117 KiB
Optimization, the central part of any Machine Learning algortithm
Almost every problem in machine learning and data science starts with
a dataset X, a model g(\beta), which is a function of the
parameters \beta and a cost function C(X, g(\beta)) that allows
us to judge how well the model g(\beta) explains the observations
X. The model is fit by finding the values of \beta that minimize
the cost function. Ideally we would be able to solve for \beta
analytically, however this is not possible in general and we must use
some approximative/numerical method to compute the minimum.
In our discussion on Logistic Regression we studied the
case of
two classes, with y_i either
0 or 1. Furthermore we assumed also that we have only two
parameters \beta in our fitting, that is we
defined probabilities
\begin{align*}
p(y_i=1|x_i,\boldsymbol{\beta}) &= \frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}},\nonumber\\
p(y_i=0|x_i,\boldsymbol{\beta}) &= 1 - p(y_i=1|x_i,\boldsymbol{\beta}),
\end{align*}
where \boldsymbol{\beta} are the weights we wish to extract from data, in our case \beta_0 and \beta_1.
Our compact equations used a definition of a vector \boldsymbol{y} with n
elements y_i, an n\times p matrix \boldsymbol{X} which contains the
x_i values and a vector \boldsymbol{p} of fitted probabilities
p(y_i\vert x_i,\boldsymbol{\beta}). We rewrote in a more compact form
the first derivative of the cost function as
\frac{\partial \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = -\boldsymbol{X}^T\left(\boldsymbol{y}-\boldsymbol{p}\right).
If we in addition define a diagonal matrix \boldsymbol{W} with elements
p(y_i\vert x_i,\boldsymbol{\beta})(1-p(y_i\vert x_i,\boldsymbol{\beta}), we can obtain a compact expression of the second derivative as
\frac{\partial^2 \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}\partial \boldsymbol{\beta}^T} = \boldsymbol{X}^T\boldsymbol{W}\boldsymbol{X}.
This defines what is called the Hessian matrix.
If we can set up these equations, Newton-Raphson's iterative method is normally the method of choice. It requires however that we can compute in an efficient way the matrices that define the first and second derivatives.
Our iterative scheme is then given by
\boldsymbol{\beta}^{\mathrm{new}} = \boldsymbol{\beta}^{\mathrm{old}}-\left(\frac{\partial^2 \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}\partial \boldsymbol{\beta}^T}\right)^{-1}_{\boldsymbol{\beta}^{\mathrm{old}}}\times \left(\frac{\partial \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}}\right)_{\boldsymbol{\beta}^{\mathrm{old}}},
or in matrix form as
\boldsymbol{\beta}^{\mathrm{new}} = \boldsymbol{\beta}^{\mathrm{old}}-\left(\boldsymbol{X}^T\boldsymbol{W}\boldsymbol{X} \right)^{-1}\times \left(-\boldsymbol{X}^T(\boldsymbol{y}-\boldsymbol{p}) \right)_{\boldsymbol{\beta}^{\mathrm{old}}}.
The right-hand side is computed with the old values of \beta.
If we can compute these matrices, in particular the Hessian, the above is often the easiest method to implement.
Let us quickly remind ourselves how we derive the above method.
Perhaps the most celebrated of all one-dimensional root-finding
routines is Newton's method, also called the Newton-Raphson
method. This method requires the evaluation of both the
function f and its derivative f' at arbitrary points.
If you can only calculate the derivative
numerically and/or your function is not of the smooth type, we
normally discourage the use of this method.
The Newton-Raphson formula consists geometrically of extending the
tangent line at a current point until it crosses zero, then setting
the next guess to the abscissa of that zero-crossing. The mathematics
behind this method is rather simple. Employing a Taylor expansion for
x sufficiently close to the solution s, we have
f(s)=0=f(x)+(s-x)f'(x)+\frac{(s-x)^2}{2}f''(x) +\dots.
\label{eq:taylornr} \tag{1}
For small enough values of the function and for well-behaved functions, the terms beyond linear are unimportant, hence we obtain
f(x)+(s-x)f'(x)\approx 0,
yielding
s\approx x-\frac{f(x)}{f'(x)}.
Having in mind an iterative procedure, it is natural to start iterating with
x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.
