I am reading the theory of SVM. In kernel trick, what I understand is, if we have a data which is not linear separable in the original dimensions n, we use the kernel to map the data to a higher space to be linear separable (we have to choose the right kernel depending on the data set, etc). However, when I watched this video of Andrew ng Kernel SVM, What I understand is we can map original data into a smaller space which make me confused!? Any explanation.
Could you explain me how does RBF kernel work to map each original data sample x1(x11,x12,x13,....,x1n) to a higher space (with dimensions m) to be X1(X11,X12,X13,...,X1m) with a concrete example. Also, what I understand is the kernel compute the inner product of the transformed data (so there is an other transformation before the RBF, which means that RBF transform implicitly the data to a higher space but How?).
other thing: the kernel is a function k(x,x1):(R^n)^2->R =g(x).g(x1), with g is a transformation function, how to define g in the case of RBF kernel?
Suppose that we are in the test set, What I understand is x is the sample to be classified and x1 is the support vector (because only the support vectors will be used to calculate the hyperplane). in the case of RBF
k(x,x1)=exp(-(x-x1)^2/2sigma), so where is the transformation?
Last question: Admit that the RBF do the mapping to a higher dimension m, it is possible to show this m? I want to see the theoretical reality.
I want to implement SVM with RBF kernel. What is the m here and how to choose it? How to implement kernel trick in practice?
Could you explain me how does RBF kernel work to map each original data sample x1(x11,x12,x13,....,x1n) to a higher space (with dimensions m) to be X1(X11,X12,X13,...,X1m) with a concrete example. Also, what I understand is the kernel compute the inner product of the transformed data (so there is an other transformation before the RBF, which means that RBF transform implicitly the data to a higher space but How?).
Exactly as you said - kernel is an inner product of the projected space, not the projection itself. The whole trick is that you do not ever transform your data, because it is computationally too expensive to do so.
other thing: the kernel is a function k(x,x1):(R^n)^2->R =g(x).g(x1), with g is a transformation function, how to define g in the case of RBF kernel?
For rbf kernel, g is actually a mapping from R^n into the space of continuous functions (L2), and each point is mapped into unnormalized gaussian distribution with mean x, and variance sigma^2. Thus (up to some normalizing constant A that we will drop)
g(x) = N(x, sigma^2)[z] / A # notice this is not a number but a function of z!
and now inner product in the space of functions is the integral of products over the whole domain thus
K(x, y) = <g(x), g(y)>
= INT_{R^n} N(x, sigma^2)[z] N(y, sigma^2)[z] / A^2 dz
= B exp(-||x-y||^2 / (2*sigma^2))
where B is some constant factor (normalization) depending solely on sigma^2, thus we can drop it (as scaling does not really matter here) for computational simplicity.
Suppose that we are in the test set, What I understand is x is the sample to be classified and x1 is the support vector (because only the support vectors will be used to calculate the hyperplane). in the case of RBF k(x,x1)=exp(-(x-x1)^2/2sigma), so where is the transformation?
as said before - transformation is never explicitly used, you simply show that inner product of your hyperplane with the transformed point can be expressed again as inner products with support vectors, thus you do not ever transform anything, just use kernels
<w, g(x)> = < SUM_{i=1}^N alpha_i y_i g(sv_i), g(x)>
= SUM_{i=1}^N alpha_i y_i <g(sv_i), g(x)>
= SUM_{i=1}^N alpha_i y_i K(sv_i, x)
where sv_i is i'th support vector, alpha_i is the per-sample weight (Lagrange multiplier) found during the optimization process and y_i is label of i'th support vector.
Last question: Admit that the RBF do the mapping to a higher dimension m, it is possible to show this m? I want to see the theoretical reality.
In this case m is infinity, as your new space is space of continuous functions in the domain of R^n -> R, thus a single vector (function) is defined as a continuum (size of the set of real numbers) values - one per each possible input value coming from R^n (it is a simple set theory result that R^n for any positive n is of size continuum). Thus in terms of pure mathematics, m = |R|, and using set theory this is so called Beth_1 (https://en.wikipedia.org/wiki/Beth_number).
I want to implement SVM with RBF kernel. What is the m here and how to choose it? How to implement kernel trick in practice?
You do not choose m, it is defined by the kernel itself. Implementing kernel trick in practise requires expressing all your optimization routines in the form, where training points are used solely in the context of inner products, and just replacing them with kernel calls. This is way too complex to describe in SO form.
