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.
Related
I am new to machine learning. I am having a question regarding polynomial regression using one feature.
My understanding is that if there is one input feature, we can create a hypothesis function by taking the squares and cubes the feature.
Suppose x1 is the input feature and our hypothesis function becomes something like this :
htheta(x) = theta0 + (theta1)x1 + (theta2)x1^2 + (theta3)x1^3.
My question is what is the use case of such scenario ? In what type of data, this type of hypothesis function will help ?
This scenario is for simple curve fitting problems. For example, you might have a spring and want to know how far the spring is stretched as a function of how much force you apply (the spring needn't be a linear spring obeying Hooke's law). You could build a model by collecting a bunch of measurements of different forces applied on the spring (measured in Newtons) and the resulting spring extension (also called displacement) in centimeters. You could then build a model of the form F(x) = theta_1 * x + theta_2 * x^3 + theta_3 * x^5 and fit the three theta parameters. You could of course do this with any other single variable problem (height vs. age, weight vs. blood pressure, current vs. voltage). In practice, you generally have many more than just a one dependent variable though.
Also worth pointing out that the transformations needn't be polynomial in the dependent variable (x in this case). You could just as well try logs, square roots, exponentials etc. If you're asking why is it always a parameter times a function of the input variable, this is more of a modeling choice than anything (specifically a linear model since it's linear in theta). It does not have to be this way and is a simple assumption that restricts the class of functions. Linear models also satisfy some intuitive statistical properties which also justify their use (see here)
Let's say I want to model the sqrt function with a neural network. But for every input x, there are two answers sqrt(x) = y and sqrt(x) = -y. (But in reality, I don't know that I have the sqrt function - I just have a lot of data - so I don't know a priori if there are 0, 1, 2, or more answers y for every input x.) How can I get the correct distribution of y?
The question you should ask yourself, is how would you train such a neural network :
If you were thinking of giving x -> {sqrt(x),-sqrt(x)} as training example i, then that means that you know the number of outputs by looking at a single example so you should encode the fact that there are 2 outputs.
I you were thinking of giving examples that are sometimes x -> sqrt(x) and other time x -> -sqrt(x), then your neural network would be very hard to train as each gradient could change the weights in a completely different way. It would probably converge after many training examples, and would simply output the y it saw the most.
A better way of doing this would probably be to use something like a reward function in reinforcement learning. The function in that case would be f(x)=x^2 and would simply say if the output is correct. That would let your network train for both positive and negative outputs. So try to find a reward function for your problem!
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.
While following the Coursera-Machine Learning class, I wanted to test what I learned on another dataset and plot the learning curve for different algorithms.
I (quite randomly) chose the Online News Popularity Data Set, and tried to apply a linear regression to it.
Note : I'm aware it's probably a bad choice but I wanted to start with linear reg to see later how other models would fit better.
I trained a linear regression and plotted the following learning curve :
This result is particularly surprising for me, so I have questions about it :
Is this curve even remotely possible or is my code necessarily flawed?
If it is correct, how can the training error grow so quickly when adding new training examples? How can the cross validation error be lower than the train error?
If it is not, any hint to where I made a mistake?
Here's my code (Octave / Matlab) just in case:
Plot :
lambda = 0;
startPoint = 5000;
stepSize = 500;
[error_train, error_val] = ...
learningCurve([ones(mTrain, 1) X_train], y_train, ...
[ones(size(X_val, 1), 1) X_val], y_val, ...
lambda, startPoint, stepSize);
plot(error_train(:,1),error_train(:,2),error_val(:,1),error_val(:,2))
title('Learning curve for linear regression')
legend('Train', 'Cross Validation')
xlabel('Number of training examples')
ylabel('Error')
Learning curve :
S = ['Reg with '];
for i = startPoint:stepSize:m
temp_X = X(1:i,:);
temp_y = y(1:i);
% Initialize Theta
initial_theta = zeros(size(X, 2), 1);
% Create "short hand" for the cost function to be minimized
costFunction = #(t) linearRegCostFunction(X, y, t, lambda);
% Now, costFunction is a function that takes in only one argument
options = optimset('MaxIter', 50, 'GradObj', 'on');
% Minimize using fmincg
theta = fmincg(costFunction, initial_theta, options);
[J, grad] = linearRegCostFunction(temp_X, temp_y, theta, 0);
error_train = [error_train; [i J]];
[J, grad] = linearRegCostFunction(Xval, yval, theta, 0);
error_val = [error_val; [i J]];
fprintf('%s %6i examples \r', S, i);
fflush(stdout);
end
Edit : if I shuffle the whole dataset before splitting train/validation and doing the learning curve, I have very different results, like the 3 following :
Note : the training set size is always around 24k examples, and validation set around 8k examples.
Is this curve even remotely possible or is my code necessarily flawed?
It's possible, but not very likely. You might be picking the hard to predict instances for the training set and the easy ones for the test set all the time. Make sure you shuffle your data, and use 10 fold cross validation.
Even if you do all this, it is still possible for it to happen, without necessarily indicating a problem in the methodology or the implementation.
If it is correct, how can the training error grow so quickly when adding new training examples? How can the cross validation error be lower than the train error?
Let's assume that your data can only be properly fitted by a 3rd degree polynomial, and you're using linear regression. This means that the more data you add, the more obviously it will be that your model is inadequate (higher training error). Now, if you choose few instances for the test set, the error will be smaller, because linear vs 3rd degree might not show a big difference for too few test instances for this particular problem.
For example, if you do some regression on 2D points, and you always pick 2 points for your test set, you will always have 0 error for linear regression. An extreme example, but you get the idea.