The above is Newton-Raphson's method. It has a simple geometric
interpretation, namely x_{n+1} is the point where the tangent from
(x_n,f(x_n)) crosses the $x$-axis. Close to the solution,
Newton-Raphson converges fast to the desired result. However, if we
are far from a root, where the higher-order terms in the series are
important, the Newton-Raphson formula can give grossly inaccurate
results. For instance, the initial guess for the root might be so far
from the true root as to let the search interval include a local
maximum or minimum of the function. If an iteration places a trial
guess near such a local extremum, so that the first derivative nearly
vanishes, then Newton-Raphson may fail totally
Newton's method can be generalized to systems of several non-linear equations and variables. Consider the case with two equations
\begin{array}{cc} f_1(x_1,x_2) &=0\\
f_2(x_1,x_2) &=0,\end{array}
which we Taylor expand to obtain
\begin{array}{cc} 0=f_1(x_1+h_1,x_2+h_2)=&f_1(x_1,x_2)+h_1
\partial f_1/\partial x_1+h_2
\partial f_1/\partial x_2+\dots\\
0=f_2(x_1+h_1,x_2+h_2)=&f_2(x_1,x_2)+h_1
\partial f_2/\partial x_1+h_2
\partial f_2/\partial x_2+\dots
\end{array}.
Defining the Jacobian matrix \boldsymbol{J} we have
\boldsymbol{J}=\left( \begin{array}{cc}
\partial f_1/\partial x_1 & \partial f_1/\partial x_2 \\
\partial f_2/\partial x_1 &\partial f_2/\partial x_2
\end{array} \right),
we can rephrase Newton's method as
\left(\begin{array}{c} x_1^{n+1} \\ x_2^{n+1} \end{array} \right)=
\left(\begin{array}{c} x_1^{n} \\ x_2^{n} \end{array} \right)+
\left(\begin{array}{c} h_1^{n} \\ h_2^{n} \end{array} \right),
where we have defined
\left(\begin{array}{c} h_1^{n} \\ h_2^{n} \end{array} \right)=
-\boldsymbol{J}^{-1}
\left(\begin{array}{c} f_1(x_1^{n},x_2^{n}) \\ f_2(x_1^{n},x_2^{n}) \end{array} \right).
We need thus to compute the inverse of the Jacobian matrix and it
is to understand that difficulties may
arise in case \boldsymbol{J} is nearly singular.
It is rather straightforward to extend the above scheme to systems of more than two non-linear equations. In our case, the Jacobian matrix is given by the Hessian that represents the second derivative of cost function.
Steepest descent
The basic idea of gradient descent is
that a function F(\mathbf{x}),
\mathbf{x} \equiv (x_1,\cdots,x_n), decreases fastest if one goes from \bf {x} in the
direction of the negative gradient -\nabla F(\mathbf{x}).
It can be shown that if
\mathbf{x}_{k+1} = \mathbf{x}_k - \gamma_k \nabla F(\mathbf{x}_k),
with \gamma_k > 0.
For \gamma_k small enough, then $F(\mathbf{x}_{k+1}) \leq
F(\mathbf{x}_k)$. This means that for a sufficiently small \gamma_k
we are always moving towards smaller function values, i.e a minimum.
The previous observation is the basis of the method of steepest
descent, which is also referred to as just gradient descent (GD). One
starts with an initial guess \mathbf{x}_0 for a minimum of F and
computes new approximations according to
\mathbf{x}_{k+1} = \mathbf{x}_k - \gamma_k \nabla F(\mathbf{x}_k), \ \ k \geq 0.
The parameter \gamma_k is often referred to as the step length or
the learning rate within the context of Machine Learning.
Ideally the sequence \{\mathbf{x}_k \}_{k=0} converges to a global
minimum of the function F. In general we do not know if we are in a
global or local minimum. In the special case when F is a convex
function, all local minima are also global minima, so in this case
gradient descent can converge to the global solution. The advantage of
this scheme is that it is conceptually simple and straightforward to
implement. However the method in this form has some severe
limitations:
In machine learing we are often faced with non-convex high dimensional cost functions with many local minima. Since GD is deterministic we will get stuck in a local minimum, if the method converges, unless we have a very good intial guess. This also implies that the scheme is sensitive to the chosen initial condition.
Note that the gradient is a function of $\mathbf{x} = (x_1,\cdots,x_n)$ which makes it expensive to compute numerically.
The gradient descent method
is sensitive to the choice of learning rate \gamma_k. This is due
to the fact that we are only guaranteed that $F(\mathbf{x}_{k+1}) \leq
F(\mathbf{x}_k)$ for sufficiently small \gamma_k. The problem is to
determine an optimal learning rate. If the learning rate is chosen too
small the method will take a long time to converge and if it is too
large we can experience erratic behavior.
