Is the model or cost function that shapes these lines? [closed] - machine-learning

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I have a question about cost functions in Machine Learning and their graphs. For instance, look at the following images. What function shapes them, the cost function or the model? I though was the cost function, like MSE in the first image. The second image I have no idea what function has that shape. All this is very confusing to me because in the book "Hands on Machine Learning... 2nd Edition" page 122 is written:
Fortunately, the MSE cost function for a Linear Regression model happens to be a convex function...
and
This implies that there are no local minima, just one global minimum.
What I don't understand is why MSE is convex only with Linear Regression model if it is quadratic? I mean, I believe that function always will have that "bowl" shape because it is quadratic. Or maybe not always because if was like that would be easy to choose MSE for any model and I would find the global minimum always since the main goal in a machine learning process is minimize the value of the cost function.

Why MSE is convex only with Linear Regression model if it is quadratic? I mean, I believe that function always will have that "bowl" shape because it is quadratic.
You're right.
The MSE cost function will be always convex over θ.
It will also be always convex over x if a model, θ = f(x), is linear.
It could be, however, non-convex over x if a model is non-linear.
For example, if a model is θ = x2
MSE(θ) = √(θ' - θ)2 = √(θ' - x2)2
will have two global minima, one at x = √|θ'| and the other at x = -√|θ'|. (Kind of "w" shape rather than "bowl" shape.)
But over the axis of θ, there is only one global minimum at θ = θ'.

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Partial derivative for mean absolute error [closed]

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What is the partial derivative for MAE? I understand that for mean squared error (MSE) the partial derivative with respect to some x1 would be -x1 * (y_pred - y_actual) assuming the the following version of MSE is used.
What is the partial derivative for x1 when the loss function is MAE instead of MSE? I've been trying to find this but I haven't had any luck. Would it just be -(y_pred - y_actual) when x1 is greater than 0, and (y_pred - y_actual) when x1 is less than 0? Or is there something else that I'm missing?
Unless you're having a single neuron, there's no fixed formula for partial derivative of loss function with respect to each weight; it depends strictly on the connections between neurons in the network. And the partial derivative formula is not only one, each weight has a different one.
For small network with kinda 2, 3 layers, apply chain rule, and sum rule to find the partial derivative of loss function manually, otherwise, dynamic programming in backpropagation is needed.

Why the hypothesis has to introduce two parameters, namely θ0 and θ1

I was learning Machine Learning from this course on Coursera taught by Andrew Ng. The instructor defines the hypothesis as a linear function of the "input" (x, in my case) like the following:
hθ(x) = θ0 + θ1(x)
In supervised learning, we have some training data and based on that we try to "deduce" a function which closely maps the inputs to the corresponding outputs. To deduce the function, we introduce the hypothesis as a linear function of input (x). My question is, why the function involving two θs is chosen? Why it can't be as simple as y(i) = a * x(i) where a is a co-efficient? Later we can go about finding a "good" value of a for a given example (i) using an algorithm? This question might look very stupid. I apologize but I'm not very good at machine learning I am just a beginner. Please help me understand this.
Thanks!
The a corresponds to θ1. Your proposed linear model is leaving out the intercept, which is θ0.
Consider an output function y equal to the constant 5, or perhaps equal to a constant plus some tiny fraction of x which never exceeds .01. Driving the error function to zero is going to be difficult if your model doesn't have a θ0 that can soak up the D.C. component.

How to find the value of theta 0 and theta 1? [closed]

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I am new to ML, I am not sure on how to solve this problem
Could someone tell me how to solve this problem of finding values in a a step by step manner?
From newcomer view point you can actually just test:
h1=0.5+0.5x
h2=0+0.5x
h3=0.5+0x
h4=1+0.5x
h5=1+x
Then which one of the hs(1..5) gives exact observed values of y(0.5,1,2,0) for a given set of dependent variables x(1,2,4,0).
You can answer that by passing sample values of x in the above equation.
I hope i made it simple enough
Here is the cache It's one of most easy problems in machine learning.
Just see that we have to create a linear regression model to fit the following data:-
STEP 1:UNDERSTANDING THE PROBLEM
And as mentioned at the last of question it should completely fit the data.
We have to find theta0 and theta1 in such a way such that given value of x Htheta(x) will give the correct value of y.
STEP 2:FINDING THETA1
In these m examples take any 2 random examples
Htheta(x2)-Htheta(x1) = theta1*(x2)-theta1*(x1)
-----Subtracting those 2 variables(eliminating theta0)
hteta(x2) = y2
(y corresponding to that x in the data as the parameters exactly fit the data provided )
(y2-y1)/(x2-x1) = theta1
----taking common and then dividing by(x2-x1) on both sides of equation
From this:
theta1 = 0.5
STEP3 :CALCULATING THETA0
Take any random example and put the values of theta1, y and x in this equation
y = theta1*x + theta0
theta0 will come out to be 0
My approach would be to view these points by plotting a graph with x,y values. Since it's a straight line, calculate tan(theta) using normal trigonometry, which in this case is y/x(Since it's mentioned they fit perfectly!!). eg:-
tan(theta1) = 0.5/1 or 1/2
Calculate arctan(1/2) // Approx 0.5
Note:- This is not a scalable approach but just some maths fun! Sorry.
In general you would execute some non-iterative algorithmic approach (probably based on solving a system of linear equations) or some iterative approach like GD (Gradient Descent), but this is more simple here, as it's already given that there is a perfect fit.
Perfect fit means: loss/error of zero.
Loss of zero implicates, that sigma0 needs to be zero or else sample 4 (last one) induces a loss
Overall loss is the sum of sample-losses and each loss/component is nonnegative -> we can't tolerate a loss here
When sigma0 is fixed, sample 4 has an infinite amount of solutions producing no loss
But sample 1 shows that it has to be 0.5 to induce no loss
Check the others, it's fitting perfectly
One assumption i made:
Gradient-descent will converge to the optimal solution (which is not always true, even for convex-optimization problems; it's depending learning-rates; one might use line-searches to proof convergence based on some assumptions about the problem; but all that is irrelevant here)

Wouldn't setting the first derivative of Cost function J to 0 gives the exact Theta values that minimize the cost?

I am currently doing Andrew NG's ML course. From my calculus knowledge, the first derivative test of a function gives critical points if there are any. And considering the convex nature of Linear / Logistic Regression cost function, it is a given that there will be a global / local optima. If that is the case, rather than going a long route of taking a miniscule baby step at a time to reach the global minimum, why don't we use the first derivative test to get the values of Theta that minimize the cost function J in a single attempt , and have a happy ending?
That being said, I do know that there is a Gradient Descent alternative called Normal Equation that does just that in one successful step unlike the former.
On a second thought, I am thinking if it is mainly because of multiple unknown variables involved in the equation (which is why the Partial Derivative comes into play?) .
Let's take an example:
Gradient simple regression cost function:
Δ[RSS(w) = [(y-Hw)T(y-Hw)]
y : output
H : feature vector
w : weights
RSS: residual sum of squares
Equating this to 0 for getting the closed form solution will give:
w = (H T H)-1 HT y
Now assuming there are D features, the time complexity for calculating transpose of matrix is around O(D3). If there are a million features, it is computationally impossible to do within reasonable amount of time.
We use these gradient descent methods since they give solutions with reasonably acceptable solutions within much less time.

What is the role of the bias in neural networks? [closed]

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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

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