I observed that as I adjust the eps parameter downwards, the model considers more variables (i.e. less variables have a coef of zero). Is there an explanation for this trend?
Figure 1: eps == 0.001
Figure 2: eps == 0.0001
ElasticNetCV calculates a regularization path [min_alpha, max_alpha] of length n_alphas and runs cross-validation to determine the best alpha. Here, max_alpha is the value for which ElasticNet produces all-zero solution, and min_alpha equals eps * max_alpha. So, for smaller alphas we expect fewer zero coefficients as the objective Loss + min_alpha * Regularization approaches ordinary least squares.
I have hypothesis function h(x) = theta0 + theta1*x.
How can I select theta0 and theta1 value for the linear regression model?
The question is unclear whether you would like to do this by hand (with the underlying math), use a program like Excel, or solve in a language like MATLAB or Python.
To start, here is a website offering a summary of the math involved for a univariate calculation: http://www.statisticshowto.com/probability-and-statistics/regression-analysis/find-a-linear-regression-equation/
Here, there is some discussion of the matrix formulation of the multivariate problem (I know you asked for univariate but some people find the matrix formulation helps them conceptualize the problem): https://onlinecourses.science.psu.edu/stat501/node/382
We should start with a bit of an intuition, based on the level of the question. The goal of a linear regression is to find a set of variables, in your case thetas, that minimize the distance between the line formed and the data points observed (often, the square of this distance). You have two "free" variables in the equation you defined. First, theta0: this is the intercept. The intercept is the value of the response variable (h(x)) when the input variable (x) is 0. This visually is the point where the line will cross the y axis. The second variable you have defined is the slope (theta1), this variable expresses how much the response variable changes when the input changes. If theta1 = 0, h(x) does not change when x changes. If theta1 = 1, h(x) increases and decreases at the same rate as x. If theta1 = -1, h(x) responds in the opposite direction: if x increases, h(x) decreases by the same amount; if x decreases, h(x) increases by the quantity.
For more information, Mathworks provides a fairly comprehensive explanation: https://www.mathworks.com/help/symbolic/mupad_ug/univariate-linear-regression.html
So after getting a handle on what we are doing conceptually, lets take a stab at the math. We'll need to calculate the standard deviation of our two variables, x and h(x). WTo calculate the standard deviation, we will calculate the mean of each variable (sum up all the x's and then divide by the number of x's, do the same for h(x)). The standard deviation captures how much a variable differs from its mean. For each x, subtract the mean of x. Sum these differences up and then divide by the number of x's minus 1. Finally, take the square root. This is your standard deviation.
Using this, we can normalize both variables. For x, subtract the mean of x and divide by the standard deviation of x. Do this for h(x) as well. You will now have two lists of normalized numbers.
For each normalized number, multiply the value by its pair (the first normalized x value with its h(x) pair, for all values). Add these products together and divide by N. This gives you the correlation. To get the least squares estimate of theta1, calculate this correlation value times the standard deviation of h(x) divided by the standard deviation of x.
Given all this information, calculating the intercept (theta0) is easy, all we'll have to do is take the mean of h(x) and subtract the product (multiply!) of our calculated theta1 and the average of x.
Phew! All taken care of! We have our least squares solution for those two variables. Let me know if you have any questions! One last excellent resource: https://people.duke.edu/~rnau/mathreg.htm
If you are asking about the hypothesis function in linear regression, then those theta values are selected by an algorithm called gradient descent. This helps in finding the theta values to minimize the cost function.
I understood the overall SVM algorithm consisting of Lagrangian Duality and all, but I am not able to understand why particularly the Lagrangian multiplier is greater than zero for support vectors.
Thank you.
This might be a late answer but I am putting my understanding here for other visitors.
Lagrangian multiplier, usually denoted by α is a vector of the weights of all the training points as support vectors.
Suppose there are m training examples. Then α is a vector of size m. Now focus on any ith element of α: αi. It is clear that αi captures the weight of the ith training example as a support vector. Higher value of αi means that ith training example holds more importance as a support vector; something like if a prediction is to be made, then that ith training example will be more important in deriving the decision.