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I'm aware of the gradient descent and the back-propagation algorithm. What I don't get is: when is using a bias important and how do you use it?
For example, when mapping the AND function, when I use two inputs and one output, it does not give the correct weights. However, when I use three inputs (one of which is a bias), it gives the correct weights.
I think that biases are almost always helpful. In effect, a bias value allows you to shift the activation function to the left or right, which may be critical for successful learning.
It might help to look at a simple example. Consider this 1-input, 1-output network that has no bias:
The output of the network is computed by multiplying the input (x) by the weight (w0) and passing the result through some kind of activation function (e.g. a sigmoid function.)
Here is the function that this network computes, for various values of w0:
Changing the weight w0 essentially changes the "steepness" of the sigmoid. That's useful, but what if you wanted the network to output 0 when x is 2? Just changing the steepness of the sigmoid won't really work -- you want to be able to shift the entire curve to the right.
That's exactly what the bias allows you to do. If we add a bias to that network, like so:
...then the output of the network becomes sig(w0*x + w1*1.0). Here is what the output of the network looks like for various values of w1:
Having a weight of -5 for w1 shifts the curve to the right, which allows us to have a network that outputs 0 when x is 2.
A simpler way to understand what the bias is: it is somehow similar to the constant b of a linear function
y = ax + b
It allows you to move the line up and down to fit the prediction with the data better.
Without b, the line always goes through the origin (0, 0) and you may get a poorer fit.
Here are some further illustrations showing the result of a simple 2-layer feed forward neural network with and without bias units on a two-variable regression problem. Weights are initialized randomly and standard ReLU activation is used. As the answers before me concluded, without the bias the ReLU-network is not able to deviate from zero at (0,0).
Two different kinds of parameters can
be adjusted during the training of an
ANN, the weights and the value in the
activation functions. This is
impractical and it would be easier if
only one of the parameters should be
adjusted. To cope with this problem a
bias neuron is invented. The bias
neuron lies in one layer, is connected
to all the neurons in the next layer,
but none in the previous layer and it
always emits 1. Since the bias neuron
emits 1 the weights, connected to the
bias neuron, are added directly to the
combined sum of the other weights
(equation 2.1), just like the t value
in the activation functions.1
The reason it's impractical is because you're simultaneously adjusting the weight and the value, so any change to the weight can neutralize the change to the value that was useful for a previous data instance... adding a bias neuron without a changing value allows you to control the behavior of the layer.
Furthermore the bias allows you to use a single neural net to represent similar cases. Consider the AND boolean function represented by the following neural network:
(source: aihorizon.com)
w0 corresponds to b.
w1 corresponds to x1.
w2 corresponds to x2.
A single perceptron can be used to
represent many boolean functions.
For example, if we assume boolean values
of 1 (true) and -1 (false), then one
way to use a two-input perceptron to
implement the AND function is to set
the weights w0 = -3, and w1 = w2 = .5.
This perceptron can be made to
represent the OR function instead by
altering the threshold to w0 = -.3. In
fact, AND and OR can be viewed as
special cases of m-of-n functions:
that is, functions where at least m of
the n inputs to the perceptron must be
true. The OR function corresponds to
m = 1 and the AND function to m = n.
Any m-of-n function is easily
represented using a perceptron by
setting all input weights to the same
value (e.g., 0.5) and then setting the
threshold w0 accordingly.
Perceptrons can represent all of the
primitive boolean functions AND, OR,
NAND ( 1 AND), and NOR ( 1 OR). Machine Learning- Tom Mitchell)
The threshold is the bias and w0 is the weight associated with the bias/threshold neuron.
The bias is not an NN term. It's a generic algebra term to consider.
Y = M*X + C (straight line equation)
Now if C(Bias) = 0 then, the line will always pass through the origin, i.e. (0,0), and depends on only one parameter, i.e. M, which is the slope so we have less things to play with.
C, which is the bias takes any number and has the activity to shift the graph, and hence able to represent more complex situations.
In a logistic regression, the expected value of the target is transformed by a link function to restrict its value to the unit interval. In this way, model predictions can be viewed as primary outcome probabilities as shown:
Sigmoid function on Wikipedia
This is the final activation layer in the NN map that turns on and off the neuron. Here also bias has a role to play and it shifts the curve flexibly to help us map the model.
A layer in a neural network without a bias is nothing more than the multiplication of an input vector with a matrix. (The output vector might be passed through a sigmoid function for normalisation and for use in multi-layered ANN afterwards, but that’s not important.)
