications to combine this information. It's really just a case of multiplying the pdf for the state estimate and the measurement and re-normalizing to a unit volume. In other words, I believe both things are true, and they are independent, so I can multiply their pdfs and re-normalize to get a combined result. The rest is just linear algebra. Now let's get to Bayes. P(A) is supposed to be the probability of an "event" or predicate. I can squint sideways and translate that as integrate my pdf for the prior estimate (Normal with mean x and covariance P) over a some unspecified bounds and convert that to a probability of my estimate being in those bounds, but I'm not sure that's what is intended. Again a similar thing applies for B (Normal with mean z and covariance R). It's frustrating that the bounds are never stated, because they could be radically different spaces in the numerator and denominator. I guess I'll give that a pass because maybe the notation would be too cumbersome if it was included, but this simplification seems to never be stated in any of the books or papers I've read on it. Next, P(B|A) has to be probability of another "event", and if I squint again, the best I'm able to come up with is that it means a new Normal pdf with mean = H'z and covariance = H'RH. However, I haven't seen that spelled out any where, and so really that's just assuming the conclusion I want, which is super questionable. It also doesn't help me understand the left side of the Bayes equation - that vertical bar there seems to mean something different. When I look to the definition of conditional probability, I don't see anything about the vertical bar applied to multivariate Normal distributions. If you translate it all to English, it reads as a coherent sentence, and that's fine. This is the "prior state estimate", that's the "observation", and this other thing is that's the "a posteriori" etc... However, if Bayes really helps with the understanding the math, the vertical bar has to mean something specific as an operator, and one would hope it meant the same thing on the left and right sides of the equation. Similarly, in order for all of those P( ... ) to become scalars so multiplication and division are well defined, you need some integration bounds to turn the event into a probability. But at this point, I'm not even sure if that's what is intended by the notation - maybe they aren't scalars, and multiplication and division mean something radically different here. I honestly don't know.
deleted 9y ago
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quietbritishjim 9y ago
> What does A|B or B|A even mean when A is a 3D Gaussian and B is 2D? A|B is not an object in its own right. P is not just a function that means "probability of this thing", as it would if you were dealing with probabilistic events [1]. Instead, P(A) means the probability density function of the random variable A. It is an abuse of notation, and maybe that's what's confusing you: A both represents the random variable, which is a complex object [2], and the particular number that's being substituted into its probability density function to give a number as a result. Mathematicians use a longer notation: P_A(a). Here A (which should be a subscript but I can't write that here) represents the random variable, P_A is a function R^3->R (the PDF of A) and a is a particular R^3, so that P_A(a) is in R. Then this formula: > P(A|B) = P(B|A) P(A) / P(B) should really be this (I wrote the subscripts in square brackets just to try and distinguish from the actual function parameters): > P_[A|B=b](a) = P_[B|A=a](b) P_[A](a) / P_[B](b) Note that A|B=b is a random variable, in this case describing the position of the object given that it was detected at b. [1] https://en.wikipedia.org/wiki/Event_(probability_theory) https://en.wikipedia.org/wiki/Event_(probability_theory) [2] https://en.wikipedia.org/wiki/Random_variable#Definition https://en.wikipedia.org/wiki/Random_variable#Definition
- civility 9y ago
  Thank you for the reply. I spent most of the day talking to a coworker about this stuff, but I didn't see your message until just now. I agree about the abuse of notation, but even with your improvements, there is still plenty to be confused about. In particular, the denominator has P_[B](b), which looks like a function from R^2->R (the PDF of B in measurement space). I guess it could also mean that PDF evaluated at the specific value of b. However, I think it's actual value is supposed to be "the integral from negative to positive infinity in all directions of whatever is in the numerator", which isn't at all obvious from the notation. However, maybe I've got that wrong. Let's go one step further. On the left side we've got a function R^3->R with parameter 'a'. On the right, we've got two functions of 'b' (R^2->R), one in the numerator, and another in the denominator, presumably evaluated at the value 'b'. Both of those evaluations give scalars, and so it looks like we just get a scaled version of P_[A](a), which is definitely not the right answer. An alternative is that we're not really evaluating those two functions at 'b', and instead leaving them as functions. If that's so, where did the free variable come from? It's not in the argument list on the left hand side. Although it is in the subscript on the left... Yeah, I'm still pretty confused :-)

