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
Related
I came up with the following result, tested on many data sets, but I do not have a formal proof yet:
Theorem: The width L of any confidence interval is asymptotically equal (as n tends to infinity) to a power function of n, namely L=A / n^B where A and B are two positive constants depending on the data set, and n is the sample size.
See here and here for details. The B exponent seems to be very similar to the Hurst exponent in time series, not only in terms of what it represents, but also in the values that it takes: B=1/2 corresponds to perfect data (no auto-correlation or undesirable features) and B=1 corresponds to "bad data" typically with strong auto-correlations.
Note that B=1/2 is what everyone uses nowadays, assuming observations are independently and identically distributed, with an underlying normal distribution. I also devised a method to make the interval width converges faster to zero: O(1/n) rather than O(1/SQRT(n)). This is also described in section 3.3. in my article on re-sampling (here) and my approach in this context seems very much related to what is called second-order accurate intervals (usually achieved with modern versions of bootstrapping, see here.)
My question is whether my theorem is original, ground-breaking, and correct, and how would someone prove it (or refute it.)
Example of Confidence Interval
Perl code to produce confidence intervals for the correlation
The first problem is, what do you mean by confidence interval?
Let's say i do non parametric estimation of a density probability function with a kernel density estimator.
Interval confidence has no meaning in this setting. however you can compute something which is the "speed" of convergence of your kernel density estimator to your target function. Depending on the choice of the distance you choose between function, you can get different speed of convergence. And for example, the best speed with $L^{\infty}$ distance depends on a $\log(n)$ factor.
By the way you give yourself a counterexample in your first article.
So for me your theorem can not exist for two reasons :
It is not clear, you need to specify exactly what you mean by confidence interval. You need to say what do you mean by depending on the dataset (does it depends on $N$ the number of observations?)
There is "counter example", since asymptotic speed of convergence of estimators can be more complicated than what you say.
i was reading about Classification Algorithm KNN and came across with one term Distance Sensitive Data. I was not able to Found what exactly is Distance Sensitive Data wha are it's classifications, How to say if our Data is Distance-Sensitive or Not?
Suppose that xi and xj are vectors of observed features in cases i and j. Then, as you probably know, kNN is based on distances ||xi-xj||, such as the Euclidean one.
Now if xi and xj contain just a single feature, individual's height in meters, we are fine, as there are no other "competing" features. Suppose that next we add annual salary in thousands. Consequently, we look at distances between vectors like (1.7, 50000) and (1.8, 100000).
Then, in the case of the Euclidean distance, clearly salary feature dominates height and it's almost like we are using the salary feature alone. That is,
||xi-xj||2 ≈ |50000-100000|.
However, if the two features actually have similar importance, then we are doing a poor job. It is even worse if salary is actually irrelevant and we should be using height alone. Interestingly, under weak conditions, our classifier still has nice properties such as universal consistency even in such bad situations. The problem is that in finite samples the performance is our classifier is very bad so that the convergence is very slow.
So, as to deal with that, one may want to consider different distances, such that do something about the scale. Commonly people standardize (set the mean to zero and variance to 1) each feature, but that's not a complete solution either. There are various proposals what could be done (see, e.g., here).
On the other hand, algorithms based on decision trees do not suffer from this. In those cases we just look for a point where to split the variable. For instance, if salary takes values in [0,100000] and the split is at 40000, then Salary/10 would be slit at 4000 so that the results would not change.
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.
For my class project, I am working on the Kaggle competition - Don't get kicked
The project is to classify test data as good/bad buy for cars. There are 34 features and the data is highly skewed. I made the following choices:
Since the data is highly skewed, out of 73,000 instances, 64,000 instances are bad buy and only 9,000 instances are good buy. Since building a decision tree would overfit the data, I chose to use kNN - K nearest neighbors.
After trying out kNN, I plan to try out Perceptron and SVM techniques, if kNN doesn't yield good results. Is my understanding about overfitting correct?