This means that you’re using a linear function and thus an input of all zeros will always be mapped to an output of all zeros. This might be a reasonable solution for some systems but in general it is too restrictive.
Using a bias, you’re effectively adding another dimension to your input space, which always takes the value one, so you’re avoiding an input vector of all zeros. You don’t lose any generality by this because your trained weight matrix needs not be surjective, so it still can map to all values previously possible.
2D ANN:
For a ANN mapping two dimensions to one dimension, as in reproducing the AND or the OR (or XOR) functions, you can think of a neuronal network as doing the following:
On the 2D plane mark all positions of input vectors. So, for boolean values, you’d want to mark (-1,-1), (1,1), (-1,1), (1,-1). What your ANN now does is drawing a straight line on the 2d plane, separating the positive output from the negative output values.
Without bias, this straight line has to go through zero, whereas with bias, you’re free to put it anywhere.
So, you’ll see that without bias you’re facing a problem with the AND function, since you can’t put both (1,-1) and (-1,1) to the negative side. (They are not allowed to be on the line.) The problem is equal for the OR function. With a bias, however, it’s easy to draw the line.
Note that the XOR function in that situation can’t be solved even with bias.
When you use ANNs, you rarely know about the internals of the systems you want to learn. Some things cannot be learned without a bias. E.g., have a look at the following data: (0, 1), (1, 1), (2, 1), basically a function that maps any x to 1.
If you have a one layered network (or a linear mapping), you cannot find a solution. However, if you have a bias it's trivial!
In an ideal setting, a bias could also map all points to the mean of the target points and let the hidden neurons model the differences from that point.
Modification of neuron WEIGHTS alone only serves to manipulate the shape/curvature of your transfer function, and not its equilibrium/zero crossing point.
The introduction of bias neurons allows you to shift the transfer function curve horizontally (left/right) along the input axis while leaving the shape/curvature unaltered.
This will allow the network to produce arbitrary outputs different from the defaults and hence you can customize/shift the input-to-output mapping to suit your particular needs.
See here for graphical explanation:
http://www.heatonresearch.com/wiki/Bias
In a couple of experiments in my masters thesis (e.g. page 59), I found that the bias might be important for the first layer(s), but especially at the fully connected layers at the end it seems not to play a big role.
This might be highly dependent on the network architecture / dataset.
If you're working with images, you might actually prefer to not use a bias at all. In theory, that way your network will be more independent of data magnitude, as in whether the picture is dark, or bright and vivid. And the net is going to learn to do it's job through studying relativity inside your data. Lots of modern neural networks utilize this.
For other data having biases might be critical. It depends on what type of data you're dealing with. If your information is magnitude-invariant --- if inputting [1,0,0.1] should lead to the same result as if inputting [100,0,10], you might be better off without a bias.
Bias determines how much angle your weight will rotate.
In a two-dimensional chart, weight and bias can help us to find the decision boundary of outputs.
Say we need to build a AND function, the input(p)-output(t) pair should be
{p=[0,0], t=0},{p=[1,0], t=0},{p=[0,1], t=0},{p=[1,1], t=1}
Now we need to find a decision boundary, and the ideal boundary should be:
See? W is perpendicular to our boundary. Thus, we say W decided the direction of boundary.
However, it is hard to find correct W at first time. Mostly, we choose original W value randomly. Thus, the first boundary may be this:
Now the boundary is parallel to the y axis.
We want to rotate the boundary. How?
By changing the W.
So, we use the learning rule function: W'=W+P:
W'=W+P is equivalent to W' = W + bP, while b=1.
Therefore, by changing the value of b(bias), you can decide the angle between W' and W. That is "the learning rule of ANN".
You could also read Neural Network Design by Martin T. Hagan / Howard B. Demuth / Mark H. Beale, chapter 4 "Perceptron Learning Rule"
In simpler terms, biases allow for more and more variations of weights to be learnt/stored... (side-note: sometimes given some threshold). Anyway, more variations mean that biases add richer representation of the input space to the model's learnt/stored weights. (Where better weights can enhance the neural net’s guessing power)
For example, in learning models, the hypothesis/guess is desirably bounded by y=0 or y=1 given some input, in maybe some classification task... i.e some y=0 for some x=(1,1) and some y=1 for some x=(0,1). (The condition on the hypothesis/outcome is the threshold I talked about above. Note that my examples setup inputs X to be each x=a double or 2 valued-vector, instead of Nate's single valued x inputs of some collection X).