ssivark 9y ago

It's Bayesian because we are estimating a (posterior) distribution for the state of the system, rather than just a "point" estimate. Restricting to only the Gaussian distribution (while a nice example) is losing the forest for the trees. > However, I'm not sure I understand a single component of the expression P(A|B) = P(B|A) P(A) / P(B) in this context. What does A|B or B|A even mean when A is a 3D Gaussian and B is 2D? What are the dimensionalities of these operations? What does it mean to divide a product of pdfs? I'm not able to understand why any of that is bothersome. Suppose A is the system's state (x,y,z) and B is the sensor reading (measures just the projection along the x-axis, such as a depth measurement). Since the sensor is not perfect, it could generate a range of readings for any given position of the object, specified by p(B[x]|A[x,y,z]) -- suppose you're give one value of B which is sampled from that distribution (say that was a Gaussian with standard deviation sigma). Given that value, you need to estimate the likelihood p(A|B). Alternatively, you can imagine performing variational inference for p(A|B) in the family of Gaussian distributions (which happens to contain the true distribution in this case). Whichever way we go about it, p(A|B) is a Gaussian with the same standard deviation sigma, centered at the measured B. (Intuitively, we can see that a measurement B will constrain the "x" component of A but not the other two coordinates) p(A) is the "prior" you get as an extrapolation from your knowledge of the system's past state. However complicated/nonlinear the system's evolution function, you can find p(A, t) either as a closed form expression, or by doing simulations, etc. (typically from p(A,t-1) but note, for example, that this general formulation can be used even if your system's dynamics does not satisfy the Markov property). If you have both p(A|B) and p(A) manifestly in the form of probability density functions, multiplying them is easy. At the end of this process, it is good enough to have an un-normalized posterior for A, which specifies the relative probability of different states. Dividing by p(B) is equivalent to normalizing the candidate posterior distribution. More generally, if the prior and the likelihood are specified by moments, or by some variational parameters, or by simulations, we can think of other ways to combine them.

civility 9y ago
> I'm not able to understand why any of that is bothersome. Ok. It's bothersome to me because Bayes gets trotted out in the context of Kalman filters as though it is shedding light on the topic or providing rigor to the approach, but I haven't found anyone who can really describe what the elements of that expression are without hand waiving. You've switched from capital 'P' to lowercase 'p' in your message (which I take to mean Probability and pdf respectively). If that was intentional, it opens a new batch of questions for what the notation means. Anyways, maybe I'm just a slow learner, too pedantic, or something else. If you're interested, read my other reply where I go into it more.
- gugagore 9y ago
  I'm not the poster you're responding to, but your observation about the distinction between P and p is crucial. The version of Bayes' rule that is relevant for the Kalman filter is the third one here: https://en.wikipedia.org/wiki/Bayes'_theorem#Random_variables https://en.wikipedia.org/wiki/Bayes'_theorem#Random_variable... ("if both X and Y are continuous").

rsp1984 9y ago

However, I'm not sure I understand a single component of the expression P(A|B) = P(B|A) P(A) / P(B) in this context. You've come to the right place :) In the context of least squares estimation you can usually ignore the denominator P(B) and only look at P(A|B) = P(B|A) P(A). Real mathematicians (!= me) cringe here because the numerator alone isn't a proper PDF but since we're only seeking the value where that PDF has its maximum, that's ok. Good, so now we have P(A|B) ~ P(B|A) P(A). What you want to know is the maximum of P(A|B), that is "where's the most probable position of my 3D point A given all the 2D measurements B"? Let's look at the solution P(B|A) P(A). The first factor is the likelihood function, the second factor is the prior. The prior is easy: It's some prior belief about the distribution of A. So if you know the position of your 3D point somewhat (even without looking at B), say you know it must be in a range of coordinates, or you know that Z must be positive or any other statement about A that you know must hold, that information would go into the prior. P(B|A) on the other hand is your likelihood. That's kind of the "meat" of Bayes. Your likelihood is a distribution over B, given A. Meaning, if you knew your 3D point A, how would the PDFs of your 2D observations of it look like? Note that the likelihood is to be understood as a function of state, meaning the domain of it is A. So to summarize argmax(P(A|B)) = argmax(P(B|A) P(A)), meaning to get the most probable location of your 3D point you need to look in your likelihood domain A (in this case the 3D space) where the 3D points are that would produce the observations B that you have and then also take into account your prior beliefs about A, to narrow down that set of candidate points. In the concrete case of least squares both likelihood and prior are assumed to be gaussian, so if you multiply them you get a new gaussian with a single maximum. Neat!

