I just realized that support vector machine can be used for regression thanks to the nice article However, I am quite confused with the definition of the hyperparameter C.
I am well aware of the slack variables \xi_i associated with each data point and the hyperparameter C in classification SVM. There, the objective function is
\min_{w, b} \frac{|w|}{2} + C\sum_{i=1}^N \xi_i, such that
y_i (w \cdot x_i + b) \ge 1 - \xi_i and \xi_i \ge 0.
In SVM, the larger C is, the larger the penalty and hence soft SVM reduces to hard SVM as C goes to infinity. (sorry for the raw latex code, i remember latex is supported but it seems not the case)
From the linked article, the objective function and the constraints are as follows
I think the equations also imply that the larger C is, the larger the penalty. However, the author of the article claims the opposite,
I noticed that someone asked the author the same question at the end of the article but there has been no response.
I guess there might be a typo in the equation, so I looked for support from any reference and then I found that SVR in Python uses the same convention that the strength of regularization is inversely proportional to C. I tried to check the source code of SVR but I can't find any formula. Can someone help resolve this? Thanks!
C is regularization parameter, which means its before the square of W, not before the relaxation, so the C parameter may be equal to 1/C.
I'm estimating a linear OLS regression using some software, and I have three variables: Y (dependent), X1 (independent), and Intercept (a column of "1"s I manually created). I created Intercept because this particular software doesn't have a function to add a constant term.
The coefficients of X and Intercept are almost perfectly inverse (i.e. Intercept-coefficient = 1.5 and X-coefficient = negative 1.51). Both Y and X are columns of very small percentage changes (i.e. 0.0001). I've tried adding some other independent variables, and quickly run into multicollinearity issues - not sure if that's simply because the variables are highly similar, though.
I'm not very experienced with stats, are the coefficients a dead giveaway from statistical issues with the regression? Any advice is much appreciated, thank you!
Yesterday, I posted a question about general concept of SVM Primal Form Implementation:
Support Vector Machine Primal Form Implementation
and "lejlot" helped me out to understand that what I am solving is a QP problem.
But I still don't understand how my objective function can be expressed as QP problem
(http://en.wikipedia.org/wiki/Support_vector_machine#Primal_form)
Also I don't understand how QP and Quasi-Newton method are related
All I know is Quasi-Newton method will SOLVE my QP problem which supposedly formulated from
my objective function (which I don't see the connection)
Can anyone walk me through this please??
For SVM's, the goal is to find a classifier. This problem can be expressed in terms of a function that you are trying to minimize.
Let's first consider the Newton iteration. Newton iteration is a numerical method to find a solution to a problem of the form f(x) = 0.
Instead of solving it analytically we can solve it numerically by the follwing iteration:
x^k+1 = x^k - DF(x)^-1 * F(x)
Here x^k+1 is the k+1th iterate, DF(x)^-1 is the inverse of the Jacobian of F(x) and x is the kth x in the iteration.
This update runs as long as we make progress in terms of step size (delta x) or if our function value approaches 0 to a good degree. The termination criteria can be chosen accordingly.
Now consider solving the problem f'(x)=0. If we formulate the Newton iteration for that, we get
x^k+1 = x - HF(x)^-1 * DF(x)
Where HF(x)^-1 is the inverse of the Hessian matrix and DF(x) the gradient of the function F. Note that we are talking about n-dimensional Analysis and can not just take the quotient. We have to take the inverse of the matrix.
Now we are facing some problems: In each step, we have to calculate the Hessian matrix for the updated x, which is very inefficient. We also have to solve a system of linear equations, namely y = HF(x)^-1 * DF(x) or HF(x)*y = DF(x).
So instead of computing the Hessian in every iteration, we start off with an initial guess of the Hessian (maybe the identity matrix) and perform rank one updates after each iterate. For the exact formulas have a look here.
So how does this link to SVM's?
When you look at the function you are trying to minimize, you can formulate a primal problem, which you can the reformulate as a Dual Lagrangian problem which is convex and can be solved numerically. It is all well documented in the article so I will not try to express the formulas in a less good quality.
But the idea is the following: If you have a dual problem, you can solve it numerically. There are multiple solvers available. In the link you posted, they recommend coordinate descent, which solves the optimization problem for one coordinate at a time. Or you can use subgradient descent. Another method is to use L-BFGS. It is really well explained in this paper.