Now coming to the OP's concern:
I am not able to understand why particularly the Lagrangian multiplier
is greater than zero for support vectors.
It is just a construct. When you say αi=0, it is just that ith training example has zero weight as a support vector. You can instead also say that that ith example is not a support vector.
Side note: One of the KKT's conditions is the complementary slackness: αigi(w)=0 for all i. For a support vector, it must lie on the margin which implies that gi(w)=0. Now αi can or cannot be zero; anyway it is satisfying the complementary slackness condition.
For αi=0, you can choose whether you want to call such points a support vector or not based on the discussion given above. But for a non-support vector, αi must be zero for satisfying the complementary slackness as gi(w) is not zero.
I can't figure this out too...
If we take a simple example, say of 3 data points, 2 of positive class (yi=1): (1,2) (3,1) and one negative (yi=-1): (-1,-1) - and we calculate using Lagrange multipliers, we will get a perfect w (0.25,0.5) and b = -0.25, but one of our alphas was negative (a1 = 6/32, a2 = -1/32, a3 = 5/32).
Hello actually i'm developing a CBIR using opencv for features extraction and for the svm.
My issues is : I'm using a ONE_CLASS classifier with a RBF Kernel. I'm using the function predict of opencvSVM with the last parameter at true ( means that if the classifieur is binary then it return the signed distance to the margin ) in order to classify my data.
But even if this parameters is set to true it's return only the label ( so not very helpful in my case).
So my question is : What is the equation ( knowing the vector support) to calcul the distance of a data to the marge ?
Thanks
In general, distance from the separating hyperplane is the exact same thing SVM is using for classification. In other words classification equation is simply a sign of the signed distance you are asking for.
Given your kernel K, support vectors SV_i, and alpha coefficients alpha_i (Lagrange multipliers) and threshold (bias) b the equation is simply, given labels associated with each support vector y_i:
sgn_dist(x) = SUM_i alpha_i y_i K(x, SV_i) - b
This is signed distance, so you get positive value when it is on positive side and negative otherwise, if you want a "true" distance (without label) simply divide by label or take absolute value
dist(x) = |sgn_dist(x)| = |SUM_i alpha_i y_i K(x, SV_i) - b|
I am taking this course on Neural networks in Coursera by Geoffrey Hinton (not current).
I have a very basic doubt on weight spaces.
https://d396qusza40orc.cloudfront.net/neuralnets/lecture_slides%2Flec2.pdf
Page 18.
If I have a weight vector (bias is 0) as [w1=1,w2=2] and training case as {1,2,-1} and {2,1,1}
where I guess {1,2} and {2,1} are the input vectors. How can it be represented geometrically?
I am unable to visualize it? Why is training case giving a plane which divides the weight space into 2? Could somebody explain this in a coordinate axes of 3 dimensions?
The following is the text from the ppt:
1.Weight-space has one dimension per weight.
2.A point in the space has particular setting for all the weights.
3.Assuming that we have eliminated the threshold each hyperplane could be represented as a hyperplane through the origin.
My doubt is in the third point above. Kindly help me understand.
It's probably easier to explain if you look deeper into the math. Basically what a single layer of a neural net is performing some function on your input vector transforming it into a different vector space.
You don't want to jump right into thinking of this in 3-dimensions. Start smaller, it's easy to make diagrams in 1-2 dimensions, and nearly impossible to draw anything worthwhile in 3 dimensions (unless you're a brilliant artist), and being able to sketch this stuff out is invaluable.
Let's take the simplest case, where you're taking in an input vector of length 2, you have a weight vector of dimension 2x1, which implies an output vector of length one (effectively a scalar)
In this case it's pretty easy to imagine that you've got something of the form:
input = [x, y]
weight = [a, b]
output = ax + by
If we assume that weight = [1, 3], we can see, and hopefully intuit that the response of our perceptron will be something like this:
With the behavior being largely unchanged for different values of the weight vector.
It's easy to imagine then, that if you're constraining your output to a binary space, there is a plane, maybe 0.5 units above the one shown above that constitutes your "decision boundary".
As you move into higher dimensions this becomes harder and harder to visualize, but if you imagine that that plane shown isn't merely a 2-d plane, but an n-d plane or a hyperplane, you can imagine that this same process happens.