How big is your test set?
Also, make sure that your test set remains constant throughout the plotting of the learning curves. Only the train set should increase.
If it is not, any hint to where I made a mistake?
Your test set might not be large enough or your train and test sets might not be properly randomized. You should shuffle the data and use 10 fold cross validation.
You might want to also try to find other research regarding that data set. What results are other people getting?
Regarding the update
That makes a bit more sense, I think. Test error is generally higher now. However, those errors look huge to me. Probably the most important information this gives you is that linear regression is very bad at fitting this data.
Once more, I suggest you do 10 fold cross validation for learning curves. Think of it as averaging all of your current plots into one. Also shuffle the data before running the process.
Is it a good practice to use sigmoid or tanh output layers in Neural networks directly to estimate probabilities?
i.e the probability of given input to occur is the output of sigmoid function in the NN
EDIT
I wanted to use neural network to learn and predict the probability of a given input to occur..
You may consider the input as State1-Action-State2 tuple.
Hence the output of NN is the probability that State2 happens when applying Action on State1..
I Hope that does clear things..
EDIT
When training NN, I do random Action on State1 and observe resultant State2; then teach NN that input State1-Action-State2 should result in output 1.0
First, just a couple of small points on the conventional MLP lexicon (might help for internet searches, etc.): 'sigmoid' and 'tanh' are not 'output layers' but functions, usually referred to as "activation functions". The return value of the activation function is indeed the output from each layer, but they are not the output layer themselves (nor do they calculate probabilities).
Additionally, your question recites a choice between two "alternatives" ("sigmoid and tanh"), but they are not actually alternatives, rather the term 'sigmoidal function' is a generic/informal term for a class of functions, which includes the hyperbolic tangent ('tanh') that you refer to.
The term 'sigmoidal' is probably due to the characteristic shape of the function--the return (y) values are constrained between two asymptotic values regardless of the x value. The function output is usually normalized so that these two values are -1 and 1 (or 0 and 1). (This output behavior, by the way, is obviously inspired by the biological neuron which either fires (+1) or it doesn't (-1)). A look at the key properties of sigmoidal functions and you can see why they are ideally suited as activation functions in feed-forward, backpropagating neural networks: (i) real-valued and differentiable, (ii) having exactly one inflection point, and (iii) having a pair of horizontal asymptotes.
In turn, the sigmoidal function is one category of functions used as the activation function (aka "squashing function") in FF neural networks solved using backprop. During training or prediction, the weighted sum of the inputs (for a given layer, one layer at a time) is passed in as an argument to the activation function which returns the output for that layer. Another group of functions apparently used as the activation function is piecewise linear function. The step function is the binary variant of a PLF:
def step_fn(x) :
if x <= 0 :
y = 0
if x > 0 :
y = 1
(On practical grounds, I doubt the step function is a plausible choice for the activation function, but perhaps it helps understand the purpose of the activation function in NN operation.)
I suppose there an unlimited number of possible activation functions, but in practice, you only see a handful; in fact just two account for the overwhelming majority of cases (both are sigmoidal). Here they are (in python) so you can experiment for yourself, given that the primary selection criterion is a practical one:
# logistic function
def sigmoid2(x) :
return 1 / (1 + e**(-x))
# hyperbolic tangent
def sigmoid1(x) :
return math.tanh(x)
what are the factors to consider in selecting an activation function?
First the function has to give the desired behavior (arising from or as evidenced by sigmoidal shape). Second, the function must be differentiable. This is a requirement for backpropagation, which is the optimization technique used during training to 'fill in' the values of the hidden layers.
For instance, the derivative of the hyperbolic tangent is (in terms of the output, which is how it is usually written) :
def dsigmoid(y) :
return 1.0 - y**2
Beyond those two requriements, what makes one function between than another is how efficiently it trains the network--i.e., which one causes convergence (reaching the local minimum error) in the fewest epochs?
#-------- Edit (see OP's comment below) ---------#
I am not quite sure i understood--sometimes it's difficult to communicate details of a NN, without the code, so i should probably just say that it's fine subject to this proviso: What you want the NN to predict must be the same as the dependent variable used during training. So for instance, if you train your NN using two states (e.g., 0, 1) as the single dependent variable (which is obviously missing from your testing/production data) then that's what your NN will return when run in "prediction mode" (post training, or with a competent weight matrix).
You should choose the right loss function to minimize.
The squared error does not lead to the maximum likelihood hypothesis here.
The squared error is derived from a model with Gaussian noise:
P(y|x,h) = k1 * e**-(k2 * (y - h(x))**2)
You estimate the probabilities directly. Your model is:
P(Y=1|x,h) = h(x)
P(Y=0|x,h) = 1 - h(x)
P(Y=1|x,h) is the probability that event Y=1 will happen after seeing x.
The maximum likelihood hypothesis for your model is:
h_max_likelihood = argmax_h product(
h(x)**y * (1-h(x))**(1-y) for x, y in examples)
This leads to the "cross entropy" loss function.
See chapter 6 in Mitchell's Machine Learning
for the loss function and its derivation.
There is one problem with this approach: if you have vectors from R^n and your network maps those vectors into the interval [0, 1], it will not be guaranteed that the network represents a valid probability density function, since the integral of the network is not guaranteed to equal 1.
E.g., a neural network could map any input form R^n to 1.0. But that is clearly not possible.
So the answer to your question is: no, you can't.
However, you can just say that your network never sees "unrealistic" code samples and thus ignore this fact. For a discussion of this (and also some more cool information on how to model PDFs with neural networks) see contrastive backprop.