Another popular algorithm for solving problems like that is ADMM (alternating direction method of multipliers). In order to use ADMM you would have to reformulate the given problem into an equal problem that would give the same solution, but has the correct format for ADMM. For that I suggest reading Boyds script on ADMM.
In general: First, understand the function you are trying to minimize and then choose the numerical method that is most suited. In this case, subgradient descent and coordinate descent are most suited, as stated in the Wikipedia link.
I have a probability / stats question related to implementing Naive Bayes Classifiers, in particular about implementing Laplace Smoothing to avoid the Zero count issue and overfitting.
From what Ive read, the basic NBC formula using MLE looks like this:
p(C│F_1 ...F_n )=(p(C)p(F_1 |C)...p(F_n |C))/(p(F_1)...p(F_n))
However if one of the p(F_i |C) is zero, the whole probability becomes 0. One solution is Lapace smooth
p(F_i│C)~(x_i+k)/(N+kd)
Where x_i is the number of times F_i appeared in class C, N is the number of times class C occurred and d is the number of distinct values F_i has been known to take on.
My question is this:
What if anything needs to be done to p(C) in the numerator, and p(F_i) in the denominator?
Let f = (f_1 ... f_n). Laplace smoothing affects the class likelihood term, p(f|C). It does not affect the class prior p(C). It does have an effect on the marginal probability of f, in the sense that:
p(f) = \sum_c p(C) * p(f|C)
Whereby the term p(f|C) is different if you're Laplace smoothing than if you're not. But since the denominator is constant for all C, you shouldn't be bothering to evaluate this anyway.
P.S. This isn't really a programming question!
I've been trying to find an answer to this for months (to be used in a machine learning application), it doesn't seem like it should be a terribly hard problem, but I'm a software engineer, and math was never one of my strengths.
Here is the scenario:
I have a (possibly) unevenly weighted coin and I want to figure out the probability of it coming up heads. I know that coins from the same box that this one came from have an average probability of p, and I also know the standard deviation of these probabilities (call it s).
(If other summary properties of the probabilities of other coins aside from their mean and stddev would be useful, I can probably get them too.)
I toss the coin n times, and it comes up heads h times.
The naive approach is that the probability is just h/n - but if n is small this is unlikely to be accurate.
Is there a computationally efficient way (ie. doesn't involve very very large or very very small numbers) to take p and s into consideration to come up with a more accurate probability estimate, even when n is small?
I'd appreciate it if any answers could use pseudocode rather than mathematical notation since I find most mathematical notation to be impenetrable ;-)
Other answers:
There are some other answers on SO that are similar, but the answers provided are unsatisfactory. For example this is not computationally efficient because it quickly involves numbers way smaller than can be represented even in double-precision floats. And this one turned out to be incorrect.
Unfortunately you can't do machine learning without knowing some basic math---it's like asking somebody for help in programming but not wanting to know about "variables" , "subroutines" and all that if-then stuff.
The better way to do this is called a Bayesian integration, but there is a simpler approximation called "maximum a postieri" (MAP). It's pretty much like the usual thinking except you can put in the prior distribution.
Fancy words, but you may ask, well where did the h/(h+t) formula come from? Of course it's obvious, but it turns out that it is answer that you get when you have "no prior". And the method below is the next level of sophistication up when you add a prior. Going to Bayesian integration would be the next one but that's harder and perhaps unnecessary.
As I understand it the problem is two fold: first you draw a coin from the bag of coins. This coin has a "headsiness" called theta, so that it gives a head theta fraction of the flips. But the theta for this coin comes from the master distribution which I guess I assume is Gaussian with mean P and standard deviation S.
What you do next is to write down the total unnormalized probability (called likelihood) of seeing the whole shebang, all the data: (h heads, t tails)
L = (theta)^h * (1-theta)^t * Gaussian(theta; P, S).
Gaussian(theta; P, S) = exp( -(theta-P)^2/(2*S^2) ) / sqrt(2*Pi*S^2)
This is the meaning of "first draw 1 value of theta from the Gaussian" and then draw h heads and t tails from a coin using that theta.
The MAP principle says, if you don't know theta, find the value which maximizes L given the data that you do know. You do that with calculus. The trick to make it easy is that you take logarithms first. Define LL = log(L). Wherever L is maximized, then LL will be too.
so
LL = hlog(theta) + tlog(1-theta) + -(theta-P)^2 / (2*S^2)) - 1/2 * log(2*pi*S^2)
By calculus to look for extrema you find the value of theta such that dLL/dtheta = 0.