If we ignore the bias, many inputs may end up being represented by a lot of the same weights (i.e. the learnt weights mostly occur close to the origin (0,0).
The model would then be limited to poorer quantities of good weights, instead of the many many more good weights it could better learn with bias. (Where poorly learnt weights lead to poorer guesses or a decrease in the neural net’s guessing power)
So, it is optimal that the model learns both close to the origin, but also, in as many places as possible inside the threshold/decision boundary. With the bias we can enable degrees of freedom close to the origin, but not limited to origin's immediate region.
In neural networks:
Each neuron has a bias
You can view bias as a threshold (generally opposite values of threshold)
Weighted sum from input layers + bias decides activation of a neuron
Bias increases the flexibility of the model.
In absence of bias, the neuron may not be activated by considering only the weighted sum from the input layer. If the neuron is not activated, the information from this neuron is not passed through rest of the neural network.
The value of bias is learnable.
Effectively, bias = — threshold. You can think of bias as how easy it is to get the neuron to output a 1 — with a really big bias, it’s very easy for the neuron to output a 1, but if the bias is very negative, then it’s difficult.
In summary: bias helps in controlling the value at which the activation function will trigger.
Follow this video for more details.
Few more useful links:
geeksforgeeks
towardsdatascience
Expanding on zfy's explanation:
The equation for one input, one neuron, one output should look:
y = a * x + b * 1 and out = f(y)
where x is the value from the input node and 1 is the value of the bias node;
y can be directly your output or be passed into a function, often a sigmoid function. Also note that the bias could be any constant, but to make everything simpler we always pick 1 (and probably that's so common that zfy did it without showing & explaining it).
Your network is trying to learn coefficients a and b to adapt to your data.
So you can see why adding the element b * 1 allows it to fit better to more data: now you can change both slope and intercept.
If you have more than one input your equation will look like:
y = a0 * x0 + a1 * x1 + ... + aN * 1
Note that the equation still describes a one neuron, one output network; if you have more neurons you just add one dimension to the coefficient matrix, to multiplex the inputs to all nodes and sum back each node contribution.
That you can write in vectorized format as
A = [a0, a1, .., aN] , X = [x0, x1, ..., 1]
Y = A . XT
i.e. putting coefficients in one array and (inputs + bias) in another you have your desired solution as the dot product of the two vectors (you need to transpose X for the shape to be correct, I wrote XT a 'X transposed')
So in the end you can also see your bias as is just one more input to represent the part of the output that is actually independent of your input.
To think in a simple way, if you have y=w1*x where y is your output and w1 is the weight, imagine a condition where x=0 then y=w1*x equals to 0.
If you want to update your weight you have to compute how much change by delw=target-y where target is your target output. In this case 'delw' will not change since y is computed as 0. So, suppose if you can add some extra value it will help y = w1x + w01, where bias=1 and weight can be adjusted to get a correct bias. Consider the example below.
In terms of line slope, intercept is a specific form of linear equations.
y = mx + b
Check the image
image
Here b is (0,2)
If you want to increase it to (0,3) how will you do it by changing the value of b the bias.
For all the ML books I studied, the W is always defined as the connectivity index between two neurons, which means the higher connectivity between two neurons.
The stronger the signals will be transmitted from the firing neuron to the target neuron or Y = w * X as a result to maintain the biological character of neurons, we need to keep the 1 >=W >= -1, but in the real regression, the W will end up with |W| >=1 which contradicts how the neurons are working.
As a result, I propose W = cos(theta), while 1 >= |cos(theta)|, and Y= a * X = W * X + b while a = b + W = b + cos(theta), b is an integer.
Bias acts as our anchor. It's a way for us to have some kind of baseline where we don't go below that. In terms of a graph, think of like y=mx+b it's like a y-intercept of this function.
output = input times the weight value and added a bias value and then apply an activation function.
The term bias is used to adjust the final output matrix as the y-intercept does. For instance, in the classic equation, y = mx + c, if c = 0, then the line will always pass through 0. Adding the bias term provides more flexibility and better generalisation to our neural network model.
The bias helps to get a better equation.
Imagine the input and output like a function y = ax + b and you need to put the right line between the input(x) and output(y) to minimise the global error between each point and the line, if you keep the equation like this y = ax, you will have one parameter for adaptation only, even if you find the best a minimising the global error it will be kind of far from the wanted value.
You can say the bias makes the equation more flexible to adapt to the best values