• Posted by Konstantin 07.03.2017 No Comments

    Ever since the "Prior Confusion" post I was planning to formulate one of its paragraphs as the following abstract puzzle, but somehow it took me 8 years to write it up.

    According to fictional statistical studies, the following is known about a fictional chronic disease "statistite":

    1. About 30% of people in the world have statistite.
    2. About 35% of men in the world have it.
    3. In Estonia, 20% of people have statistite.
    4. Out of people younger than 20 years, just 5% have the disease.
    5. A recent study of a random sample of visitors to the Central Hospital demonstrated that 40% of them suffer from statistite.

    Mart, a 19-year Estonian male medical student is standing in the foyer of the Central Hospital, reading these facts from an information sheet and wondering: what are his current chances of having statistite? How should he model himself: should he consider himself as primarily "an average man", "a typical Estonian", "just a young person", or "an average visitor of the hospital"? Could he combine the different aspects of his personality to make better use of the available information? How? In general, what would be the best possible probability estimate, given the data?

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  • Posted by Konstantin 23.11.2016 3 Comments

    Basic linear algebra, introductory statistics and some familiarity with core machine learning concepts (such as PCA and linear models) are the prerequisites of this post. Otherwise it will probably make no sense. An abridged version of this text is also posted on Quora.

    Most textbooks on statistics cover covariance right in their first chapters. It is defined as a useful "measure of dependency" between two random variables:

        \[\mathrm{cov}(X,Y) = E[(X - E[X])(Y - E[Y])].\]

    The textbook would usually provide some intuition on why it is defined as it is, prove a couple of properties, such as bilinearity, define the covariance matrix for multiple variables as {\bf\Sigma}_{i,j} = \mathrm{cov}(X_i, X_j), and stop there. Later on the covariance matrix would pop up here and there in seeminly random ways. In one place you would have to take its inverse, in another - compute the eigenvectors, or multiply a vector by it, or do something else for no apparent reason apart from "that's the solution we came up with by solving an optimization task".

    In reality, though, there are some very good and quite intuitive reasons for why the covariance matrix appears in various techniques in one or another way. This post aims to show that, illustrating some curious corners of linear algebra in the process.

    Meet the Normal Distribution

    The best way to truly understand the covariance matrix is to forget the textbook definitions completely and depart from a different point instead. Namely, from the the definition of the multivariate Gaussian distribution:

    We say that the vector \bf x has a normal (or Gaussian) distribution with mean \bf \mu and covariance \bf \Sigma if:

        \[\Pr({\bf x}) =|2\pi{\bf\Sigma}|^{-1/2} \exp\left(-\frac{1}{2}({\bf x} - {\bf\mu})^T{\bf\Sigma}^{-1}({\bf x} - {\bf \mu})\right).\]

    To simplify the math a bit, we will limit ourselves to the centered distribution (i.e. {\bf\mu} = {\bf 0}) and refrain from writing out the normalizing constant |2\pi{\bf\Sigma}|^{-1/2}. Now, the definition of the (centered) multivariate Gaussian looks as follows:

        \[\Pr({\bf x}) \propto \exp\left(-0.5{\bf x}^T{\bf\Sigma}^{-1}{\bf x}\right).\]

    Much simpler, isn't it? Finally, let us define the covariance matrix as nothing else but the parameter of the Gaussian distribution. That's it. You will see where it will lead us in a moment.

    Transforming the Symmetric Gaussian

    Consider a symmetric Gaussian distribution, i.e. the one with {\bf \Sigma = \bf I} (the identity matrix). Let us take a sample from it, which will of course be a symmetric, round cloud of points:

    We know from above that the likelihood of each point in this sample is

    (1)   \[P({\bf x}) \propto \exp(-0.5 {\bf x}^T {\bf x}).\]

    Now let us apply a linear transformation {\bf A} to the points, i.e. let {\bf y} ={\bf Ax}. Suppose that, for the sake of this example, {\bf A} scales the vertical axis by 0.5 and then rotates everything by 30 degrees. We will get the following new cloud of points {\bf y}:

    What is the distribution of {\bf y}? Just substitute {\bf x}={\bf A}^{-1}{\bf y} into (1), to get:

    (2)   \begin{align*} P({\bf y}) &\propto \exp(-0.5 ({\bf A}^{-1}{\bf y})^T({\bf A}^{-1}{\bf y}))\\ &=\exp(-0.5{\bf y}^T({\bf AA}^T)^{-1}{\bf y}) \end{align*}

    This is exactly the Gaussian distribution with covariance {\bf \Sigma} = {\bf AA}^T. The logic works both ways: if we have a Gaussian distribution with covariance \bf \Sigma, we can regard it as a distribution which was obtained by transforming the symmetric Gaussian by some {\bf A}, and we are given {\bf AA}^T.

    More generally, if we have any data, then, when we compute its covariance to be \bf\Sigma, we can say that if our data were Gaussian, then it could have been obtained from a symmetric cloud using some transformation \bf A, and we just estimated the matrix {\bf AA}^T, corresponding to this transformation.

