Related
Where does the graph of the loss function in machine learning come from?
I am studying about machine learning. I sometimes don't understand models that have been optimized using regularization terms.
In the explanation of regularization, the following figure may appear.
Here is an example of the L1 regularization term. I have assumed that the model has two weight parameters w1, w2. That is, the equation of model y is expressed by the following equation.
y = w1x1 + w2x2
For simplicity, I ignored the bias term.
The red squares represent regularization terms. And the blue ellipses are represents the loss function without the regularization term.
The regularization term is given by
| w1 | ^ q + | w2 | ^ q = r ^ q (r is const.)
Therefore, the equation of the graph at w1> 0 and w2> 0 is expressed as follows.
w2 = (r ^ q-| w1 | ^ q) ^ (1 / q)
By substituting w1 for this equation (q = 0 at Lasso), you can draw a graph of the regularized term.
On the other hand, I could not draw a graph of the loss function. Perhaps you need more than one piece of data to draw this graph. For simplicity, I have assumed that I have only two pieces of data. I define them as (x11, x12, t1), (x21, x22, t2). When the loss function is MSE, it is expressed by the following equation.
Ed = 1/2 * {(t1-w1x11-w2x12) + (t1-w1x21-w2x22)}
If I simplify this, it is expressed as
Ed = a*w1^2 + b*w1 + c*w2^2 + d*w2 + e*w1*w2 + f
Here, a, b, c, d, e, and f are functions represented by all or part of x11, x12, x21, and x22. After finding a, b, c, d, e, and f, I thought that if we substitute w1 for this equation, we could draw a graph of the loss function. However, I cannot draw well.
Is the above understanding correct? Thank you.
To visualize the loss function, Ed which is a function of w1 and w2, we should visualize it as a 3 dimensional plot. For example, you can use Geogebra to visualize a 3 dimensional surface plot.
Here is an example, where a=3, b=-1, c=1, d =-1 , e=2.
The 2D plot that you see is called a countor plot. This link enables you to draw it online.
To draw a contour plot manually, you fix the value of Ed, then you obtained a quadratic equation, after which, as you varies w1, you can solve for your w2, for each w1, you can obtain up to 2 w2 as it is quadratic.
Remark: If you are looking for closed form expression in terms of arbitrary q, that could be more challenging.
As I understand it, in a deep neural network, we use an activation function (g) after applying the weights (w) and bias(b) (z := w * X + b | a := g(z)). So there is a composition function of (g o z) and the activation function makes so our model can learn function other than linear functions. I see that Sigmoid and Tanh activation function makes our model non-linear, but I have some trouble seeing that a ReLu (which takes the max out of 0 and z) can make a model non-linear...
Let's say if every Z is always positive, then it would be as if there was no activation function...
So why does ReLu make a neural network model non-linear?
Deciding if a function is linear or not is of course not a matter of opinion or debate; there is a very simple definition of a linear function, which is roughly:
f(a*x + b*y) = a*f(x) + b*f(y)
for every x & y in the function domain and a & b constants.
The requirement "for every" means that, if we are able to find even a single example where the above condition does not hold, then the function is nonlinear.
Assuming for simplicity that a = b = 1, let's try x=-5, y=1 with f being the ReLU function:
f(-5 + 1) = f(-4) = 0
f(-5) + f(1) = 0 + 1 = 1
so, for these x & y (in fact for every x & y with x*y < 0) the condition f(x + y) = f(x) + f(y) does not hold, hence the function is nonlinear...
The fact that we may be able to find subdomains (e.g. both x and y being either negative or positive here) where the linearity condition holds is what defines some functions (such as ReLU) as piecewise-linear, which are still nonlinear nevertheless.
Now, to be fair to your question, if in a particular application the inputs happened to be always either all positive or all negative, then yes, in this case the ReLU would in practice end up behaving like a linear function. But for neural networks this is not the case, hence we can rely on it indeed to provide our necessary non-linearity...
While doing MOOC on ML by Andrew Ng, he in theory explains theta'*X gives us hypothesis and while doing coursework we use theta*X. Why it's so?
theta'*X is used to calculate the hypothesis for a single training example when X is a vector. Then you have to calculate theta' to get to the h(x) definition.
In the practice, since you have more than one training example, X is a Matrix (your training set) with "m x n" dimension where m is the number of your training examples and n your number of features.
Now, you want to calculate h(x) for all your training examples with your theta parameter in just one move right?
Here is the trick: theta has to be a n x 1 vector then when you do Matrix-Vector Multiplication (X*theta) you will obtain an m x 1 vector with all your h(x)'s training examples in your training set (X matrix). Matrix multiplication will create the vector h(x) row by row making the corresponding math and this will be equal to the h(x) definition at each training example.
