I am reading the mathematical formulation of SVM and on many sources I found this idea that "max-margin hyperplane is completely determined by those \vec{x}_i which lie nearest to it. These \vec{x}_i are called support vectors."
Could an expert explain mathematically this consequence? please.
This is true only if you use representation theorem
Your separation plane W can be represented as Sum(i= 1 to m) Alpha_i*phi(x_i)
Where m - number of examples in your train sample and x_i is example i. And Phi is function that maps x_i to some feature space.
So, your SVM algorithm will find vector Alpha=(Alpha_1...Alpha_m), - alpha_i for x_i. Every x_i(example in train sample) that his alpha_i is NOT zero is support vector of W.
Hence the name - SVM.
What happens if your data is separable, is that you need only support vectors that are close to separation margin W, and all rest of the training set can be discarded(its alphas is 0). Amount of support vectors that algorithm will use is depends on complexity of data and kerenl you using.
Related
I have a question on inverse prediction in Machine Learning/Data Science. Here I give a example to illustrate my question: I have 20 input features X = (x0, x1, ... x19) and 3 output variables Y = (y0, y1, y2). The number of training/test data usually small, such as <1000 items or even <100 in the training set.
In general, by using the machine learning toolbox (such as scikit learn), I can train the models (such as random forest, linear/polynomial regression and neural network) from X --> Y. But what I actually want to know is, for example, how should I set X, so that I can have y1 values in a specific range (for example y1 > 100).
Does anyone know how to solve this kind of "inverse prediction"? There are two ways in my mind:
Train the model in the normal way: X-->Y, then set a dense mesh in the high dimension X space. In this example, it is 20 dimensions. Then use all the point in this mesh as input data and throw them to the trained model. Select all the input points where the predicted y1 > 100. Finally, use some methods, such as clustering to look for some patterns in the selected data points.
Direct learn models from Y to X. Then, set a dense mesh in the high dimension Y space, where let y1 > 100. Then use the trained models to calculate the X data points.
The second method might be OK when the Y also have high dimensions. But usually, in my application, Y is very low-dimension and X is very high-dimension, which makes me think method 2 is not very practical.
Does anyone have any new thoughts? I think this should be somehow very common in industry and maybe some people meet similar situation before.
Thank you!
From what I understand of your needs, #1 is an excellent fit for this problem. I recommend that you use a simple binary classifier SVM to discriminate good/bad X vectors. SVM works well with high-dimensional spaces, and reading out the coefficients is easy in most SVM interfaces.
Similar note that may be useful:
In inverse/backward prediction, we can predict inversely with similar accuracy of direct/forward prediction of X--->Y and backward of Y--->X only just with solving the systems of equations X<---->Y assuming weights and intercepts. Also, usually, it is better for linear problems AX=B. Note that it is usually possible the Python code for inverse prediction has a considerable error while solving the system of equations (n*n) is better choice with suitable accuracy for that.
Regards
Consider a parametric binary classifier (such as Logistic Regression, SVM etc.) trained on a dataset (say containing two features for e.g. Blood Pressure and Cholesterol level). The dataset is thrown away and the trained model can only be used as a black box (no tweaks and inside information can be gathered from the trained model). Only a set of data points can be provided and their labels predicted.
Is it possible to get information about the mean and/or standard deviation and/or range of the features of the dataset on which this model was trained? If yes, how so? and If no, then why can't we?
Thank you for your response! :)
SVM does not provide any information about the data statistics, it is a maximum margin classifier and it finds the best separating hyperplane between two datasets in the feature space, as a linear combination of "support vectors". If you use kernel functions, then this combination is in the kernel space, it is not even in the original feature space. SVM does not have a straightforward probabilistic interpretation whatsoever.
Logistic regression is a discriminative classifer and models the conditional probability p (y|x,w) where y is your label, x is your data and w are the features. After maximum likelihood training you are left with w and it is again a discriminator (hyperplane) in the feature space, so you don't have the features again.
