I am trying to implement Naive Bayes Gaussian classifier on the number classification data. Where each feature represents a pixel.
When trying to implement this, I hit a bump, I've noticed that some the feature variance equate to 0.
This is an issue because I would not be able to divide by 0 when trying to solve the probability.
What can I do to work around this?
Very short answer is you cannot - even though you can usually try to fit Gaussian distribution to any data (no matter its true distribution) there is one exception - the constant case (0 variance). So what can you do? There are three main solutions:
Ignore 0-variance pixels. I do not recommend this approach as it loses information, but if it is 0 variance for each class (which is a common case for MNIST - some pixels are black, independently from class) then it is actually fully justified mathematically. Why? The answer is really simple, if for each class, given feature is constant (equal to some single value) then it brings literally no information for classification, thus ignoring it will not affect the model which assumes conditional independence of features (such as NB).
Instead of doing MLE estimate (so using N(mean(X), std(X))) use the regularised estimator, for example of form N(mean(X), std(X) + eps), which is equivalent to adding eps-noise independently to each pixel. This is a very generic approach that I would recommend.
Use better distribution class, if your data is images (and since you have 0 variance I assume these are binary images, maybe even MNIST) you have K features, each in [0, 1] interval. You can use multinomial distribution with bucketing, so P(x e Bi|y) = #{ x e Bi | y } / #{ x | y }. Finally this is usually the best thing to do (however requires some knowledge of your data), as the problem is you are trying to use a model which is not suited for the data provided, and I can assure you, that proper distribution will always give better results with NB. So how can you find a good distribution? Plot conditonal marginals P(xi|y) for each feature, and look how they look like, based on that - choose distribution class which matches the behaviour, I can assure you these will not look like Gaussians.
Related
I am using matrix factorization as a recommender system algorithm based on the user click behavior records. I try two matrix factorization method:
The first one is the basic SVD whose prediction is just the product of user factor vector u and item factor i: r = u * i
The second one I used is the SVD with bias component.
r = u * i + b_u + b_i
where b_u and b_i represents the bias of preference of users and items.
One of the models I use has a very low performance, and the other one is reasonable. I really do not understand why the latter one performs worse, and I doubt that it is overfitting.
I googled methods to detect overfitting, and found the learning curve is a good way. However, the x-axis is the size of the training set and y-axis is the accuracy. This make me quite confused. How can I change the size of the training set? Pick out some of the records out of the data set?
Another problem is, I tried to plot the iteration-loss curve (The loss is the ). And it seems the curve is normal:
But I am not sure whether this method is correct because the metrics I use are precision and recall. Shall I plot the iteration-precision curve??? Or this one already tells that my model is correct?
Can anybody please tell me whether I am going in the right direction? Thank you so much. :)
I will answer in reverse:
So you are trying two different models, one that uses straight matrix factorization r = u * i and the other which enters the biases, r = u * i + b_u + b_i.
You mentioned you are doing Matrix Factorization for a recommender system which looks at user's clicks. So my question here is: Is this an Implicit ratings case? or Explicit one? I believe is an Implicit ratings problem if it is about clicks.
This is the first important thing you need to be very aware of, whether your problem is about Explicit or Implicit ratings. Because there are some differences about the way they are used and implemented.
If you check here:
http://yifanhu.net/PUB/cf.pdf
Implicit ratings are treated in a way that the number of times someone clicked or bought a given item for example is used to infer a confidence level. If you check the error function you can see that the confidence levels are used almost as a weight factor. So the whole idea is that in this scenario the biases have no meaning.
In the case of Explicit Ratings, where one has ratings as a score for example from 1-5, one can calculate those biases for users and products (averages of these bounded scores) and introduce them in the ratings formula. They make sense int his scenario.
The whole point is, depending whether you are in one scenario or the other you can use the biases or not.
On the other hand, your question is about over fitting, for that you can plot training errors with test errors, depending on the size of your data you can have a holdout test data, if the errors differ a lot then you are over fitting.
Another thing is that matrix factorization models usually include regularization terms, see the article posted here, to avoid over fitting.
So I think in your case you are having a different problem the one I mentioned before.
I want to create a synthetic dataset consisting of 2 classes and 3 features for testing a hyperparameter optimization technique for a SVM classifier with a RBF kernel. The hyperparameters are gamma and C (the cost).
I have created my current 3D synthetic dataset as follows:
I have created 10 based points for each class by sampling from a multivariate normal distribution with mean (1,0,0) and (0,1,0), respectively, and unit variance.
I have added more points to each class by picking a base point at random and then sampling a new point from a normal distribution with mean equal to the chosen base point and variance I/5.
