<jupyter_start><jupyter_text>Data Analysis and Machine Learning  Lectures: Optimization and  Gradient Methods  **<NAME>**, Department of Physics, University of Oslo and Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State UniversityDate: **Sep 20, 2018**Copyright 1999-2018, <NAME>. Released under CC Attribution-NonCommercial 4.0 license Optimization, the central part of any Machine Learning algortithmAlmost every problem in machine learning and data science starts witha dataset $X$, a model $g(\beta)$, which is a function of theparameters $\beta$ and a cost function $C(X, g(\beta))$ that allowsus to judge how well the model $g(\beta)$ explains the observations$X$. The model is fit by finding the values of $\beta$ that minimizethe cost function. Ideally we would be able to solve for $\beta$analytically, however this is not possible in general and we must usesome approximative/numerical method to compute the minimum. Steepest descentThe method of steepest descent The basic idea of gradient descent isthat a function $F(\mathbf{x})$, $\mathbf{x} \equiv (x_1,\cdots,x_n)$, decreases fastest if one goes from $\bf {x}$ in thedirection of the negative gradient $-\nabla F(\mathbf{x})$.It can be shown that if $$\mathbf{x}_{k+1} = \mathbf{x}_k - \gamma_k \nabla F(\mathbf{x}_k), \ \ \gamma_k &gt; 0$$ for $\gamma_k$ small enough, then $F(\mathbf{x}_{k+1}) \leqF(\mathbf{x}_k)$. This means that for a sufficiently small $\gamma_k$we are always moving towards smaller function values, i.e a minimum. More on Steepest descentThe previous observation is the basis of the method of steepestdescent, which is also referred to as just gradient descent (GD). Onestarts with an initial guess $\mathbf{x}_0$ for a minimum of $F$ andcomputes 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 orthe learning rate within the context of Machine Learning. The idealIdeally the sequence $\{ \mathbf{x}_k \}_{k=0}$ converges to a globalminimum of the function $F$. In general we do not know if we are in aglobal or local minimum. In the special case when $F$ is a convexfunction, all local minima are also global minima, so in this casegradient descent can converge to the global solution. The advantage ofthis scheme is that it is conceptually simple and straightforward toimplement. However the method in this form has some severelimitations:In machine learing we are often faced with non-convex high dimensionalcost functions with many local minima. Since GD is deterministic wewill get stuck in a local minimum, if the method converges, unless wehave a very good intial guess. This also implies that the scheme issensitive 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 sensitiveness of the gradient descentGD is sensitive to the choice of learning rate $\gamma_k$. This is dueto the fact that we are only guaranteed that $F(\mathbf{x}_{k+1}) \leqF(\mathbf{x}_k)$ for sufficiently small $\gamma_k$. The problem is todetermine an optimal learning rate. If the learning rate is chosen toosmall the method will take a long time to converge and if it is toolarge we can experience erratic behavior.Many of these shortcomings can be alleviated by introducingrandomness. One such method is that of Stochastic Gradient Descent(SGD), see below. Gradient Descent ExampleWe revisit now our simple linear regression example with a linear polynomial.<jupyter_code>%matplotlib inline


# Importing various packages
from random import random, seed
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
from matplotlib.ticker import LinearLocator, FormatStrFormatter
import sys

x = 2*np.random.rand(100,1)
y = 4+3*x+np.random.randn(100,1)

xb = np.c_[np.ones((100,1)), x]
beta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
print(beta_linreg)
beta = np.random.randn(2,1)

eta = 0.1
Niterations = 1000
m = 100

for iter in range(Niterations):
    gradients = 2.0/m*xb.T.dot(xb.dot(beta)-y)
    beta -= eta*gradients

print(beta)
xnew = np.array([[0],[2]])
xbnew = np.c_[np.ones((2,1)), xnew]
ypredict = xbnew.dot(beta)
ypredict2 = xbnew.dot(beta_linreg)
plt.plot(xnew, ypredict, "r-")
plt.plot(xnew, ypredict2, "b-")
plt.plot(x, y ,'ro')
plt.axis([0,2.0,0, 15.0])
plt.xlabel(r'$x$')
plt.ylabel(r'$y$')
plt.title(r'Gradient descent example')
plt.show()<jupyter_output>[[3.79420435]
 [3.13985512]]
[[3.79420435]
 [3.13985512]]<jupyter_text>And a corresponding example using **scikit-learn**<jupyter_code># Importing various packages
from random import random, seed
import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import SGDRegressor

x = 2*np.random.rand(100,1)
y = 4+3*x+np.random.randn(100,1)