You can do the math by hand, I did it and now is clear. Hope i can help someone. :)
In mathematics, a 'vector' is always defined as a vertically-stacked array, e.g. , and signifies a single point in a 3-dimensional space.
A 'horizontal' vector, typically signifies an array of observations, e.g. is a tuple of 3 scalar observations.
Equally, a matrix can be thought of as a collection of vectors. E.g., the following is a collection of four 3-dimensional vectors:
A scalar can be thought of as a matrix of size 1x1, and therefore its transpose is the same as the original.
More generally, an n-by-m matrix W can also be thought of as a transformation from an m-dimensional vector x to an n-dimensional vector y, since multiplying that matrix with an m-dimensional vector will yield a new n-dimensional one. If your 'matrix' W is '1xn', then this denotes a transformation from an n-dimensional vector to a scalar.
Therefore, notationally, it is customary to introduce the problem from the mathematical notation point of view, e.g. y = Wx.
However, for computational reasons, sometimes it makes more sense to perform the calculation as a "vector times a matrix" rather than "matrix times a vector". Since (Wx)' === x'W', sometimes we solve the problem like that, and treat x' as a horizontal vector. Also, if W is not a matrix, but a scalar, then Wx denotes scalar multiplication, and therefore in this case Wx === xW.
I don't know the exercises you speak of, but my assumption would be that in the course he introduced theta as a proper, vertical vector, but then transposed it to perform proper calculations, i.e. a transformation from a vector of n-dimensions to a scalar (which is your prediction).
Then in the exercises, presumably you were either dealing with a scalar 'theta' so there was no point transposing it, and was left as theta for convenience or, theta was now defined as a horizontal (i.e. transposed) vector to begin with for some reason (e.g. printing convenience), and then was left in that state when performing the necessary transformation.
I don't know what the dimensions for your theta and X are (you haven't provided anything) but actually it all depends on the X, theta and hypothesis dimensions. Let's say m is the number of features and n - the number of examples. Then, if theta is a mx1 vector and X is a nxm matrix then X*theta is a nx1 hypothesis vector.
But you will get the same result if calculate theta'*X. You can also get the same result with theta*X if theta is 1xm and X - mxn
Edit:
As #Tasos Papastylianou pointed out the same result will be obtained if X is mxn then (theta.'*X).' or X.'*theta are answers. If the hypothesis should be a 1xn vector then theta.'*X is an answer. If theta is 1xm, X - mxn and the hypothesis is 1xn then theta*X is also a correct answer.
i had the same problem for me. (ML course, linear regression)
after spending time on it, here is how i see it: there is a confusion between the x(i) vector and the X matrix.
About the hypothesis h(xi) for a xi vector (xi belongs to R3x1), theta belongs to R3x1
theta = [to;t1;t2] #R(3x1)
theta' = [to t1 t2] #R(1x3)
xi = [1 ; xi1 ; xi2] #(R3x1)
theta' * xi => to + t1.xi,1 +t2.xi,2
= h(xi) (which is a R1x1 => a real number)
to the theta'*xi works here
About the vectorization equation
in this case X is not the same thing as x (vector). it is a matrix with m rows and n+1 col (m =number of examples and n number of features on which we add the to term)
therefore from the previous example with n= 2,
the matrix X is a m x 3 matrix
X = [1 xo,1 xo,2 ; 1 x1,1 x1,2 ; .... ; 1 xi,1 xi,2 ; ...; 1 xm,1 xm,2]
if you want to vectorize the equation for the algorithm, you need to consider for that for each row i, you will have the h(xi) (a real number)
so you need to implement X * theta
that will give you for each row i
[ 1 xi,1 xi,2] * [to ; t1 ; t2] = to + t1.xi,1 + t2.xi,2
Hope it helps
I have used octave notation and syntax for writing matrices: 'comma' for separating column items, 'semicolon' for separating row items and 'single quote' for Transpose.
In the course theory under discussion, theta = [theta0; theta1; theta2; theta3; .... thetaf].
'theta' is therefore a column vector or '(f+1) x 1' matrix. Here 'f' is the number of features. theta0 is the intercept term.
With just one training example, x is a '(f+1) x 1' matrix or a column vector. Specifically x = [x0; x1; x2; x3; .... xf]
x0 is always '1'.
In this special case the '1 x (f+1)' matrix formed by taking theta' and x could be multiplied to give the correct '1x1' hypothesis matrix or a real number.
h = theta' * x is a valid expression.
But the coursework deals with multiple training examples. If there are 'm' training examples, X is a 'm x (f+1)' matrix.
To simplify, let there be two training examples each with 'f' features.
X = [ x1; x2].
(Please note 1 and 2 inside the brackets are not exponential terms but indexes for the training examples).
Here, x1 = [ x01, x11, x21, x31, .... xf1 ]
and
x2 = [ x02, x12, x22, x32, .... xf2].