It would be a very cool thing if I could determine the best C and gamma from the dataset (before running SVM), so that I can see if my optimization technique provides me the best parameters in the end.
Is there a possibility to calculate the best gamma and C parameter from the synthetic dataset described above?
Or else is there a way to create a synthetic dataset where the best gamma and C parameters are known?
Very interesting question, but the answer is no. It is completely data specific, even knowing exactly the distributions, unless you have an infinite sample, it is mathematicaly impossible to prove best C/gamma as SVM in the end is purely point-based method (as opposed to density estimation based). Typical comparison is done in a different scenario - you take real data, and fit hyperparams using other techniques, like gaussian processes (bayesian optimization) etc, which generate baseline (and probably will get to optimal C and gamma too, or at least realy close to them). In the end looking for best C and gamma is not complex problem, thus simply run good techniqe (like bayesopt) for a longer time, and you will get your optimas to compare against. Furthermore, remember that the task of hyperparams optimization is not to find a particular C and gamma, it is to find hyperparams yielding best results, and in fact, even for SVM, there might be many sets of "optimal" C and gammas, all yielding the same results (in terms of your finite dataset) despite being very far away from each other.
How can I test/check whether a given kernel (example: RBF/ polynomial) does really separate my data?
I would like to know if there is a method (not plotting the data of course) which can allow me to check if a given data set (labeled with two classes) can be separated in high dimensional space?
In short - no, there is no general way. However, for some kernels you can easily say that... everything is separable. This property, proved in many forms (among other by Schoenberg) says for example that if your kernel is of form K(x,y) = f(||x-y||^2) and f is:
ifinitely differentible
completely monotonic (which more or less means that if you take derivatives, then the first one is negative, next positive, next negative, ... )
positive
then it will always be able to separate every binary labeled, consistent dataset (there are no two points of the exact same label). Actually it says even more - that you can exactly interpolate, meaning, that even if it is a regression problem - you will get zero error. So in particular multi-class, multi-label problems also will be linearly solvable (there exists linear/multi-linear model which gives you a correct interpolation).
However, if the above properties do not hold, it does not mean that your data cannot be perfectly separated. This is only "one way" proof.
In particular, this class of kernels include RBF kernel, thus it will always be able to separate any training set (this is why it overfits so easily!)
So what about in the other way? Here you have to first fix hyperparameters of the kernel and then you can also answer it through optimization - solve hard-margin SVM problem (C=inf) and it will find a solution iff data is separable.
When we are training our model we usually use MLE to estimate our model. I know it means that the most probable data for such a learned model is our training set. But I'm wondering if its probability match 1 exactly or not?
You almost have it right. The Likelihood of a model (theta) for the observed data (X) is the probability of observing X, given theta:
L(theta|X) = P(X|theta)
For Maximum Likelihood Estimation (MLE), you choose the value of theta that provides the greatest value of P(X|theta). This does not necessarily mean that the observed value of X is the most probable for the MLE estimate of theta. It just means that there is no other value of theta that would provide a higher probability for the observed value of X.
In other words, if T1 is the MLE estimate of theta, and if T2 is any other possible value of theta, then P(X|T1) > P(X|T2). However, there still could be another possible value of the data (Y) different than the observed data (X) such that P(Y|T1) > P(X|T1).
The probability of X for the MLE estimate of theta is not necessarily 1 (and probably never is except for trivial cases). This is expected since X can take multiple values that have non-zero probabilities.
To build on what bogatron said with an example, the parameters learned from MLE are the ones that explain the data you see (and nothing else) the best. And no, the probability is not 1 (except in trivial cases).
As an example (that has been used billions of times) of what MLE does is:
If you have a simple coin-toss problem, and you observe 5 results of coin tosses (H, H, H, T, H) and you do MLE, you will end up giving p(coin_toss == H) a high probability (0.80) because you see Heads way too many times. There are good and bad things about MLE obviously...
Pros: It is an optimization problem, so it is generally quite fast to solve (even if there isn't an analytical solution).
Cons: It can overfit when there isn't a lot of data (like our coin-toss example).
The example I got in my stat classes was as follows:
A suspect is on the run ! Nothing is known about them, except that they're approximatively 1m80 tall. Should the police look for a man or a woman ?
The idea here is that you have a parameter for your model (M/F), and probabilities given that parameter. There are tall men, tall women, short men and short women. However, in the absence of any other information, the probability of a man being 1m80 is larger than the probability of a woman being 1m80. Likelihood (as bogatron very well explained) is a formalisation of that, and maximum likelihood is the estimation method based on favouring parameters which are more likely to result in the actual observations.