Many of these shortcomings can be alleviated by introducing randomness. One such method is that of Stochastic Gradient Descent (SGD), see below.
Convex functions
Ideally we want our cost/loss function to be convex(concave).
First we give the definition of a convex set: A set C in
\mathbb{R}^n is said to be convex if, for all x and y in C and
all t \in (0,1) , the point (1 − t)x + ty also belongs to
C. Geometrically this means that every point on the line segment
connecting x and y is in C as discussed below.
The convex subsets of \mathbb{R} are the intervals of
\mathbb{R}. Examples of convex sets of \mathbb{R}^2 are the
regular polygons (triangles, rectangles, pentagons, etc...).
Convex function: Let X \subset \mathbb{R}^n be a convex
set. Assume that the function f: X \rightarrow \mathbb{R} is
continuous, then f is said to be convex if
f(tx_1 + (1-t)x_2) \leq tf(x_1) + (1-t)f(x_2)
for all
x_1, x_2 \in X and for all t \in [0,1].
If \leq is replaced with a strict inequality in the
definition, we demand x_1 \neq x_2 and t\in(0,1) then f is said
to be strictly convex. For a single variable function, convexity means
that if you draw a straight line connecting f(x_1) and f(x_2), the
value of the function on the interval [x_1,x_2] is always below the
line as discussed below.
In the following we state first and second-order conditions which
ensures convexity of a function f. We write D_f to denote the
domain of f, i.e the subset of R^n where f is defined. For more
details and proofs we refer to: [S. Boyd and L. Vandenberghe. Convex Optimization. Cambridge University Press](http://stanford.edu/boyd/cvxbook/, 2004).
First order condition: Suppose f is differentiable (i.e \nabla f(x) is well defined for
all x in the domain of f). Then f is convex if and only if D_f
is a convex set and f(y) \geq f(x) + \nabla f(x)^T (y-x) holds
for all x,y \in D_f. This condition means that for a convex function
the first order Taylor expansion (right hand side above) at any point
is a global under estimator of the function. To convince yourself you can
make a drawing of f(x) = x^2+1 and draw the tangent line to f(x) and
note that it is always below the graph.
Second order condition: Assume that f is twice
differentiable, i.e the Hessian matrix exists at each point in
D_f. Then f is convex if and only if D_f is a convex set and its
Hessian is positive semi-definite for all x\in D_f. For a
single-variable function this reduces to f''(x) \geq 0. Geometrically this means that f has nonnegative curvature
everywhere.
This condition is particularly useful since it gives us an procedure for determining if the function under consideration is convex, apart from using the definition.
The next result is of great importance to us and the reason why we are going on about convex functions. In machine learning we frequently have to minimize a loss/cost function in order to find the best parameters for the model we are considering.
Ideally we want the global minimum (for high-dimensional models it is hard to know if we have local or global minimum). However, if the cost/loss function is convex the following result provides invaluable information:
Any minimum is global for convex functions.
Consider the problem of finding x \in \mathbb{R}^n such that f(x)
is minimal, where f is convex and differentiable. Then, any point
x^* that satisfies \nabla f(x^*) = 0 is a global minimum.
This result means that if we know that the cost/loss function is convex and we are able to find a minimum, we are guaranteed that it is a global minimum.
Some simple problems
-
Show that
f(x)=x^2is convex forx \in \mathbb{R}using the definition of convexity. Hint: If you re-write the definition,fis convex if the following holds for allx,y \in D_fand any\lambda \in [0,1]\lambda f(x)+(1-\lambda)f(y)-f(\lambda x + (1-\lambda) y ) \geq 0. -
Using the second order condition show that the following functions are convex on the specified domain.
-
f(x) = e^xis convex forx \in \mathbb{R}. -
g(x) = -\ln(x)is convex forx \in (0,\infty).
-
Let
f(x) = x^2andg(x) = e^x. Show thatf(g(x))andg(f(x))is convex forx \in \mathbb{R}. Also show that iff(x)is any convex function thanh(x) = e^{f(x)}is convex. -
A norm is any function that satisfy the following properties
-
f(\alpha x) = |\alpha| f(x)for all\alpha \in \mathbb{R}. -
f(x+y) \leq f(x) + f(y) -
f(x) \leq 0for allx \in \mathbb{R}^nwith equality if and only ifx = 0
Using the definition of convexity, try to show that a function satisfying the properties above is convex (the third condition is not needed to show this).