    Note that we do not know the actual \bf A, and it is mathematically totally fair. There can be many different transformations of the symmetric Gaussian which result in the same distribution shape. For example, if \bf A is just a rotation by some angle, the transformation does not affect the shape of the distribution at all. Correspondingly, {\bf AA}^T = {\bf I} for all rotation matrices. When we see a unit covariance matrix we really do not know, whether it is the “originally symmetric” distribution, or a “rotated symmetric distribution”. And we should not really care - those two are identical.

    There is a theorem in linear algebra, which says that any symmetric matrix \bf \Sigma can be represented as:

    (3)   \[{\bf \Sigma} = {\bf VDV}^T,\]

    where {\bf V} is orthogonal (i.e. a rotation) and {\bf D} is diagonal (i.e. a coordinate-wise scaling). If we rewrite it slightly, we will get:

    (4)   \[{\bf \Sigma} = ({\bf VD}^{1/2})({\bf VD}^{1/2})^T = {\bf AA}^T,\]

    where {\bf A} = {\bf VD}^{1/2}. This, in simple words, means that any covariance matrix \bf \Sigma could have been the result of transforming the data using a coordinate-wise scaling {\bf D}^{1/2} followed by a rotation \bf V. Just like in our example with \bf x and \bf y above.

    Principal Component Analysis

    Given the above intuition, PCA already becomes a very obvious technique. Suppose we are given some data. Let us assume (or “pretend”) it came from a normal distribution, and let us ask the following questions:

    1. What could have been the rotation \bf V and scaling {\bf D}^{1/2}, which produced our data from a symmetric cloud?
    2. What were the original, “symmetric-cloud” coordinates \bf x before this transformation was applied.
    3. Which original coordinates were scaled the most by \bf D and thus contribute most to the spread of the data now. Can we only leave those and throw the rest out?

    All of those questions can be answered in a straightforward manner if we just decompose \bf \Sigma into \bf V and \bf D according to (3). But (3) is exactly the eigenvalue decomposition of \bf\Sigma. I’ll leave you to think for just a bit and you’ll see how this observation lets you derive everything there is about PCA and more.

    The Metric Tensor

    Bear me for just a bit more. One way to summarize the observations above is to say that we can (and should) regard {\bf\Sigma}^{-1} as a metric tensor. A metric tensor is just a fancy formal name for a matrix, which summarizes the deformation of space. However, rather than claiming that it in some sense determines a particular transformation \bf A (which it does not, as we saw), we shall say that it affects the way we compute angles and distances in our transformed space.

    Namely, let us redefine, for any two vectors \bf v and \bf w, their inner product as:

    (5)   \[\langle {\bf v}, {\bf w}\rangle_{\Sigma^{-1}} = {\bf v}^T{\bf \Sigma}^{-1}{\bf w}.\]

    To stay consistent we will also need to redefine the norm of any vector as

    (6)   \[|{\bf v}|_{\Sigma^{-1}} = \sqrt{{\bf v}^T{\bf \Sigma}^{-1}{\bf v}},\]

    and the distance between any two vectors as

    (7)   \[|{\bf v}-{\bf w}|_{\Sigma^{-1}} = \sqrt{({\bf v}-{\bf w})^T{\bf \Sigma}^{-1}({\bf v}-{\bf w})}.\]

    Those definitions now describe a kind of a “skewed world” of points. For example, a unit circle (a set of points with “skewed distance” 1 to the center) in this world might look as follows:

    And here is an example of two vectors, which are considered “orthogonal”, a.k.a. “perpendicular” in this strange world:

    Although it may look weird at first, note that the new inner product we defined is actually just the dot product of the “untransformed” originals of the vectors:

    (8)   \[{\bf v}^T{\bf \Sigma}^{-1}{\bf w} = {\bf v}^T({\bf AA}^T)^{-1}{\bf w}=({\bf A}^{-1}{\bf v})^T({\bf A}^{-1}{\bf w}),\]

    The following illustration might shed light on what is actually happening in this \Sigma-“skewed” world. Somehow “deep down inside”, the ellipse thinks of itself as a circle and the two vectors behave as if they were (2,2) and (-2,2).

    Getting back to our example with the transformed points, we could now say that the point-cloud \bf y is actually a perfectly round and symmetric cloud “deep down inside”, it just happens to live in a skewed space. The deformation of this space is described by the tensor {\bf\Sigma}^{-1} (which is, as we know, equal to ({\bf AA}^T)^{-1}. The PCA now becomes a method for analyzing the deformation of space, how cool is that.

    The Dual Space

    We are not done yet. There’s one interesting property of “skewed” spaces worth knowing about. Namely, the elements of their dual space have a particular form. No worries, I’ll explain in a second.

    Let us forget the whole skewed space story for a moment, and get back to the usual inner product {\bf w}^T{\bf v}. Think of this inner product as a function f_{\bf w}({\bf v}), which takes a vector \bf v and maps it to a real number, the dot product of \bf v and \bf w. Regard the \bf w here as the parameter (“weight vector”) of the function. If you have done any machine learning at all, you have certainly come across such linear functionals over and over, sometimes in disguise. Now, the set of all possible linear functionals f_{\bf w} is known as the dual space to your “data space”.