Since the last term with the log has no theta in it you can ignore it.
dLL/dtheta = 0 = (h/theta) + (P-theta)/S^2 - (t/(1-theta)) = 0.
If you can solve this equation for theta you will get an answer, the MAP estimate for theta given the number of heads h and the number of tails t.
If you want a fast approximation, try doing one step of Newton's method, where you start with your proposed theta at the obvious (called maximum likelihood) estimate of theta = h/(h+t).
And where does that 'obvious' estimate come from? If you do the stuff above but don't put in the Gaussian prior: h/theta - t/(1-theta) = 0 you'll come up with theta = h/(h+t).
If your prior probabilities are really small, as is often the case, instead of near 0.5, then a Gaussian prior on theta is probably inappropriate, as it predicts some weight with negative probabilities, clearly wrong. More appropriate is a Gaussian prior on log theta ('lognormal distribution'). Plug it in the same way and work through the calculus.
You can use p as a prior on your estimated probability. This is basically the same as doing pseudocount smoothing. I.e., use
(h + c * p) / (n + c)
as your estimate. When h and n are large, then this just becomes h / n. When h and n are small, this is just c * p / c = p. The choice of c is up to you. You can base it on s but in the end you have to decide how small is too small.
You don't have nearly enough info in this question.
How many coins are in the box? If it's two, then in some scenarios (for example one coin is always heads, the other always tails) knowing p and s would be useful. If it's more than a few, and especially if only some of the coins are only slightly weighted then it is not useful.
What is a small n? 2? 5? 10? 100? What is the probability of a weighted coin coming up heads/tail? 100/0, 60/40, 50.00001/49.99999? How is the weighting distributed? Is every coin one of 2 possible weightings? Do they follow a bell curve? etc.
It boils down to this: the differences between a weighted/unweighted coin, the distribution of weighted coins, and the number coins in your box will all decide what n has to be for you to solve this with a high confidence.
The name for what you're trying to do is a Bernoulli trial. Knowing the name should be helpful in finding better resources.
Response to comment:
If you have differences in p that small, you are going to have to do a lot of trials and there's no getting around it.
Assuming a uniform distribution of bias, p will still be 0.5 and all standard deviation will tell you is that at least some of the coins have a minor bias.
How many tosses, again, will be determined under these circumstances by the weighting of the coins. Even with 500 tosses, you won't get a strong confidence (about 2/3) detecting a .51/.49 split.
In general, what you are looking for is Maximum Likelihood Estimation. Wolfram Demonstration Project has an illustration of estimating the probability of a coin landing head, given a sample of tosses.
Well I'm no math man, but I think the simple Bayesian approach is intuitive and broadly applicable enough to put a little though into it. Others above have already suggested this, but perhaps if your like me you would prefer more verbosity.
In this lingo, you have a set of mutually-exclusive hypotheses, H, and some data D, and you want to find the (posterior) probabilities that each hypothesis Hi is correct given the data. Presumably you would choose the hypothesis that had the largest posterior probability (the MAP as noted above), if you had to choose one. As Matt notes above, what distinguishes the Bayesian approach from only maximum likelihood (finding the H that maximizes Pr(D|H)) is that you also have some PRIOR info regarding which hypotheses are most likely, and you want to incorporate these priors.
So you have from basic probability Pr(H|D) = Pr(D|H)*Pr(H)/Pr(D). You can estimate these Pr(H|D) numerically by creating a series of discrete probabilities Hi for each hypothesis you wish to test, eg [0.0,0.05, 0.1 ... 0.95, 1.0], and then determining your prior Pr(H) for each Hi -- above it is assumed you have a normal distribution of priors, and if that is acceptable you could use the mean and stdev to get each Pr(Hi) -- or use another distribution if you prefer. With coin tosses the Pr(D|H) is of course determined by the binomial using the observed number of successes with n trials and the particular Hi being tested. The denominator Pr(D) may seem daunting but we assume that we have covered all the bases with our hypotheses, so that Pr(D) is the summation of Pr(D|Hi)Pr(H) over all H.
Very simple if you think about it a bit, and maybe not so if you think about it a bit more.