Since some features are numeric, I can directly use the Euclid distance as a measure, but there are other attributes which are categorical. To aptly use these features, I need to come up with my own distance measure. I read about Hamming distance, but I am still unclear on how to merge 2 distance measures so that each feature gets equal weight.
Is there a way to find a good approximate for value of k? I understand that this depends a lot on the use-case and varies per problem. But, if I am taking a simple vote from each neighbor, how much should I set the value of k? I'm currently trying out various values, such as 2,3,10 etc.
I researched around and found these links, but these are not specifically helpful -
a) Metric for nearest neighbor, which says that finding out your own distance measure is equivalent to 'kernelizing', but couldn't make much sense from it.
b) Distance independent approximation of kNN talks about R-trees, M-trees etc. which I believe don't apply to my case.
c) Finding nearest neighbors using Jaccard coeff
Please let me know if you need more information.
Since the data is unbalanced, you should either sample an equal number of good/bad (losing lots of "bad" records), or use an algorithm that can account for this. I think there's an SVM implementation in RapidMiner that does this.
You should use Cross-Validation to avoid overfitting. You might be using the term overfitting incorrectly here though.
You should normalize distances so that they have the same weight. By normalize I mean force to be between 0 and 1. To normalize something, subtract the minimum and divide by the range.
The way to find the optimal value of K is to try all possible values of K (while cross-validating) and chose the value of K with the highest accuracy. If a "good" value of K is fine, then you can use a genetic algorithm or similar to find it. Or you could try K in steps of say 5 or 10, see which K leads to good accuracy (say it's 55), then try steps of 1 near that "good value" (ie 50,51,52...) but this may not be optimal.
I'm looking at the exact same problem.
Regarding the choice of k, it's recommended be an odd value to avoid getting "tie votes".
I hope to expand this answer in the future.
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Why do we have to normalize the input for a neural network?
I understand that sometimes, when for example the input values are non-numerical a certain transformation must be performed, but when we have a numerical input? Why the numbers must be in a certain interval?
What will happen if the data is not normalized?
It's explained well here.
If the input variables are combined linearly, as in an MLP [multilayer perceptron], then it is
rarely strictly necessary to standardize the inputs, at least in theory. The
reason is that any rescaling of an input vector can be effectively undone by
changing the corresponding weights and biases, leaving you with the exact
same outputs as you had before. However, there are a variety of practical
reasons why standardizing the inputs can make training faster and reduce the
chances of getting stuck in local optima. Also, weight decay and Bayesian
estimation can be done more conveniently with standardized inputs.
In neural networks, it is good idea not just to normalize data but also to scale them. This is intended for faster approaching to global minima at error surface. See the following pictures:
Pictures are taken from the coursera course about neural networks. Author of the course is Geoffrey Hinton.
Some inputs to NN might not have a 'naturally defined' range of values. For example, the average value might be slowly, but continuously increasing over time (for example a number of records in the database).
In such case feeding this raw value into your network will not work very well. You will teach your network on values from lower part of range, while the actual inputs will be from the higher part of this range (and quite possibly above range, that the network has learned to work with).
You should normalize this value. You could for example tell the network by how much the value has changed since the previous input. This increment usually can be defined with high probability in a specific range, which makes it a good input for network.
There are 2 Reasons why we have to Normalize Input Features before Feeding them to Neural Network:
Reason 1: If a Feature in the Dataset is big in scale compared to others then this big scaled feature becomes dominating and as a result of that, Predictions of the Neural Network will not be Accurate.
Example: In case of Employee Data, if we consider Age and Salary, Age will be a Two Digit Number while Salary can be 7 or 8 Digit (1 Million, etc..). In that Case, Salary will Dominate the Prediction of the Neural Network. But if we Normalize those Features, Values of both the Features will lie in the Range from (0 to 1).
Reason 2: Front Propagation of Neural Networks involves the Dot Product of Weights with Input Features. So, if the Values are very high (for Image and Non-Image Data), Calculation of Output takes a lot of Computation Time as well as Memory. Same is the case during Back Propagation. Consequently, Model Converges slowly, if the Inputs are not Normalized.