    Note that each linear functional is determined uniquely by the parameter vector \bf w, which has the same dimensionality as \bf v, so apparently the dual space is in some sense equivalent to your data space - just the interpretation is different. An element \bf v of your “data space” denotes, well, a data point. An element \bf w of the dual space denotes a function f_{\bf w}, which projects your data points on the direction \bf w (recall that if \bf w is unit-length, {\bf w}^T{\bf v} is exactly the length of the perpendicular projection of \bf v upon the direction \bf w). So, in some sense, if \bf v-s are “vectors”, \bf w-s are “directions, perpendicular to these vectors”. Another way to understand the difference is to note that if, say, the elements of your data points numerically correspond to amounts in kilograms, the elements of \bf w would have to correspond to “units per kilogram”. Still with me?

    Let us now get back to the skewed space. If \bf v are elements of a skewed Euclidean space with the metric tensor {\bf\Sigma}^{-1}, is a function f_{\bf w}({\bf v}) = {\bf w}^T{\bf v} an element of a dual space? Yes, it is, because, after all, it is a linear functional. However, the parameterization of this function is inconvenient, because, due to the skewed tensor, we cannot interpret it as projecting vectors upon \bf w nor can we say that \bf w is an “orthogonal direction” (to a separating hyperplane of a classifier, for example). Because, remember, in the skewed space it is not true that orthogonal vectors satisfy {\bf w}^T{\bf v}=0. Instead, they satisfy {\bf w}^T{\bf \Sigma}^{-1}{\bf v} = 0. Things would therefore look much better if we parameterized our dual space differently. Namely, by considering linear functionals of the form f^{\Sigma^{-1}}_{\bf z}({\bf v}) = {\bf z}^T{\bf \Sigma}^{-1}{\bf v}. The new parameters \bf z could now indeed be interpreted as an “orthogonal direction” and things overall would make more sense.

    However when we work with actual machine learning models, we still prefer to have our functions in the simple form of a dot product, i.e. f_{\bf w}, without any ugly \bf\Sigma-s inside. What happens if we turn a “skewed space” linear functional from its natural representation into a simple inner product?

    (9)   \[f^{\Sigma^{-1}}_{\bf z}({\bf v}) = {\bf z}^T{\bf\Sigma}^{-1}{\bf v} = ({\bf \Sigma}^{-1}{\bf z})^T{\bf v} = f_{\bf w}({\bf v}),\]

    where {\bf w} = {\bf \Sigma}^{-1}{\bf z}. (Note that we can lose the transpose because \bf \Sigma is symmetric).

    What it means, in simple terms, is that when you fit linear models in a skewed space, your resulting weight vectors will always be of the form {\bf \Sigma}^{-1}{\bf z}. Or, in other words, {\bf\Sigma}^{-1} is a transformation, which maps from “skewed perpendiculars” to “true perpendiculars”. Let me show you what this means visually.

    Consider again the two “orthogonal” vectors from the skewed world example above:

    Let us interpret the blue vector as an element of the dual space. That is, it is the \bf z vector in a linear functional {\bf z}^T{\bf\Sigma}^{-1}{\bf v}. The red vector is an element of the “data space”, which would be mapped to 0 by this functional (because the two vectors are “orthogonal”, remember).

    For example, if the blue vector was meant to be a linear classifier, it would have its separating line along the red vector, just like that:

    If we now wanted to use this classifier, we could, of course, work in the “skewed space” and use the expression {\bf z}^T{\bf\Sigma}^{-1}{\bf v} to evaluate the functional. However, why don’t we find the actual normal \bf w to that red separating line so that we wouldn’t need to do an extra matrix multiplication every time we use the function?

    It is not too hard to see that {\bf w}={\bf\Sigma}^{-1}{\bf z} will give us that normal. Here it is, the black arrow:

    Therefore, next time, whenever you see expressions like {\bf w}^T{\bf\Sigma}^{-1}{\bf v} or ({\bf v}-{\bf w})^T{\bf\Sigma}^{-1}({\bf v}-{\bf w}), remember that those are simply inner products and (squared) distances in a skewed space, while {\bf \Sigma}^{-1}{\bf z} is a conversion from a skewed normal to a true normal. Also remember that the “skew” was estimated by pretending that the data were normally-distributed.

    Once you see it, the role of the covariance matrix in some methods like the Fisher’s discriminant or Canonical correlation analysis might become much more obvious.

    The Dual Space Metric Tensor

    “But wait”, you should say here. “You have been talking about expressions like {\bf w}^T{\bf\Sigma}^{-1}{\bf v} all the time, while things like {\bf w}^T{\bf\Sigma}{\bf v} are also quite common in practice. What about those?”

    Hopefully you know enough now to suspect that {\bf w}^T{\bf\Sigma}{\bf v} is again an inner product or a squared norm in some deformed space, just not the “internal data metric space”, that we considered so far. Which space is it? It turns out it is the “internal dual metric space”. That is, whilst the expression {\bf w}^T{\bf\Sigma}^{-1}{\bf v} denoted the “new inner product” between the points, the expression {\bf w}^T{\bf\Sigma}{\bf v} denotes the “new inner product” between the parameter vectors. Let us see why it is so.

