FYS-STK4155/doc/LectureNotes/_build/jupyter_execute/chapteroptimization.ipynb

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

1. Show that f(x)=x^2 is convex for x \in \mathbb{R} using the definition of convexity. Hint: If you re-write the definition, f is convex if the following holds for all x,y \in D_f and any \lambda \in [0,1] \lambda f(x)+(1-\lambda)f(y)-f(\lambda x + (1-\lambda) y ) \geq 0.

2. Using the second order condition show that the following functions are convex on the specified domain.

• f(x) = e^x is convex for x \in \mathbb{R}.

• g(x) = -\ln(x) is convex for x \in (0,\infty).

3. Let f(x) = x^2 and g(x) = e^x. Show that f(g(x)) and g(f(x)) is convex for x \in \mathbb{R}. Also show that if f(x) is any convex function than h(x) = e^{f(x)} is convex.

4. 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 0 for all x \in \mathbb{R}^n with equality if and only if x = 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},

In [1]:

%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)

Out [1]:

/var/folders/td/3yk470mj5p931p9dtkk0y6jw0000gn/T/ipykernel_31672/483257001.py:18: MatplotlibDeprecationWarning: Calling gca() with keyword arguments was deprecated in Matplotlib 3.4. Starting two minor releases later, gca() will take no keyword arguments. The gca() function should only be used to get the current axes, or if no axes exist, create new axes with default keyword arguments. To create a new axes with non-default arguments, use plt.axes() or plt.subplot().
  ax = fig.gca(projection="3d")

<mpl_toolkits.mplot3d.art3d.Poly3DCollection at 0x1183f2640>

And then as countor plot

In [2]:

pt.axis("equal")
pt.contour(xmesh, ymesh, fmesh)
guesses = [np.array([2, 2./5])]

Find guesses

In [3]:

x = guesses[-1]
s = -df(x)

Run it!

In [4]:

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)

[ 1.33333333 -0.26666667]

What happened?

In [5]:

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-")

Out [5]:

[<matplotlib.lines.Line2D at 0x118c9b1c0>]

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},

Warning:

Output truncated. This notebook contains too many cells to display efficiently.