xb = np.c_[np.ones((100,1)), x]
beta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
print(beta_linreg)
sgdreg = SGDRegressor(n_iter = 50, penalty=None, eta0=0.1)
sgdreg.fit(x,y.ravel())
print(sgdreg.intercept_, sgdreg.coef_)<jupyter_output>[[4.20374363]
 [2.83470272]]
[4.21610176] [2.82633808]<jupyter_text>Convex functionsIdeally 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 $C$ andall $t \in (0,1)$ , the point $(1 − t)x + ty$ also belongs toC. Geometrically this means that every point on the line segmentconnecting $x$ 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 theregular polygons (triangles, rectangles, pentagons, etc...). Convex function**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 inequaltiy 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 illustrated below. Conditions on convex functionsIn the following we state first and second-order conditions whichensures convexity of a function $f$. We write $D_f$ to denote thedomain of $f$, i.e the subset of $R^n$ where $f$ is defined. For moredetails and proofs we refer to: [<NAME> and <NAME>. 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 forall $x$ in the domain of $f$). Then $f$ is convex if and only if $D_f$is a convex set and $$f(y) \geq f(x) + \nabla f(x)^T (y-x) $$ holdsfor all $x,y \in D_f$. This condition means that for a convex functionthe first order Taylor expansion (right hand side above) at any pointa global under estimator of the function. To convince yourself you canmake a drawing of f(x) = x^2+1 and draw the tangent line to $f(x)$ andnote that it is always below the graph.**Second order condition.**Assume that $f$ is twicedifferentiable, 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 itsHessian is positive semi-definite for all $x\in D_f$. For asingle-variable function this reduces to $f''(x) \geq0$. Geometrically this means that $f$ has nonnegative curvatureeverywhere.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. More on convex functionsThe next result is of great importance to us and the reason why we aregoing on about convex functions. In machine learning we frequentlyhave to minimize a loss/cost function in order to find the bestparameters for the model we are considering. Ideally we want theglobal minimum (for high-dimensional models it is hard to knowif we have local or global minimum). However, if the cost/loss functionis 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 problems1. 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). Revisiting our first homeworkWe will use linear regression as a case study for the gradient descentmethods. Linear regression is a great test case for the gradientdescent methods discussed in the lectures since it has severaldesirable properties such as:1. An analytical solution (recall homework set 1).2. The gradient can be computed analytically.3. The cost function is convex which guarantees that gradient descent converges for small enough learning ratesWe revisit the example from homework set 1 where we had $$y_i = 5x_i^2 + 0.1\xi_i, \ i=1,\cdots,100$$ with $x_i \in [0,1] $ chosen randomly with a uniform distribution. Additionally $\xi_i$ represents stochastic noise chosen according to a normal distribution $\cal {N}(0,1)$. The linear regression model is given by $$h_\beta(x) = \hat{y} = \beta_0 + \beta_1 x,$$ such that $$\hat{y}_i = \beta_0 + \beta_1 x_i.$$  Gradient descent exampleLet $\mathbf{y} = (y_1,\cdots,y_n)^T$, $\mathbf{\hat{y}} = (\hat{y}_1,\cdots,\hat{y}_n)^T$ and $\beta = (\beta_0, \beta_1)^T$t is convenient to write $\mathbf{\hat{y}} = X\beta$ where $X \in \mathbb{R}^{100 \times 2} $ is the design matrix given by $$\begin{equation}X \equiv \begin{bmatrix}1 &amp; x_1  \\\vdots &amp; \vdots  \\1 &amp; x_{100} &amp;  \\\end{bmatrix}.\label{_auto1} \tag{1}\end{equation}$$ The loss function is given by $$C(\beta) = ||X\beta-\mathbf{y}||^2 = ||X\beta||^2 - 2 \mathbf{y}^T X\beta + ||\mathbf{y}||^2 = \sum_{i=1}^{100} (\beta_0 + \beta_1 x_i)^2 - 2 y_i (\beta_0 + \beta_1 x_i) + y_i^2$$ and we want to find $\beta$ such that $C(\beta)$ is minimized. The derivative of the cost/loss functionComputing $\partial C(\beta) / \partial \beta_0$ and $\partial C(\beta) / \partial \beta_1$ we can show  that the gradient can be written as $$\nabla_\beta C(\beta) = (\partial C(\beta) / \partial \beta_0, \partial C(\beta) / \partial \beta_1)^T = 2\begin{bmatrix} \sum_{i=1}^{100} \left(\beta_0+\beta_1x_i-y_i\right) \\\sum_{i=1}^{100}\left( x_i (\beta_0+\beta_1x_i)-y_ix_i\right) \\\end{bmatrix} = 2X^T(X\beta - \mathbf{y}),$$ where $X$ is the design matrix defined above. The Hessian matrixThe Hessian matrix of $C(\beta)$ is given by $$\hat{H} \equiv \begin{bmatrix}\frac{\partial^2 C(\beta)}{\partial \beta_0^2} &amp; \frac{\partial^2 C(\beta)}{\partial \beta_0 \partial \beta_1}  \\\frac{\partial^2 C(\beta)}{\partial \beta_0 \partial \beta_1} &amp; \frac{\partial^2 C(\beta)}{\partial \beta_1^2} &amp;  \\\end{bmatrix} = 2X^T X.$$ This result implies that $C(\beta)$ is a convex function since the matrix $X^T X$ always is positive semi-definite. Simple programWe can now write a program that minimizes $C(\beta)$ using the gradient descent method with a constant learning rate $\gamma$ according to $$\beta_{k+1} = \beta_k - \gamma \nabla_\beta C(\beta_k), \ k=0,1,\cdots$$ We can use the expression we computed for the gradient and let use a$\beta_0$ be chosen randomly and let $\gamma = 0.001$. Stop iteratingwhen $||\nabla_\beta C(\beta_k) || &lt; \epsilon = 10^{-8}$. And finally we can compare our solution for $\beta$ with the analytic result given by $\beta= (X^TX)^{-1} X^T \mathbf{y}$.<jupyter_code>import numpy as np