Standard steepest descent
Before we proceed, we would like to discuss the approach called the standard Steepest descent (different from the above steepest descent discussion), which again leads to us having to be able to compute a matrix. It belongs to the class of Conjugate Gradient methods (CG).
The success of the CG method for finding solutions of non-linear problems is based on the theory of conjugate gradients for linear systems of equations. It belongs to the class of iterative methods for solving problems from linear algebra of the type
\boldsymbol{A}\boldsymbol{x} = \boldsymbol{b}.
In the iterative process we end up with a problem like
\boldsymbol{r}= \boldsymbol{b}-\boldsymbol{A}\boldsymbol{x},
where \boldsymbol{r} is the so-called residual or error in the iterative process.
When we have found the exact solution, \boldsymbol{r}=0.
The residual is zero when we reach the minimum of the quadratic equation
P(\boldsymbol{x})=\frac{1}{2}\boldsymbol{x}^T\boldsymbol{A}\boldsymbol{x} - \boldsymbol{x}^T\boldsymbol{b},
with the constraint that the matrix \boldsymbol{A} is positive definite and
symmetric. This defines also the Hessian and we want it to be positive definite.
We denote the initial guess for \boldsymbol{x} as \boldsymbol{x}_0.
We can assume without loss of generality that
\boldsymbol{x}_0=0,
or consider the system
\boldsymbol{A}\boldsymbol{z} = \boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_0,
instead.
One can show that the solution \boldsymbol{x} is also the unique minimizer of the quadratic form
f(\boldsymbol{x}) = \frac{1}{2}\boldsymbol{x}^T\boldsymbol{A}\boldsymbol{x} - \boldsymbol{x}^T \boldsymbol{x} , \quad \boldsymbol{x}\in\mathbf{R}^n.
This suggests taking the first basis vector \boldsymbol{r}_1 (see below for definition)
to be the gradient of f at \boldsymbol{x}=\boldsymbol{x}_0,
which equals
\boldsymbol{A}\boldsymbol{x}_0-\boldsymbol{b},
and
\boldsymbol{x}_0=0 it is equal -\boldsymbol{b}.
We can compute the residual iteratively as
\boldsymbol{r}_{k+1}=\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_{k+1},
which equals
\boldsymbol{b}-\boldsymbol{A}(\boldsymbol{x}_k+\alpha_k\boldsymbol{r}_k),
or
(\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_k)-\alpha_k\boldsymbol{A}\boldsymbol{r}_k,
which gives
\alpha_k = \frac{\boldsymbol{r}_k^T\boldsymbol{r}_k}{\boldsymbol{r}_k^T\boldsymbol{A}\boldsymbol{r}_k}
leading to the iterative scheme
\boldsymbol{x}_{k+1}=\boldsymbol{x}_k-\alpha_k\boldsymbol{r}_{k},
%matplotlib inline
import numpy as np
import numpy.linalg as la
import scipy.optimize as sopt
import matplotlib.pyplot as pt
from mpl_toolkits.mplot3d import axes3d
def f(x):
return 0.5*x[0]**2 + 2.5*x[1]**2
def df(x):
return np.array([x[0], 5*x[1]])
fig = pt.figure()
ax = fig.gca(projection="3d")
xmesh, ymesh = np.mgrid[-2:2:50j,-2:2:50j]
fmesh = f(np.array([xmesh, ymesh]))
ax.plot_surface(xmesh, ymesh, fmesh)And then as countor plot
pt.axis("equal")
pt.contour(xmesh, ymesh, fmesh)
guesses = [np.array([2, 2./5])]Find guesses
x = guesses[-1]
s = -df(x)Run it!
def f1d(alpha):
return f(x + alpha*s)
alpha_opt = sopt.golden(f1d)
next_guess = x + alpha_opt * s
guesses.append(next_guess)
print(next_guess)What happened?
pt.axis("equal")
pt.contour(xmesh, ymesh, fmesh, 50)
it_array = np.array(guesses)
pt.plot(it_array.T[0], it_array.T[1], "x-")Conjugate gradient method
In the CG method we define so-called conjugate directions and two vectors
\boldsymbol{s} and \boldsymbol{t}
are said to be
conjugate if
\boldsymbol{s}^T\boldsymbol{A}\boldsymbol{t}= 0.