civility 9y ago
I appreciate the kind reply, but I don't think you understood the gist of my complaint. I have a degree (just undergrad) in math, and I've implemented Kalman filters, Kalman smoothers, information filters, particle filters and so on at least a dozen times. I know what operations to perform, and I even have an intuition about why they work. When I complain about the Bayes theorem way of describing or thinking about Kalman filters, I really mean that I don't understand what almost anything in that expression means. The notation is intuitive if read as English, but opaque to what operations are being performed on what mathematical objects. Capital P means probability of an event or predicate, and A is some abstraction of the thing I'm trying to estimate, and B is the abstraction of a new observation or measurement. Think of it this way: I have some knowledge of my prior estimate, and I'm able to characterize it as a Normally distributed random variable. It has a mean and a covariance in 3D space (let's call those x and P). There is a multivariate function for that pdf (lowercase p); it's an exponential of the negative square of the Mahalanobis distance normalized to have a unit volume. The same thing applies for my measurement, except it's only a 2D mean and covariance (let's call those z and R). I already know that I can use my H matrix, some matrix inverses and multiplications to combine this information. It's really just a case of multiplying the pdf for the state estimate and the measurement and re-normalizing to a unit volume. In other words, I believe both things are true, and they are independent, so I can multiply their pdfs and re-normalize to get a combined result. The rest is just linear algebra. Now let's get to Bayes. P(A) is supposed to be the probability of an "event" or predicate. I can squint sideways and translate that as integrate my pdf for the prior estimate (Normal with mean x and covariance P) over a some unspecified bounds and convert that to a probability of my estimate being in those bounds, but I'm not sure that's what is intended. Again a similar thing applies for B (Normal with mean z and covariance R). It's frustrating that the bounds are never stated, because they could be radically different spaces in the numerator and denominator. I guess I'll give that a pass because maybe the notation would be too cumbersome if it was included, but this simplification seems to never be stated in any of the books or papers I've read on it. Next, P(B|A) has to be probability of another "event", and if I squint again, the best I'm able to come up with is that it means a new Normal pdf with mean = H'z and covariance = H'RH. However, I haven't seen that spelled out any where, and so really that's just assuming the conclusion I want, which is super questionable. It also doesn't help me understand the left side of the Bayes equation - that vertical bar there seems to mean something different. When I look to the definition of conditional probability, I don't see anything about the vertical bar applied to multivariate Normal distributions. If you translate it all to English, it reads as a coherent sentence, and that's fine. This is the "prior state estimate", that's the "observation", and this other thing is that's the "a posteriori" etc... However, if Bayes really helps with the understanding the math, the vertical bar has to mean something specific as an operator, and one would hope it meant the same thing on the left and right sides of the equation. Similarly, in order for all of those P( ... ) to become scalars so multiplication and division are well defined, you need some integration bounds to turn the event into a probability. But at this point, I'm not even sure if that's what is intended by the notation - maybe they aren't scalars, and multiplication and division mean something radically different here. I honestly don't know.

deleted 9y ago

[deleted]

quietbritishjim 9y ago

> What does A|B or B|A even mean when A is a 3D Gaussian and B is 2D? A|B is not an object in its own right. P is not just a function that means "probability of this thing", as it would if you were dealing with probabilistic events [1]. Instead, P(A) means the probability density function of the random variable A. It is an abuse of notation, and maybe that's what's confusing you: A both represents the random variable, which is a complex object [2], and the particular number that's being substituted into its probability density function to give a number as a result. Mathematicians use a longer notation: P_A(a). Here A (which should be a subscript but I can't write that here) represents the random variable, P_A is a function R^3->R (the PDF of A) and a is a particular R^3, so that P_A(a) is in R. Then this formula: > P(A|B) = P(B|A) P(A) / P(B) should really be this (I wrote the subscripts in square brackets just to try and distinguish from the actual function parameters): > P_[A|B=b](a) = P_[B|A=a](b) P_[A](a) / P_[B](b) Note that A|B=b is a random variable, in this case describing the position of the object given that it was detected at b. [1] https://en.wikipedia.org/wiki/Event_(probability_theory) https://en.wikipedia.org/wiki/Event_(probability_theory) [2] https://en.wikipedia.org/wiki/Random_variable#Definition https://en.wikipedia.org/wiki/Random_variable#Definition

civility 9y ago
Thank you for the reply. I spent most of the day talking to a coworker about this stuff, but I didn't see your message until just now. I agree about the abuse of notation, but even with your improvements, there is still plenty to be confused about. In particular, the denominator has P_[B](b), which looks like a function from R^2->R (the PDF of B in measurement space). I guess it could also mean that PDF evaluated at the specific value of b. However, I think it's actual value is supposed to be "the integral from negative to positive infinity in all directions of whatever is in the numerator", which isn't at all obvious from the notation. However, maybe I've got that wrong. Let's go one step further. On the left side we've got a function R^3->R with parameter 'a'. On the right, we've got two functions of 'b' (R^2->R), one in the numerator, and another in the denominator, presumably evaluated at the value 'b'. Both of those evaluations give scalars, and so it looks like we just get a scaled version of P_[A](a), which is definitely not the right answer. An alternative is that we're not really evaluating those two functions at 'b', and instead leaving them as functions. If that's so, where did the free variable come from? It's not in the argument list on the left hand side. Although it is in the subscript on the left... Yeah, I'm still pretty confused :-)