So X is a '2 x (f+1)' matrix.
Now to answer the question, theta' is a '1 x (f+1)' matrix and X is a '2 x (f+1)' matrix. With this, the following expressions are not valid.
theta' * X
theta * X
The expected hypothesis matrix, 'h', should have two predicted values (two real numbers), one for each of the two training examples. 'h' is a '2 x 1' matrix or column vector.
The hypothesis can be obtained only by using the expression, X * theta which is valid and algebraically correct. Multiplying a '2 x (f+1)' matrix with a '(f+1) x 1' matrix resulting in a '2 x 1' hypothesis matrix.
When Andrew Ng first introduced x in the cost function J(theta), x is a column vector
aka
[x0; x1; ... ; xn]
i.e.
x0;
x1;
...;
xn
However, in the first programming assignment, we are given X, which is an (m * n) matrix, (# training examples * features per training example). The discrepancy comes with the fact that from file the individual x vectors(training samples) are stored as horizontal row vectors rather than the vertical column vectors!!
This means the X matrix you see is actually an X' (X Transpose) matrix!!
Since we have X', we need to make our code work given our equation is looking for h(theta) = theta' * X(when the vectors in matrix X are column vectors)
we have the linear algebra identity for matrix and vector multiplication:
(A*B)' == (B') * (A') as shown here Properties of Transposes
let t = theta,
given, h(t) = t' * X
h(t)' = (t' X)'
= X' * t
Now we have our variables in the format they were actually given to us. What I mean is that our input file really holds X' and theta is normal, so multiplying them in the order specified above will give a practically equivilant output to that he taught us to use which was theta' * X. Since we are summing all the elements of h(t)' at the end it doesn't matter that it is transposed for the final calculation. However, if you wanted h(t), not h(t)', you could always take your computed result and transpose it because
(A')' == A
However, for the coursera machine learning programming assignment 1, this is unnecessary.
This is because the computer has the coordinate (0,0) positioned on the top left, while geometry has the coordinate (0,0) positioned on the bottom left.
enter image description here
I was reading about non-parametric kernel density estimation.
http://en.wikipedia.org/wiki/Kernel_density_estimation
For uni-variate where D = 1, we can write like
For Multivariate Kernel density estimation (KDE), more preciously for d=3 and X = (x,y,z) can we write:
Is this technically correct? Can any one help with this?
This is very difficult to do on your own, and you really should do this through some package. Nevertheless, the definition is:
fH(x)= 1 / n \sum{i=1}n KH (x - xi), where
x = (x1, x2, …, xd)T, xi = (xi1, xi2, …, xid)T, i = 1, 2, …, n are d-vectors;
H is the bandwidth (or smoothing) d×d matrix which is symmetric and positive definite;
K is the kernel function which is a symmetric multivariate density;
KH(x) = |H|−1/n K(H−1/2x).
When I learn Logistic Regression, we use negative log likelihood to optimize the parameters w for us.
SO the loss function(negative log likelihood) is L(w).
There is an assertion that: the magnitude of the optimal w can go to infi nity when the training samples are linearly seperable.
I get very confused:
1. what does the magnitude of optimal w mean?
2. Could you explain why w can go infinity?
It is the norm (euclidean, for example) what is usually understood as a magnitude of a vector.
Assume that we do binary classification and classes are linearly separable. That means
that there exists w' such that (x1, w') ≥ 0 for x1 from one class and (x2, w') < 0 otherwise. Then consider z = a w' for some positive a. It's clear that (x1, z) ≥ 0 and (x2, z) < 0 (we can multiply equations for w' by a and use linearity of dot product), so as you can see there are separating hyperplanes (zs) of unbounded norm (magnitude).
That's why one should add regularization term.
Short answer:
This is fundamental characteristic of the log function.
consider:
log(x), where x spans (0,1)
Range of values log(x) can take:
is (-Inf, 0)
More specifically to your question -
Log likelihood is given by: ( see image )
l(w) = y * log( h(x)) + (1 - y) * log (1 - h(x) )
where,
h(x) is a sigmoid function parameters by w:
h(x) = ( 1 + exp{-wx} )^-1
For simplicity consider the case of a training example where y = 1,
the equation becomes :
likelihood (l) :
= y * log ( h(x) );
= log ( h(x) )
h(x) in logistic regression maybe represented by the sigmoid function.
it has a range (0,1)
Hence,
range of (l):
(log (0), log(1) ) = (-Inf, 0)
(l) spans the range (-Inf, 0)
The above simplification only considered the (y = 1) case. If you consider the entire log likelihood function (i.e for y=1 & y=0), you will see a inverted bowl shaped cost function. Hence there is a optimum weight that will maximize log likelihood (l) or minimize negative log likelihood (-l)