    Consider an example again. Suppose that our space transformation \bf A scaled all points by 2 along the x axis. The point (1,0) became (2,0), the point (3, 1) became (6, 1), etc. Think of it as changing the units of measurement - before we measured the x axis in kilograms, and now we measure it in pounds. Consequently, the norm of the point (2,0) according to the new metric, |(2,0)|_{\Sigma^{-1}} will be 1, because 2 pounds is still just 1 kilogram “deep down inside”.

    What should happen to the parameter ("direction") vectors due to this transformation? Can we say that the parameter vector (1,0) also got scaled to (2,0) and that the norm of the parameter vector (2,0) is now therefore also 1? No! Recall that if our initial data denoted kilograms, our dual vectors must have denoted “units per kilogram”. After the transformation they will be denoting “units per pound”, correspondingly. To stay consistent we must therefore convert the parameter vector (”1 unit per kilogram”, 0) to its equivalent (“0.5 units per pound”,0). Consequently, the norm of the parameter vector (0.5,0) in the new metric will be 1 and, by the same logic, the norm of the dual vector (2,0) in the new metric must be 4. You see, the “importance of a parameter/direction” gets scaled inversely to the “importance of data” along that parameter or direction.

    More formally, if {\bf x}'={\bf Ax}, then

    (10)   \begin{align*} f_{\bf w}({\bf x}) &= {\bf w}^T{\bf x} = {\bf w}^T{\bf A}^{-1}{\bf x}'\\ & =(({\bf A}^{-1})^T{\bf w})^T{\bf x}'=f_{({\bf A}^{-1})^T{\bf w}}({\bf x}'). \end{align*}

    This means, that the transformation \bf A of the data points implies the transformation {\bf B}:=({\bf A}^{-1})^T of the dual vectors. The metric tensor for the dual space must thus be:

    (11)   \[({\bf BB}^T)^{-1}=(({\bf A}^{-1})^T{\bf A}^{-1})^{-1}={\bf AA}^T={\bf \Sigma}.\]

    Remember the illustration of the “unit circle” in the {\bf \Sigma}^{-1} metric? This is how the unit circle looks in the corresponding \bf\Sigma metric. It is rotated by the same angle, but it is stretched in the direction where it was squished before.

    Intuitively, the norm (“importance”) of the dual vectors along the directions in which the data was stretched by \bf A becomes proportionally larger (note that the “unit circle” is, on the contrary, “squished” along those directions).

    But the “stretch” of the space deformation in any direction can be measured by the variance of the data. It is therefore not a coincidence that {\bf w}^T{\bf \Sigma}{\bf w} is exactly the variance of the data along the direction \bf w (assuming |{\bf w}|=1).

    The Covariance Estimate

    Once we start viewing the covariance matrix as a transformation-driven metric tensor, many things become clearer, but one thing becomes extremely puzzling: why is the inverse covariance of the data a good estimate for that metric tensor? After all, it is not obvious that {\bf X}^T{\bf X}/n (where \bf X is the data matrix) must be related to the \bf\Sigma in the distribution equation \exp(-0.5{\bf x}^T{\bf\Sigma}^{-1}{\bf x}).

    Here is one possible way to see the connection. Firstly, let us take it for granted that if \bf X is sampled from a symmetric Gaussian, then {\bf X}^T{\bf X}/n is, on average, a unit matrix. This has nothing to do with transformations, but just a consequence of pairwise independence of variables in the symmetric Gaussian.

    Now, consider the transformed data, {\bf Y}={\bf XA}^T (vectors in the data matrix are row-wise, hence the multiplication on the right with a transpose). What is the covariance estimate of \bf Y?

    (12)   \[{\bf Y}^T{\bf Y}/n=({\bf XA}^T)^T{\bf XA}^T/n={\bf A}({\bf X}^T{\bf X}){\bf A}^T/n\approx {\bf AA}^T,\]

    the familiar tensor.

    This is a place where one could see that a covariance matrix may make sense outside the context of a Gaussian distribution, after all. Indeed, if you assume that your data was generated from any distribution P with uncorrelated variables of unit variance and then transformed using some matrix \bf A, the expression {\bf X}^T{\bf X}/n will still be an estimate of {\bf AA}^T, the metric tensor for the corresponding (dual) space deformation.

    However, note that out of all possible initial distributions P, the normal distribution is exactly the one with the maximum entropy, i.e. the “most generic”. Thus, if you base your analysis on the mean and the covariance matrix (which is what you do with PCA, for example), you could just as well assume your data to be normally distributed. In fact, a good rule of thumb is to remember, that whenever you even mention the word "covariance matrix", you are implicitly fitting a Gaussian distribution to your data.

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  • Posted by Konstantin 09.01.2016 No Comments

    This is a (slightly updated) repost of my quora answer to the corresponding question.

    There are many ways in which smart people tend to explain Bayesian statistics and contrast it with a "non-Bayesian" one. One usually highlights that the primary concept of a Bayesian approach is the the desire to model everything as a probability distribution. Once this is fact is clear, many smart people would proceed to claim that this is, in fact, what fundamentally sets Bayesian statistics aside from the "classical" one. However, I feel that this kind of explanation is somewhat incomplete. It is not like classical statisticians do not use complete probability distributions. The difference is in general somewhat more subtle and philosophical.