The following can be considered. Use a Gaussian classifier. Assume that your class is produced by the prior class probability p (y). Then a class conditional density p (x|y,w) produces your data. Then by the Bayes rule, you will have: p (y|x,w) = (p (y)p (x|y,w))/p (x). If you define the class conditional density p (x|y,w) as Gaussian, its parameter set w will consists of the mean vector m and covariance matrix C of x, assuming it is being produced by the class y. But remember that, this will work only based on the assumption that the current data vector belongs to a specific class. Conditioned on w, a better option would be for mean vector: E [x|w]. This the expectation of x with respect to p (x|w). It comes down to a weighted average of mean vectors for the class y=0 and y=1, with respect to their prior class probabilities. Same should work for covariance as well, but it needs to be derived properly, I am not %100 sure right now.
How do I do the transformation in the test data when I have the trained SVM model in hand? I am trying to simulate the SVM output from mathematical equations and the trained SVM model (using RBF kernel). How do I do that?
In SVM, some of the common kernels used are:
Here xi and xj represent two samples. Now if the data has, say 5 samples, does this transformation include all the combination of two samples to generate the transformed feature space, like, x1 and x1, x1 and x2, x1 and x3,..., x4 and x5, x5 and x5.
If data has two features, then a polynomial transformation of order 2 transforms the input to 3 dimensions, as explained her in slide 15 http://www.robots.ox.ac.uk/~az/lectures/ml/lect3.pdf
Now how can be find the similar explantion for the transformation using the RBF kernel? I am trying to write a code for transforming the test data so that i can apply the trained SVM model on it.
This is way more complex than that. In short - you do not map your data directly into feature space. You simply change the dot product to the one induced by the kernel. What happens "inside" SVM when you work with polynomial kernel, each point is actually (indirectly) transformed to O(d^p) dimensional space (where d-input data dimension, p-degree of polynomial kernel). From mathematical perspective you work with some (often unknown) projection phi_K(x) which has the property that K(x, y) = <phi_K(x), phi_K(y)>, and nothing more. In SVM implementations, you do not need actual data representation (as phi_K(x) is usually huge, sometimes even infinite, like in RBF case) but instead it needs vector of dot product of your point will each element of the training set.
Thus what you do (in implementations, not from math perspective) is you provide:
During training whole Gram matrix, G defined as G_ij = K(x_i, x_j) where x_i is i'th training sample
During testing, when you get new point y you provide it to SVM as a vector of dot products H such that H_i = K(y, x_i), where again x_i are your training points (in fact you just need values for support vectors, but many implementations, like libsvm, actually require vector of the size of the training set - you can simply put 0's for K(y, x_j) if x_j is not a training vector)
Just remember, that this is not the same as training linear SVM "on top" of the above representation. This is just a way implementations usually accept your data, as they need a definition of dot product (function) and it is often easier to pass numbers than functions (but some of them, like scikit-learn SVC module, actually accepts functions as kernel parameter).
So what is RBF kernel? It is actually a mapping from points to functions space of normal distributions with means in your training points. And then dot product is just an integral from -inf to +inf from the product of such two functions. Sounds complex? It is at first sight, but it is a really nice trick, worth understanding!
When using SVMlight or LIBSVM in order to classify phrases as positive or negative (Sentiment Analysis), is there a way to determine which are the most influential words that affected the algorithms decision? For example, finding that the word "good" helped determine a phrase as positive, etc.
If you use the linear kernel then yes - simply compute the weights vector:
w = SUM_i y_i alpha_i sv_i
Where:
sv - support vector
alpha - coefficient found with SVMlight
y - corresponding class (+1 or -1)
(in some implementations alpha's are already multiplied by y_i and so they are positive/negative)
Once you have w, which is of dimensions 1 x d where d is your data dimension (number of words in the bag of words/tfidf representation) simply select the dimensions with high absolute value (no matter positive or negative) in order to find the most important features (words).
If you use some kernel (like RBF) then the answer is no, there is no direct method of taking out the most important features, as the classification process is performed in completely different way.