"""
The following setup is just a suggestion, feel free to write it the way you like.
"""

#Setup problem described in the exercise
N  = 100 #Nr of datapoints
M  = 2 #Nr of features
x  = np.random.rand(N) #Uniformly generated x-values in [0,1]
y  = 5*x**2 + 0.1*np.random.randn(N)
X  = np.c_[np.ones(N),x] #Construct design matrix

#Compute beta according to normal equations to compare with GD solution
Xt_X_inv = np.linalg.inv(np.dot(X.T,X))
Xt_y     = np.dot(X.transpose(),y)
beta_NE = np.dot(Xt_X_inv,Xt_y)
print(beta_NE)<jupyter_output>[-0.72135172  4.78496149]<jupyter_text>Gradient descent and RidgeWe have also discussed Ridge regression where the loss function contains a regularized given by the $L_2$ norm of $\beta$, $$C_{\text{ridge}}(\beta) = ||X\beta -\mathbf{y}||^2 + \lambda ||\beta||^2, \ \lambda \geq 0.$$ In order to minimize $C_{\text{ridge}}(\beta)$ using GD we only have adjust the gradient as follows $$\nabla_\beta C_{\text{ridge}}(\beta)  = 2\begin{bmatrix} \sum_{i=1}^{100} \left(\beta_0+\beta_1x_i-y_i\right) \\\sum_{i=1}^{100}\left( x_i (\beta_0+\beta_1x_i)-y_ix_i\right) \\\end{bmatrix} + 2\lambda\begin{bmatrix} \beta_0 \\ \beta_1\end{bmatrix} = 2 (X^T(X\beta - \mathbf{y})+\lambda \beta).$$ We can now extend our program to minimize $C_{\text{ridge}}(\beta)$ using gradient descent and compare with the analytical solution given by $$\beta_{\text{ridge}} = \left(X^T X + \lambda I_{2 \times 2} \right)^{-1} X^T \mathbf{y},$$ for $\lambda = {0,1,10,50,100}$ ($\lambda = 0$ corresponds to ordinary least squares). We can then compute $||\beta_{\text{ridge}}||$ for each $\lambda$.<jupyter_code>import numpy as np

"""
The following setup is just a suggestion, feel free to write it the way you like.
"""

#Setup problem described in the exercise
N  = 100 #Nr of datapoints
M  = 2   #Nr of features
x  = np.random.rand(N)
y  = 5*x**2 + 0.1*np.random.randn(N)


#Compute analytic beta for Ridge regression 
X    = np.c_[np.ones(N),x]
XT_X = np.dot(X.T,X)

l  = 0.1 #Ridge parameter lambda
Id = np.eye(XT_X.shape[0])

Z = np.linalg.inv(XT_X+l*Id)
beta_ridge = np.dot(Z,np.dot(X.T,y))