    Consider the question "what is your height?". For a classical statistician there exists some abstract "true answer", say "180cm", which is a fixed number - your one and only height. The problem is, of course, you do not know this number because every measurement is slightly different, so the classical statistician will add that "there is a normally-distributed measurement error". In the world of a pure Bayesian there are almost no "fixed numbers" - everything is a probability distribution, and so is your height! That is, a Bayesian should say that "your height is a Normal distribution centered around 180cm".

    Note that from the mathematical perspective there is no difference between the two representations - in both cases the number 180cm is mentioned, and the normal distribution. However, from a philosophical, syntactical, methodological and "mental" perspectives this tends to have serious implications, and there has been historically a kind of an ongoing intellectual feud between the statisticians who lend more towards the first or the second approach (it is somewhat resemblant of how there is a divide among the physicists with regard to their support of the Copenhagen interpretation of quantum mechanics).

    One of the implications of denying the fact that things in the world are mostly fixed (and are all pure distributions instead) is that you may not use many of the common sense inference methods directly. What is my height if I stand on a chair? "Well, it is your height plus the height of a chair", a classical statistician would say. He can keep in mind the measurement errors, if necessary, but those could be dealt with later. In the Bayesian world heights are not numbers, so the procedure of adding heights implies convoluting two distributions to get the resulting distribution. If both distributions are Gaussian, the result will match that of the "common sense", but note that now the common sense somehow became "just one special case". Moreover, a Bayesian might even keep the possibility that "your height and the height of the chair are dependent" in the back of his mind, just in case. Because when you speak about two numbers in the Bayesian world, you must immediately start thinking about their joint distribution.

    On the other hand, modeling everything in probabilities lets you use probability theory inference methods (Bayes rule, convolutions, marginalizations, etc) everywhere, without the need to differentiate between "fixed numbers" and "random measurement errors" and this adds peace of mind as well as tends to make your explanations clearer. A Bayesian confidence interval, for example, is a "fixed interval such that 95% of height measurements fall into it". A classical confidence interval, on the other hand, is "a random interval such that the true height may fall into it with 95% probability". Again, mathematically and numerically those may often be the same, but think how different the two explanations are.

    Bayesian "thinking" tends to be more flexible for complex models. Many classical statistics models would stick to fixed parameters, point or "interval" inferences, and try to "hide" the complexity of probability distributions as much as possible. As a result, reasoning about a system with many highly interconnected concepts becomes flawed. Consider a sequence of three questions:

    • What the height of this truck?
    • Will it fit under this 3m bridge?
    • Do we need pick another route?

    In the "classical" mindset you would tend to give fixed answers to the questions.

    • "Height of the truck is 297".
    • "Yes, 297<300, hence it will fit".
    • "No, we do not need".

    Sometimes you may be more careful and work with confidence intervals, but it still feels unwieldy:

    • "The confidence interval on the height of the truck is 290..310"
    • ".. aahm, it might not fit..."
    • "let's pick another route, just in case"

    Note, if a followup question appears that depends on the previous inferences (e.g. "do we need to remodel the truck") answering it becomes even harder because the true uncertainty is "lost" in the intermediate steps. Such problems are never present if you are disciplined as a Bayesian. Note the answers:

    • "The height of the truck is a normal distribution N(297, 10)"
    • "It will fit under the bridge with probability 60%"
    • "We need another route with probability 40%"

    At any point is information about the uncertainty is preserved in the distributions and you are free to combine it further, or apply a decision-theoretic utility model. This makes Bayesian networks possible, for example.

    It is interesting to see how this largely philosophical preference leads to two completely different (albeit complementary) sets of techniques. Indeed, if you are a true classical statistician, your work revolves around parameterized probability distributions. You write them down like P_\alpha(x), where x is the "truly random" value from some probability space, and \alpha is the "fixed but unknown" parameter. Your whole "school of thought" is now focused on clever ad-hoc techniques for computing estimates of this fixed parameter from the provided distribution.

    For a pure Bayesian, however, there is no "fixed" \alpha that has to be treated somehow separately. Instead, \alpha is also a part of some probability space, and instead of writing P_\alpha(x) he would safely write P(x| \alpha), P(\alpha | x), or P(x, \alpha). As a result, the probability distribution he works with are not parameterized any more, and all of the clever techniques that the classical statisticians have invented over the centuries for estimating parameters become seemingly useless. At this point a classical statistician puts his hands down and goes home, as there is nothing to do for him - there are no "unknowns". The Bayesian is, however, left to struggle with mathematically trivial, yet computationally incredibly heavy methods for extracting essentially the same values that the classical statistician could have obtained using his "parameter estimation" approaches. That's why the Bayesian "school of thought" is mostly focused on computationally-efficient methods for marginalization and sampling.

    In reality, of course, a Bayesian would quite often give up and "cheat", at least partially parameterizing his models and making use of the classical estimation methods, while a "classical" statistician might happen to write P(x|\alpha) and apply the Bayes rule here and there, whenever it seems appropriate. A number of computations derived from the two theoretical backgrounds end up exactly the same.