As #lejlot mentioned, with linear kernel in SVM, one of the feature ranking strategies is based on the absolute values of weights in the model. Another simple and effective strategy is based on F-score. It considers each feature separately and therefore cannot reveal mutual information between features. You can also determine how important a feature is by removing that feature and observe the classification performance.
You can see this article for more details on feature ranking.
With other kernels in SVM, the feature ranking is not that straighforward, yet still feasible. You can construct an orthogonal set of basis vectors in the kernel space, and calculate the weights by kernel relief. Then the implicit feature ranking can be done based on the absolute value of weights. Finally the data is projected into the learned subspace.
I have read through a lot of papers and understand the basic concept of a support vector machine at a very high level. You give it a training input vector which has a set of features and bases on how the "optimization function" evaluates this input vector lets call it x, (lets say we're talking about text classification), the text associated with the input vector x is classified into one of two pre-defined classes, this is only in the case of binary classification.
So my first question is through this procedure described above, all the papers say first that this training input vector x is mapped to a higher (maybe infinite) dimensional space. So what does this mapping achieve or why is this required? Lets say the input vector x has 5 features so who decides which "higher dimension" x is going to be mapped to?
Second question is about the following optimization equation:
min 1/2 wi(transpose)*wi + C Σi = 1..n ξi
so I understand that w has something to do with the margins of the hyperplane from the support vectors in the graph and I know that C is some sort of a penalty but I dont' know what it is a penalty for. And also what is ξi representing in this case.
A simple explanation of the second question would be much appreciated as I have not had much luck understanding it by reading technical papers.
When they talk about mapping to a higher-dimensional space, they mean that the kernel accomplishes the same thing as mapping the points to a higher-dimensional space and then taking dot products there. SVMs are fundamentally a linear classifier, but if you use kernels, they're linear in a space that's different from the original data space.
To be concrete, let's talk about the kernel
K(x, y) = (xy + 1)^2 = (xy)^2 + 2xy + 1,
where x and y are each real numbers (one-dimensional). Note that
(x^2, sqrt(2) x, 1) • (y^2, sqrt(2) y, 1) = x^2 y^2 + 2 x y + 1
has the same value. So K(x, y) = phi(x) • phi(y), where phi(a) = (a^2, sqrt(2), 1), and doing an SVM with this kernel (the inhomogeneous polynomial kernel of degree 2) is the same as if you first mapped your 1d points into this 3d space and then did a linear kernel.
The popular Gaussian RBF kernel function is equivalent to mapping your points into an infinite-dimensional Hilbert space.
You're the one who decides what feature space it's mapped into, when you pick a kernel. You don't necessarily need to think about the explicit mapping when you do that, though, and it's important to note that the data is never actually transformed into that high-dimensional space explicitly - then infinite-dimensional points would be hard to represent. :)
The ξ_i are the "slack variables". Without them, SVMs would never be able to account for training sets that aren't linearly separable -- which most real-world datasets aren't. The ξ in some sense are the amount you need to push data points on the wrong side of the margin over to the correct side. C is a parameter that determines how much it costs you to increase the ξ (that's why it's multiplied there).
1) The higher dimension space happens through the kernel mechanism. However, when evaluating the test sample, the higher dimension space need not be explicitly computed. (Clearly this must be the case because we cannot represent infinite dimensions on a computer.) For instance, radial basis function kernels imply infinite dimensional spaces, yet we don't need to map into this infinite dimension space explicitly. We only need to compute, K(x_sv,x_test), where x_sv is one of the support vectors and x_test is the test sample.
The specific higher dimensional space is chosen by the training procedure and parameters, which choose a set of support vectors and their corresponding weights.
2) C is the weight associated with the cost of not being able to classify the training set perfectly. The optimization equation says to trade-off between the two undesirable cases of non-perfect classification and low margin. The ξi variables represent by how much we're unable to classify instance i of the training set, i.e., the training error of instance i.
See Chris Burges' tutorial on SVM's for about the most intuitive explanation you're going to get of this stuff anywhere (IMO).