Example: If we perform Image Classification, Size of Image will be very huge, as the Value of each Pixel ranges from 0 to 255. Normalization in this case is very important.
Mentioned below are the instances where Normalization is very important:
K-Means
K-Nearest-Neighbours
Principal Component Analysis (PCA)
Gradient Descent
When you use unnormalized input features, the loss function is likely to have very elongated valleys. When optimizing with gradient descent, this becomes an issue because the gradient will be steep with respect some of the parameters. That leads to large oscillations in the search space, as you are bouncing between steep slopes. To compensate, you have to stabilize optimization with small learning rates.
Consider features x1 and x2, where range from 0 to 1 and 0 to 1 million, respectively. It turns out the ratios for the corresponding parameters (say, w1 and w2) will also be large.
Normalizing tends to make the loss function more symmetrical/spherical. These are easier to optimize because the gradients tend to point towards the global minimum and you can take larger steps.
Looking at the neural network from the outside, it is just a function that takes some arguments and produces a result. As with all functions, it has a domain (i.e. a set of legal arguments). You have to normalize the values that you want to pass to the neural net in order to make sure it is in the domain. As with all functions, if the arguments are not in the domain, the result is not guaranteed to be appropriate.
The exact behavior of the neural net on arguments outside of the domain depends on the implementation of the neural net. But overall, the result is useless if the arguments are not within the domain.
I believe the answer is dependent on the scenario.
Consider NN (neural network) as an operator F, so that F(input) = output. In the case where this relation is linear so that F(A * input) = A * output, then you might choose to either leave the input/output unnormalised in their raw forms, or normalise both to eliminate A. Obviously this linearity assumption is violated in classification tasks, or nearly any task that outputs a probability, where F(A * input) = 1 * output
In practice, normalisation allows non-fittable networks to be fittable, which is crucial to experimenters/programmers. Nevertheless, the precise impact of normalisation will depend not only on the network architecture/algorithm, but also on the statistical prior for the input and output.
What's more, NN is often implemented to solve very difficult problems in a black-box fashion, which means the underlying problem may have a very poor statistical formulation, making it hard to evaluate the impact of normalisation, causing the technical advantage (becoming fittable) to dominate over its impact on the statistics.
In statistical sense, normalisation removes variation that is believed to be non-causal in predicting the output, so as to prevent NN from learning this variation as a predictor (NN does not see this variation, hence cannot use it).
The reason normalization is needed is because if you look at how an adaptive step proceeds in one place in the domain of the function, and you just simply transport the problem to the equivalent of the same step translated by some large value in some direction in the domain, then you get different results. It boils down to the question of adapting a linear piece to a data point. How much should the piece move without turning and how much should it turn in response to that one training point? It makes no sense to have a changed adaptation procedure in different parts of the domain! So normalization is required to reduce the difference in the training result. I haven't got this written up, but you can just look at the math for a simple linear function and how it is trained by one training point in two different places. This problem may have been corrected in some places, but I am not familiar with them. In ALNs, the problem has been corrected and I can send you a paper if you write to wwarmstrong AT shaw.ca
On a high level, if you observe as to where normalization/standardization is mostly used, you will notice that, anytime there is a use of magnitude difference in model building process, it becomes necessary to standardize the inputs so as to ensure that important inputs with small magnitude don't loose their significance midway the model building process.
example:
√(3-1)^2+(1000-900)^2 ≈ √(1000-900)^2
Here, (3-1) contributes hardly a thing to the result and hence the input corresponding to these values is considered futile by the model.
Consider the following:
Clustering uses euclidean or, other distance measures.
NNs use optimization algorithm to minimise cost function(ex. - MSE).
Both distance measure(Clustering) and cost function(NNs) use magnitude difference in some way and hence standardization ensures that magnitude difference doesn't command over important input parameters and the algorithm works as expected.
Hidden layers are used in accordance with the complexity of our data. If we have input data which is linearly separable then we need not to use hidden layer e.g. OR gate but if we have a non linearly seperable data then we need to use hidden layer for example ExOR logical gate.
Number of nodes taken at any layer depends upon the degree of cross validation of our output.