    Thus, in practice, labeling things as "Bayesian" or "non-Bayesian" is still largely a philosophical choice. For example, there are methods in machine learning, ensemble learners, that are somewhy never labeled/marketed as being "Bayesian" nor were they probably invented by someone "Bayesian", although at their core those would be among the best examples of where a Bayesian approach is different from a classical one. Those are also among the best performant models quite often, by the way.

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  • Posted by Konstantin 04.01.2016 5 Comments

    Collecting large amounts of data and then using it to "teach" computers to automatically recognize patterns is pretty much standard practice nowadays. It seems that, given enough data and the right methods, computers can get quite precise at detecting or predicting nearly anything, whether it is face recognition, fraud detection or movie recommendations.

    Whenever a new classification system is created, it is taken for granted that the system should be as precise as possible. Of course, classifiers that never make mistakes are rare, but if it possible, we should strive to have them make as few mistakes as possible, right? Here is a fun example, where things are not as obvious.

    risk

    Consider a bank, which, as is normal for a bank, makes money by giving loans to its customers. Of course, there is always a risk that a customer will default (i.e. not repay the loan). To account for that, the bank has a risk scoring system which, for a given loan application, assesses the probability that the corresponding customer may default. This probability is later used to compute the interest rate offered for the customer. To simplify a bit, the issued interest on a loan might be computed as the sum of customer's predicted default risk probability and a fixed profit margin. For example, if a customer is expected to default with probability 10% and the bank wants 5% profit on its loans on average, the loan might be issued at slightly above 15% interest. This would cover both the expected losses due to non-repayments as well as the profit margin.

    Now, suppose the bank managed to develop a perfect scoring algorithm. That is, each application gets a rating of either having 0% or 100% risk. Suppose as well that within a month the bank processes 1000 applications, half of which are predicted to be perfectly good, and half - perfectly bad. This means that 500 loans get issued with a 5% interest rate, while 500 do not get issued at all.

    Think what would happen, if the system would not do such a great job and confused 50 of the bad applications with the good ones? In this case 450 applications would be classified as "100%" risk, while 550 would be assigned a risk score of "9.1%" (we still require the system to provide valid risk probability estimates). In this case the bank would issue a total of 550 loans at 15%. Of course, 50 of those would not get repaid, yet this loss would be covered from the increased interest paid by the honest lenders. The financial returns are thus exactly the same as with the perfect classifier. However, the bank now has more clients. More applications were signed, and more contract fees were received.

    True, the clients might be a bit less happy for getting a higher interest rate, but assuming they were ready to pay it anyway, the bank does not care. In fact, the bank would be more than happy to segment its customers by offering higher interest rates to low-risk customers anyway. It cannot do it openly, though. The established practices usually constrain banks to make use of "reasonable" scorecards and offer better interest rates to low-risk customers.

    Hence, at least in this particular example, a "worse" classifier is in fact better for business. Perfect precision is not really the ultimately desired feature. Instead, the system is much more useful when it provides a relevant and "smooth" distribution of predicted risk scores, making sure the scores themselves are decently precise estimates for the probability of a default.

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  • Posted by Konstantin 22.03.2015 4 Comments

    This is a repost of my quora answer to the question: In layman's terms, how does Naive Bayes work?

    Suppose that you are a working as a security guard at the airport. Your task is to look at people who pass the security line and pick some of them as being worthy of a more detailed screening. Now, of course, telling whether a person is a potential criminal or not by just looking at him/her is hard, if at all possible, but you need to do something. You have been put there for some reason, after all.

    One of the simplest ways to approach the problem, mentally, is the following. You assign a "risk value" for each person. At the beginning (when you don't have any information about the person at all) you set this value to zero.

    Now you start studying various features of the person in front of you: is it a male or a female? Is it a kid? Is he behaving nervously? Is he carrying a big bag? Is he alone? Did the metal detector beep? Is he a foreigner? etc. For each of those features you know (subconsciously due to your presuppositions, or from actual statistics) the average increase or decrease in risk of the person being a criminal that it entails. For example, if you know that the proportion of males among criminals is the same as the proportion of males among non-criminals, observing that a person is male will not affect his risk value at all. If, however, there are more males among criminals (suppose the percentage is, say, 70%) than among decent people (where the proportion is around 50%), observing that a person in front of you is a male will increase the "risk level" by some amount (the value is log(70%/50%) ~ 0.3, to be precise). Then you see that a person is nervous. OK, you think, 90% of criminals are nervous, but only 50% of normal people are. This means that nervousness should entail a further risk increase (of log(0.9/0.5) ~ 0.6, to be technical again, so by now you have counted a total risk value of 0.9). Then you notice it is a kid. Wow, there is only 1% of kids among criminals, but around 10% among normal people. Therefore, the risk value change due to this observation will be negative (log(0.01/0.10) ~ -2.3, so your totals are around -1.4 by now).