The philosophy of the CG method is to perform searches in various conjugate directions
of our vectors \boldsymbol{x}_i obeying the above criterion, namely
\boldsymbol{x}_i^T\boldsymbol{A}\boldsymbol{x}_j= 0.
Two vectors are conjugate if they are orthogonal with respect to
this inner product. Being conjugate is a symmetric relation: if \boldsymbol{s} is conjugate to \boldsymbol{t}, then \boldsymbol{t} is conjugate to \boldsymbol{s}.
An example is given by the eigenvectors of the matrix
\boldsymbol{v}_i^T\boldsymbol{A}\boldsymbol{v}_j= \lambda\boldsymbol{v}_i^T\boldsymbol{v}_j,
which is zero unless i=j.
Assume now that we have a symmetric positive-definite matrix \boldsymbol{A} of size
n\times n. At each iteration i+1 we obtain the conjugate direction of a vector
\boldsymbol{x}_{i+1}=\boldsymbol{x}_{i}+\alpha_i\boldsymbol{p}_{i}.
We assume that \boldsymbol{p}_{i} is a sequence of n mutually conjugate directions.
Then the \boldsymbol{p}_{i} form a basis of R^n and we can expand the solution
\boldsymbol{A}\boldsymbol{x} = \boldsymbol{b} in this basis, namely
\boldsymbol{x} = \sum^{n}_{i=1} \alpha_i \boldsymbol{p}_i.
The coefficients are given by
\mathbf{A}\mathbf{x} = \sum^{n}_{i=1} \alpha_i \mathbf{A} \mathbf{p}_i = \mathbf{b}.
Multiplying with \boldsymbol{p}_k^T from the left gives
\boldsymbol{p}_k^T \boldsymbol{A}\boldsymbol{x} = \sum^{n}_{i=1} \alpha_i\boldsymbol{p}_k^T \boldsymbol{A}\boldsymbol{p}_i= \boldsymbol{p}_k^T \boldsymbol{b},
and we can define the coefficients \alpha_k as
\alpha_k = \frac{\boldsymbol{p}_k^T \boldsymbol{b}}{\boldsymbol{p}_k^T \boldsymbol{A} \boldsymbol{p}_k}
If we choose the conjugate vectors \boldsymbol{p}_k carefully,
then we may not need all of them to obtain a good approximation to the solution
\boldsymbol{x}.
We want to regard the conjugate gradient method as an iterative method.
This will us to solve systems where n is so large that the direct
method would take too much time.
We denote the initial guess for \boldsymbol{x} as \boldsymbol{x}_0.
We can assume without loss of generality that
\boldsymbol{x}_0=0,
or consider the system
\boldsymbol{A}\boldsymbol{z} = \boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_0,
instead.
One can show that the solution \boldsymbol{x} is also the unique minimizer of the quadratic form
f(\boldsymbol{x}) = \frac{1}{2}\boldsymbol{x}^T\boldsymbol{A}\boldsymbol{x} - \boldsymbol{x}^T \boldsymbol{x} , \quad \boldsymbol{x}\in\mathbf{R}^n.
This suggests taking the first basis vector \boldsymbol{p}_1
to be the gradient of f at \boldsymbol{x}=\boldsymbol{x}_0,
which equals
\boldsymbol{A}\boldsymbol{x}_0-\boldsymbol{b},
and
\boldsymbol{x}_0=0 it is equal -\boldsymbol{b}.
The other vectors in the basis will be conjugate to the gradient,
hence the name conjugate gradient method.
Let \boldsymbol{r}_k be the residual at the $k$-th step:
\boldsymbol{r}_k=\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_k.
Note that \boldsymbol{r}_k is the negative gradient of f at
\boldsymbol{x}=\boldsymbol{x}_k,
so the gradient descent method would be to move in the direction \boldsymbol{r}_k.
Here, we insist that the directions \boldsymbol{p}_k are conjugate to each other,
so we take the direction closest to the gradient \boldsymbol{r}_k
under the conjugacy constraint.
This gives the following expression
\boldsymbol{p}_{k+1}=\boldsymbol{r}_k-\frac{\boldsymbol{p}_k^T \boldsymbol{A}\boldsymbol{r}_k}{\boldsymbol{p}_k^T\boldsymbol{A}\boldsymbol{p}_k} \boldsymbol{p}_k.
We can also compute the residual iteratively as
\boldsymbol{r}_{k+1}=\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_{k+1},