print(beta_ridge)
print(np.linalg.norm(beta_ridge)) #||beta||<jupyter_output>[-0.76259836  4.89939464]
4.958389265060428<jupyter_text>Stochastic Gradient DescentStochastic gradient descent (SGD) and variants thereof address some ofthe shortcomings of the Gradient descent method discussed above.The underlying idea of SGD comes from the observation that the costfunction, which we want to minimize, can almost always be written as asum over $n$ datapoints $\{\mathbf{x}_i\}_{i=1}^n$, $$C(\mathbf{\beta}) = \sum_{i=1}^n c_i(\mathbf{x}_i,\mathbf{\beta}).$$  Computation of gradientsThis in turn means that the gradient can becomputed as a sum over $i$-gradients $$\nabla_\beta C(\mathbf{\beta}) = \sum_i^n \nabla_\beta c_i(\mathbf{x}_i,\mathbf{\beta}).$$ Stochasticity/randomness is introduced by only taking thegradient on a subset of the data called minibatches.  If there are $n$datapoints and the size of each minibatch is $M$, there will be $n/M$minibatches. We denote these minibatches by $B_k$ where$k=1,\cdots,n/M$. SGD exampleAs an example, suppose we have $10$ datapoints $( \mathbf{x}_1,\cdots, \mathbf{x}_{10} )$ and we choose to have $M=5$ minibathces,then each minibatch contains two datapoints. In particular we have$B_1 = (\mathbf{x}_1,\mathbf{x}_2), \cdots, B_5 =(\mathbf{x}_9,\mathbf{x}_{10})$. Note that if you choose $M=1$ youhave only a single batch with all datapoints and on the other extreme,you may choose $M=n$ resulting in a minibatch for each datapoint, i.e$B_k = \mathbf{x}_k$.The idea is now to approximate the gradient by replacing the sum overall datapoints with a sum over the datapoints in one the minibatchespicked at random in each gradient descent step $$\nabla_\betaC(\mathbf{\beta}) = \sum_{i=1}^n \nabla_\beta c_i(\mathbf{x}_i,\mathbf{\beta}) \rightarrow \sum_{i \in B_k}^n \nabla_\betac_i(\mathbf{x}_i, \mathbf{\beta}).$$  The gradient stepThus a gradient descent step now looks like $$\beta_{j+1} = \beta_j - \gamma_j \sum_{i \in B_k}^n \nabla_\beta c_i(\mathbf{x}_i,\mathbf{\beta})$$ where $k$ is picked at random with equalprobability from $[1,n/M]$. An iteration over the number ofminibathces (n/M) is commonly referred to as an epoch. Thus it istypical to choose a number of epochs and for each epoch iterate overthe number of minibatches, as exemplified in the code below. Simple example code<jupyter_code>import numpy as np 

n = 100 #100 datapoints 
M = 5   #size of each minibatch
m = int(n/M) #number of minibatches
n_epochs = 10 #number of epochs

j = 0
for epoch in range(1,n_epochs+1):
    for i in range(m):
        k = np.random.randint(m) #Pick the k-th minibatch at random
        #Compute the gradient using the data in minibatch Bk
        #Compute new suggestion for 
        j += 1<jupyter_output><empty_output><jupyter_text>Taking the gradient only on a subset of the data has two importantbenefits. First, it introduces randomness which decreases the chancethat our opmization scheme gets stuck in a local minima. Second, ifthe size of the minibatches are small relative to the number ofdatapoints ($M &lt; n$), the computation of the gradient is muchcheaper since we sum over the datapoints in the k-th minibatch and notall $n$ datapoints. When do we stop?A natural question is when do we stop the search for a new minimum?One possibility is to compute the full gradient after a given numberof epochs and check if the norm of the gradient is smaller than somethreshold and stop if true. However, the condition that the gradientis zero is valid also for local minima, so this would only tell usthat we are close to a local/global minimum. However, we could alsoevaluate the cost function at this point, store the result andcontinue the search. If the test kicks in at a later stage we cancompare the values of the cost function and keep the $\beta$ thatgave the lowest value. Slightly different approachAnother approach is to let the step length $\gamma_j$ depend on thenumber of epochs in such a way that it becomes very small after areasonable time such that we do not move at all.As an example, let $e = 0,1,2,3,\cdots$ denote the current epoch and let $t_0, t_1 &gt; 0$ be two fixed numbers. Furthermore, let $t = e \cdot m + i$ where $m$ is the number of minibatches and $i=0,\cdots,m-1$. Then the function $$\gamma_j(t; t_0, t_1) = \frac{t_0}{t+t_1} $$ goes to zero as the number of epochs gets large. I.e. we start with a step length $\gamma_j (0; t_0, t_1) = t_0/t_1$ which decays in *time* $t$.In this way we can fix the number of epochs, compute $\beta$ andevaluate the cost function at the end. Repeating the computation willgive a different result since the scheme is random by design. Then wepick the final $\beta$ that gives the lowest value of the costfunction.<jupyter_code>import numpy as np 

def step_length(t,t0,t1):
    return t0/(t+t1)

n = 100 #100 datapoints 
M = 5   #size of each minibatch
m = int(n/M) #number of minibatches
n_epochs = 500 #number of epochs
t0 = 1.0
t1 = 10

gamma_j = t0/t1
j = 0
for epoch in range(1,n_epochs+1):
    for i in range(m):
        k = np.random.randint(m) #Pick the k-th minibatch at random
        #Compute the gradient using the data in minibatch Bk
        #Compute new suggestion for beta
        t = epoch*m+i
        gamma_j = step_length(t,t0,t1)
        j += 1

print("gamma_j after %d epochs: %g" % (n_epochs,gamma_j))<jupyter_output><empty_output>
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