    You can continue this as long as you want, including more and more features, each of which will modify your total risk value by either increasing it (if you know this particular feature is more representative of a criminal) or decreasing (if the features is more representative of a decent person). When you are done collecting the features, all is left for you is to compare the result with some threshold level. Say, if the total risk value exceeds 10, you declare the person in front of you to be potentially dangerous and take it into a detailed screening.

    The benefit of such an approach is that it is rather intuitive and simple to compute. The drawback is that it does not take the cross-play of features into account. It may very well be the case that while the feature "the person is a kid" on its own greatly reduces the risk value, and the feature "has a moustache" on its own has close to no effect, a combination of the two ("a kid with a moustache") would actually have to increase the risk by a lot. This would not happen when you simply add the separate feature contributions, as described above.

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  • Posted by Konstantin 25.02.2013 No Comments

    If anyone tells you he or she understands probability theory, do not believe them. That person simply does not know enough of it to admit, that probability theory is riddled with paradoxes, where common sense must step aside and wait in silence, or your brain will hurt. Substring statistics is probably among the lesser-known, yet magically unintuitive examples.

    Consider a sequence of random coin flips. Each coin flip is either a "heads" or a "tails", hence the sequence might written down as a sequence of H and T-s: HTHTHHTT...

    It is easy to show that the probability of the sequence to begin with, say, HHH is equal to P(HHH) = 1/8th, as is the case with any other three-letter combination: P(HHT) = P(THH) = P(THT) = 1/8, etc. Moreover, by symmetry, the probability of seeing a particular three-letter combination at any fixed position in the sequence is still always 1/8-th. All three-letter substrings seem to be equivalent here.

    But let us now play a game, where we throw a coin until we see a particular three-letter combination. To be more specific, let us wait until either HHT or HHH comes up. Suppose I win in the first case and you win in the second one. Obviously, the game first proceeds until two heads are flipped. Then, whichever coin flip comes up next determines the winner. Sounds pretty fair, doesn't it? Well, it turns out that, surprisingly, if you count carefully the expected number of coin flips to obtain HHT, it happens to be 8, whereas for HHH it is 14! Ha! Does it mean I have an advantage? Suprisingly again, no. The probability of HHT occuring before HHH in any given sequence is still precisely 0.5 and, as we reasoned initially, the game is still fair.

    We can, however, construct even more curious situations with four-letter combinations. Suppose I bet on HTHT and you bet on THTT.  The expected number of coin flips to obtain my combination can be computed to be 20. The expected number of flips to get your combination is smaller: 18 flips. However, it is still more probable (64%) that my combination will happen before yours!

    If this is not amusing enough, suppose that four of us are playing such a game. Player A bets on the string THH, Player B bets on HHT, player C on HTT and player D on TTH. It turns out that A's combination will occur earlier than B's with probability 75%. B's combination, however, wins over C's with probability 66.7%. C's combination, though, wins over D's with probability 75%. And, to close the loop, D wins over A with probability 66.7%! This is just like the nontransitive dice.

    Hopefully, you are amazed enough at this point to require an explanation for how this all might happen. Let me leave it to you as a small puzzle:

    • Explain in simple terms, how can it happen so that the expected time to first occurrence of otherwise equiprobable substrings may be different?
    • Explain in simple terms, how can it be so that one substring has higher than 50% chance of occuring earlier than some other substring.
    • Finally, explain why the two effects above are not strictly related to each other.

    PS: The theory used to compute actual probabilities and expected times to occurrence of a substring is elegant yet somewhat involved. For the practically-minded, here is the code to check the calculations.

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  • Posted by Konstantin 12.11.2012 No Comments

    This relates nicely to several previous posts here.

    Frequentists vs Bayesians

    Copyright © xkcd.

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  • Posted by Konstantin 04.12.2011 1 Comment

    There is one rule of thumb that I find quite useful and happen to use fairly often. It is probably not widely known nor described in textbooks (I stumbled upon it on my own), so I regularly have to explain it.  Next time I'll just point out to this post.

    The rule is the following: a proportion estimate obtained on a sample of n points should only be trusted up to an error of \frac{1}{\sqrt{n}}.

    For example, suppose that you read in the newspaper that "25% of students like statistics". Now, if this result has been obtained from a survey of 64 participants, you should actually interpret the answer as 0.25\pm\frac{1}{\sqrt{64}}, that is, 0.25\pm 0.125, which means that the actual percentage lies somewhere between 12.5% and 37.5%.

    As another example, in machine learning, you often get to see cases where someone evaluates two classification algorithms on a test set of, say, 400 instances, measures that the first algorithm has a precision of 90%, the second a precision of, say, 92%, and boldly claims the dominance of the second algorithm. At this point, without going deeply into statistics, it is easy to figure that 1/\sqrt{400} should be somewhere around 5%, hence the difference between 90% and 92% is not too significant to celebrate.