Since actually creating the hyperplane requires either the input or output to be fixed, you can think of giving your perceptron a single training value as creating a "fixed" [x,y] value. This can be used to create a hyperplane. Sadly, this cannot be effectively be visualized as 4-d drawings are not really feasible in browser.
Hope that clears things up, let me know if you have more questions.
I have encountered this question on SO while preparing a large article on linear combinations (it's in Russian, https://habrahabr.ru/post/324736/). It has a section on the weight space and I would like to share some thoughts from it.
Let's take a simple case of linearly separable dataset with two classes, red and green:
The illustration above is in the dataspace X, where samples are represented by points and weight coefficients constitutes a line. It could be conveyed by the following formula:
w^T * x + b = 0
But we can rewrite it vice-versa making x component a vector-coefficient and w a vector-variable:
x^T * w + b = 0
because dot product is symmetrical. Now it could be visualized in the weight space the following way:
where red and green lines are the samples and blue point is the weight.
More possible weights are limited to the area below (shown in magenta):
which could be visualized in dataspace X as:
Hope it clarifies dataspace/weightspace correlation a bit. Feel free to ask questions, will be glad to explain in more detail.
The "decision boundary" for a single layer perceptron is a plane (hyper plane)
where n in the image is the weight vector w, in your case w={w1=1,w2=2}=(1,2) and the direction specifies which side is the right side. n is orthogonal (90 degrees) to the plane)
A plane always splits a space into 2 naturally (extend the plane to infinity in each direction)
you can also try to input different value into the perceptron and try to find where the response is zero (only on the decision boundary).
Recommend you read up on linear algebra to understand it better:
https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces
For a perceptron with 1 input & 1 output layer, there can only be 1 LINEAR hyperplane. And since there is no bias, the hyperplane won't be able to shift in an axis and so it will always share the same origin point. However, if there is a bias, they may not share a same point anymore.
I think the reason why a training case can be represented as a hyperplane because...
Let's say
[j,k] is the weight vector and
[m,n] is the training-input
training-output = jm + kn
Given that a training case in this perspective is fixed and the weights varies, the training-input (m, n) becomes the coefficient and the weights (j, k) become the variables.
Just as in any text book where z = ax + by is a plane,
training-output = jm + kn is also a plane defined by training-output, m, and n.
Equation of a plane passing through origin is written in the form:
ax+by+cz=0
If a=1,b=2,c=3;Equation of the plane can be written as:
x+2y+3z=0
So,in the XYZ plane,Equation: x+2y+3z=0
Now,in the weight space;every dimension will represent a weight.So,if the perceptron has 10 weights,Weight space will be 10 dimensional.
Equation of the perceptron: ax+by+cz<=0 ==> Class 0
ax+by+cz>0 ==> Class 1
In this case;a,b & c are the weights.x,y & z are the input features.
In the weight space;a,b & c are the variables(axis).
So,for every training example;for eg: (x,y,z)=(2,3,4);a hyperplane would be formed in the weight space whose equation would be:
2a+3b+4c=0
passing through the origin.
I hope,now,you understand it.
Consider we have 2 weights. So w = [w1, w2]. Suppose we have input x = [x1, x2] = [1, 2]. If you use the weight to do a prediction, you have z = w1*x1 + w2*x2 and prediction y = z > 0 ? 1 : 0.
Suppose the label for the input x is 1. Thus, we hope y = 1, and thus we want z = w1*x1 + w2*x2 > 0. Consider vector multiplication, z = (w ^ T)x. So we want (w ^ T)x > 0. The geometric interpretation of this expression is that the angle between w and x is less than 90 degree. For example, the green vector is a candidate for w that would give the correct prediction of 1 in this case. Actually, any vector that lies on the same side, with respect to the line of w1 + 2 * w2 = 0, as the green vector would give the correct solution. However, if it lies on the other side as the red vector does, then it would give the wrong answer.
However, suppose the label is 0. Then the case would just be the reverse.
The above case gives the intuition understand and just illustrates the 3 points in the lecture slide. The testing case x determines the plane, and depending on the label, the weight vector must lie on one particular side of the plane to give the correct answer.