But that's just a toy example, with a single binary variable... Let's expand it a bit: I threw two identical die, and the sum of their value is 7. How many side did my die have ? Well, we all know that the probability of two D6 summing to 7 is quite high. But it might as well be D4, D20, D100, ... However, P(7 | 2D6) > P(7 | 2D20), and P(7 | 2D6) > P(7 | 2D100) ..., so you might estimate that my die are 6-faced. That doesn't mean it's true, but its a reasonable estimation, in the absence of any additional information.
That's better, but we're not in machine-learning territory yet... Let's get there: if you want to fit your umptillion-layer neural network on some empirical data, you can consider all possible parameterisations, and how likely each of them is to return the empirical data. That's exploring an umptillion-dimensional space, each dimensions having infinitely many possibilities, but you can map every single one of these points to a likelihood. It is then reasonable to fit your network using these parameters: given that the empirical data did occur, it is reasonable to assume that they should be likely under your model.
That doesn't mean that your parameters are likely ! Just that under these parameters, the observed value is likely. Statistical estimation is usually not a closed problem with a single solution (like solving an equation might be, and where you would have a probability of 1), but we need to find a best solution, according to some metric. Likelihood is such a metric, and is used widely because it has some interesting properties:
It makes intuitive sense
It's reasonably simple to compute, fit and optimise, for a large family of models
For normal variables (which tend to crop up everywhere) MLE gives the same results as other methods, such as least-squares estimations
Its formulation in terms of conditional probabilities makes it easy to use/manipulate it in Bayesian frameworks
Description of classification problem:
Assume a regular dataset X with n samples and d features.
This classification problem is somewhat hard (many features, few samples, low overall AUC ~70%).
It might be useful to mention that feature selection/extraction, dimension reduction, kernels, many classifiers have been applied. So I am not interested in trying these.
I am not looking forward to see an improvement in overall AUC. The goal is to find relevant features in haystack of features.
Description of my approach:
I select all pairwise combination of d features and create many two dimensional sub-datasets x with n samples.
On each sub-dataset x, I perform a 10-fold cross-validation (using all samples of the main dataset X). A very long process, assume weeks of computation.
I select top k pairs (according to highest AUC for example) and label them as +. All other pairs are labeled as -.
For each pair, I can compute several properties (e.g. relations between each pair using Expert's knowledge). These properties can be calculated without using the labels in main dataset X.
Now I have pairs which are labeled as + or -. In addition, each pair has many properties calculated based on Expert's knowledge (i.e. features). Hence, I have a new classification problem. Lets call this newly generated dataset Y.
I train a classifier on Y while following cross-validation rules. Surprisingly, I can predict the + and - labels with 90% AUC.
As far as I can see, it means that I am able to select relevant features. However, seeing a 90% AUC makes me worried about information leakage somewhere in this long process. Specially in step 3.
I was wondering if anyone can see any leakage in this approach.
Information Leakage:
Incorporation of target labels in the actual features. Your classifier will produce good prediction while did not learn anything.
Showing your test set to you classifier during the training phase. Your classifier will "memorize" the test set and its corresponding labels without "learning" anything.
Update 1:
I want to stress that indeed I am using all data points of X in step 1. However, I am not using them ever again (even for testing). The final 90% AUC is obtained from predicting labels of dataset Y.
On the other hand, it would be useful to note that, even if I randomize the values of my main dataset X, the computed features for dataset Y is going to be the same. However, the sample labels in Y would change because the previous + pairs might not be a good one anymore. Therefore they will be labeled as -.
Update 2:
Although I haven't got any opinion, I am going to state what I have got during 4 days of talking with pattern recognition researchers. Briefly I became confident that there is no information leakage (as long as I wont go back to the first dataset X and using its labels). Later on, in case I wanted to check to see if I could have better performance in X (i.e. predicting sample labels), I need to use only a part of dataset X for pairwise comparison (as training set). Then I can use the rest of samples in X as test set while using positively predicted pairs of Y as features.
I will set this as an answer in case no one could reject this method.
If your processes in step 1 uses all data. then the features you are learning have information from the whole data set. Since you selected based on the whole dataset and THEN validation, you are leaking serious information.
You should probably stick with tools that are well known / already done for you before running out and trying weird strategies like this. Try using a model with L1 regularization to do feature selection for your, or start with some of the simpler searches like Sequential Backward Selection.
If you do cross validation correctly in the end, each training will perform its own independent feature selection. If you do one global feature selection and then do CV, you are going to be doing it wrong and probably leaking information.