    The Explanation

    The derivation of the rule is fairly straightforward. Consider a Bernoulli-distributed random variable with parameter p. We then take an i.i.d. sample of size n, and use it to estimate \hat p:

        \[\hat p = \frac{1}{n}\sum_i X_i\]

    The 95% confidence interval for this estimate, computed using the normal approximation is then:

        \[\hat p \pm 1.96\sqrt{\frac{p(1-p)}{n}}\]

    What remains is to note that 1.96\approx 2 and that \sqrt{p(1-p)} \leq 0.5. By substituting those two approximations we immediately get that the interval is at most

        \[\hat p \pm \frac{1}{\sqrt{n}}\]

    Limitations

    It is important to understand the limitations of the rule. In the cases where the true proportion estimate is p=0.5 and n is large enough for the normal approximation to make sense (20 is already good), the one-over-square-root-of-n rule is very close to a true 95% confidence interval.

    When the true proportion estimate is closer to 0 or 1, however, \sqrt{p(1-p)} is not close to 0.5 anymore, and the rule of thumb results in a conservatively large interval.

    In particular, the true 95% confidence interval for p=0.9 will be nearly two times smaller (\approx 0.6/\sqrt{n}). For p=0.99 the actual interval is five times smaller (\approx 0.2/\sqrt{n}). However, given the simplicity of the rule, the fact that the true p is rarely so close to 1, and the idea that it never hurts to be slightly conservative in statistical estimates, I'd say the one-over-a-square-root-of-n rule is a practically useful tool in most situations.

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  • Posted by Konstantin 28.10.2011 No Comments

    I do not know who is the author, but I think this is great:

    Self-referential question

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  • Posted by Konstantin 12.02.2010 4 Comments

    Statistics is mainly about using the observed data to make conclusions about the "real state of affairs", underlying that data. A classical and most widely spread technique for making these conclusions is based on significance testing. In simple terms, the idea of significance testing is to ask the question: "if the real state of affairs were X, how probable would it be for us to obtain the data D we are currently observing?". If the answer is "rather unprobable" (e.g. p < 0.05), the common decision is to reject the proposition X in favor of the alternative "not X". Otherwise the researcher claims to "see no reason to reject X".

    The logic behind that reasoning seems quite solid from the first glance, yet it is well known to be faulty. Naturally, the fact that the likelihood of the data P(D | X) is low need not imply that the underlying hypothesis is wrong - it might very well be the case that the data by itself is already rare enough to make this value low. The only correct way of making sound judgments is to consider the a-posteriori probability of the hypothesis P(X | D) instead. However, the latter can be quite inconvenient to compute. Besides, the wild popularity of significance tests and p-values seems to indicate that the issue is not at all that serious. Really, P(X | D) looks so similar to P(D | X), who cares?

    Book cover

    The book "What If There Were No Significance Tests?", which I stumbled upon recently while browsing a stray library shelf, makes it clear that this issue is not a joke. It is a collection of chapters written by renowned statisticians (most of which some-why work in the field of psychology), that quite convincingly condemns the widespread overuse of p-values and the related significance-based hypothesis testing in favor of other approaches. The main point is nailed quite precisely in the very first essay by Jacob Cohen, which I strongly advise you to read right now in order to get rid of any illusions you might still have regarding significance testing. And when you're done with that, you can continue reading this post.

    In the following I shall provide my personal summary of the marvelous "Member of Congress" example from J.Cohen's essay. So far it is the best illustration I know of, about why exactly it is dangerous to use significance tests blindly.

    Improbable does not mean impossible

    Consider the following situation. We have observed a person which we know to be a Member of the US Congress. We are interested in testing the hypothesis, that this person is an American citizen. To apply the significance testing methodology, we proceed by estimating the p-value:

    P(Congressman | American) ~ 535/300 000 000.

    This is clearly below the popular 0.05 threshold. As a result, we are forced to reject the null-hypothesis and conclude that the person is not an American citizen. Bummer.

    What is the problem here? Well, one thing is worth noting - while the probability for an American to be a congressman is low, it is even lower (precisely, zero), for a non-American. So maybe we would have been better off if we expanded the procedure above to the following "maximum-likelihood"-style reasoning:

    Considering that the likelihood P(Congressman | American) is greater than the likelihood P(Congressman | non-American), we must conclude that the person in front of us is an American rather than not.

    Did we just solve the problem? Is it enough to consider "p-values both ways" to clear things up? No!

    Maximum likelihood does not work

    Let us now consider a reversed situation. We are faced with a person, which, we know, is an American. We are interested in the hypothesis that he is a congressman. Compute the two likelihoods:

    P(American | Congressman) = 1

    P(American | not Congressman) ~ 300 000 000 / 6 700 000 000

    Observing that the first likelihood is greater than the second we are forced to conclude that the person in front of us is indeed a congressman. Bummer, again!

    Only by multiplying the likelihood with the marginal probability P(Congressman) could we have obtained the correct decision. Which is, to say, we must have been estimating the probabilities the other way around from the start.

    To summarize, be wary of these pitfalls. I would not agree with the strong negative opinion of the authors of the book, though. After all, a lot of stuff is quite fruitfully done nowadays using p-values only. However, each time you use them, do it sensibly and keep in mind the following two aspects:

    1. If your p-value is low, can this be solely due to low marginal probability of the data? What is the "reversed" p-value? What is the power of your test?
    2. If you suspect that your hypotheses might be subject to a highly non-uniform prior probabilities, do not use bare p-values. You must consider the prior!

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