While reading linear regression in Ch-2 of book "The elements of Statistical Learning", I came across 2 equations and I failed to understand how the 2nd was derived from the first.
Background:
How do we fit the linear model to a set of training data? There are
many different methods, but by far the most popular is the method of
least squares. In this approach, we pick the coefficients β to minimize the
residual sum of squares
Equation 1
RSS(β) is a quadratic function of the parameters, and hence its minimum
always exists, but may not be unique. The solution is easiest to characterize
in matrix notation. We can write
Equation 2
where X is an N × p matrix with each row an input vector, and y is an
N-vector of the outputs in the training set.
1st equation:
2nd equation:
I got it. The RHS of the 2nd equation is in the matrix form and to get the 1st equation, you have to transpose one part of the RHS of 2nd equation(this is how matrix multiplication is done)
I was reading tom Mitchell's machine learning book and he mentioned the formula for perceptron training rule is
where
: training rate
: expected output
: actual output
: ith input
This implies that if is very large then so is , but I don't understand the purpose of a large update when is large
on the contrary, I feel like if there is a large then the update should be small since a small fluctuation in will result in a big change in the final output (due to )
The adjustments are vector additions and subtractions, which can be thought as rotating a hyperplane such that class 0 falls on one part and class 1 falls on the other part.
Consider a 1xd weight vector indicating the weights of the perceptron model. Also, consider a 1xd datapoint . Then the predicted value of the perceptron model, considering a linear threshold without a loss of generality, will be
-- Eq. 1
Here '.' is a dot product, or
The hyperplane above equation is
(Ignoring the iteration indices for the weight updates for simplicity)
Let us consider we have two classes 0 and 1, again without a loss of generality, datapoints labelled 0 fall on one side where Eq.1 <= 0 of the hyperplane, and the datapoints labelled 1 fall on the other side where Eq.1 > 0.
The vector which is normal to this hyperplane is . The angle between the datapoints with label 0 should be more that 90 degrees and the datapoints between the datapoints with label 1 should be less than 90 degrees.
There are three possibilities of (ignoring the training rate)
: implying that this example is classified correctly by the present set of weights. Therefore we do not need any changes for the specific datapoint.
implying that the target was 1, but the present set of weights classified it as 0. The Eq1. which was supposed to be . Eq1. in this case is , which indicates that the angle between and is greater that 90 degrees, which should have been lesser. The update rule is . If you imagine a vector addition in 2d, this will rotate the hyperplane so that the angle between and is closer than before and less than 90 degrees.
implying that the target was 0, but the present set of weights classified it as 1. The eq1. which was supposed to be . Eq1. in this case is indicates that the angle between and is lesser that 90 degrees, which should have been greater. The update rule is . Similarly this will rotate the hyperplane so that the angle between and is greater than 90 degrees.
This is iterated over and overe and the hyperplane is rotated and adjusted so that the angle of the hyperplane's normal has less than 90 degrees with the datapoint with class labeled 1 and greater than 90 degrees with the datapoints of class labelled 0.
If the magnitude of is huge there will be big changes, and therefore it will cause problems in the process, and may take more iterations to converge depending on the magnitude of the initial weights. Therefore it is a good idea to normalise or standardise the datapoints. From this perspective it is easy to visualise what exactly the update rules are doing visually (consider the bias as a part of the hyperplane Eq.1). Now extend this to more complicated networks and or with thresholds.
Recommended reading and reference: Neural Network, A Systematic Introduction by Raul Rojas: Chapter 4
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).
I am trying to implement convolutional neural network by Lecun. I have two questions.
1) Do i have to multiply activation function on the (max_value * weight_value) in the maxpooling layer.
2) if yes than in backpropagating the error as i selected only one value from 2x2 receptive field. How can i distribute the error for the other 3 values in the receptive field. Should i replicate one error for all 2x2 window.
3) If no than in backpropagation How can i take determinant of the output i.e. (x*(1-x)? As to find gradient, we need derivative of the activated weighted sum i.e f'(x)
4) USing stochastic diagonal livenberg marquardt method, What should i take the value for eeta, and meeo in equation (21) page 2319 of Lecun Paper http://enpub.fulton.asu.edu/cseml/summer08/papers/cnn-appendix.pdf
I will be thankful for any explanation, or code sample etc.
Regards
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.