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Deep Learning

Lecture 1: Fundamentals of machine learning



Prof. Gilles Louppe
g.louppe@uliege.be

???

R: over-parametrization https://arxiv.org/pdf/2109.02355.pdf


Today

A recap on statistical learning:

  • Supervised learning
  • Empirical risk minimization
  • Under-fitting and over-fitting
  • Bias-variance dilemma

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Statistical learning


Supervised learning

Consider an unknown joint probability distribution $p_{X,Y}$.

Assume training data $$(\mathbf{x}_i,y_i) \sim p_{X,Y},$$ with $\mathbf{x}_i \in \mathcal{X}$, $y_i \in \mathcal{Y}$, $i=1, ..., N$.

  • In most cases,
    • $\mathbf{x}_i$ is a $p$-dimensional vector of features or descriptors,
    • $y_i$ is a scalar (e.g., a category or a real value).
  • The training data is generated i.i.d.
  • The training data can be of any finite size $N$.
  • In general, we do not have any prior information about $p_{X,Y}$.

???

In most cases, x is a vector, but it could be an image, a piece of text or a sample of sound.


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Supervised learning is usually concerned with the two following inference problems:

  • Classification: Given $(\mathbf{x}_i, y_i) \in \mathcal{X}\times\mathcal{Y} = \mathbb{R}^p \times \bigtriangleup^C$, for $i=1, ..., N$, we want to estimate for any new $\mathbf{x}$, $$\arg \max_y p(Y=y|X=\mathbf{x}).$$
  • Regression: Given $(\mathbf{x}_i, y_i) \in \mathcal{X}\times\mathcal{Y} = \mathbb{R}^p \times \mathbb{R}$, for $i=1, ..., N$, we want to estimate for any new $\mathbf{x}$, $$\mathbb{E}\left[ Y|X=\mathbf{x} \right].$$

???

$\bigtriangleup^C$ is the $C-1$-dimensional probability simplex $\{\mathbf{p} \in \mathbb{R}^C_+ : ||\mathbf{p}||_1 = 1\}$.


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Classification consists in identifying
a decision boundary between objects of distinct classes.


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Regression aims at estimating relationships among (usually continuous) variables.


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Probabilistic perspective

Supervised learning can be framed as probabilistic inference, where the goal is to estimate the conditional distribution $$p(Y=y|X=\mathbf{x})$$ for any new $(\mathbf{x},y)$.

This is the framing we will adopt in this course (starting from Lecture 2).


Empirical risk minimization

The traditional perspective on supervised learning is empirical risk minimization.

Consider a function $f : \mathcal{X} \to \mathcal{Y}$ produced by some learning algorithm. The predictions of this function can be evaluated through a loss $$\ell : \mathcal{Y} \times \mathcal{Y} \to \mathbb{R},$$ such that $\ell(y, f(\mathbf{x})) \geq 0$ measures how close the prediction $f(\mathbf{x})$ from $y$ is.


## Examples of loss functions

.grid[ .kol-1-3[Classification:] .kol-2-3[$\ell(y,f(\mathbf{x})) = \mathbf{1}_{y \neq f(\mathbf{x})}$] ] .grid[ .kol-1-3[Regression:] .kol-2-3[$\ell(y,f(\mathbf{x})) = (y - f(\mathbf{x}))^2$] ]


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Let $\mathcal{F}$ denote the hypothesis space, i.e. the set of all functions $f$ than can be produced by the chosen learning algorithm.

We are looking for a function $f \in \mathcal{F}$ with a small expected risk (or generalization error) $$R(f) = \mathbb{E}_{(\mathbf{x},y)\sim p_{X,Y}}\left[ \ell(y, f(\mathbf{x})) \right].$$

This means that for a given data generating distribution $p_{X,Y}$ and for a given hypothesis space $\mathcal{F}$, the optimal model is $$f_* = \arg \min_{f \in \mathcal{F}} R(f).$$


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Since $p_{X,Y}$ is unknown, the expected risk cannot be evaluated and the optimal model cannot be determined.

However, if we have i.i.d. training data $\mathbf{d} = \{(\mathbf{x}_i, y_i) | i=1,\ldots,N\}$, we can compute an estimate, the empirical risk (or training error) $$\hat{R}(f, \mathbf{d}) = \frac{1}{N} \sum_{(\mathbf{x}_i, y_i) \in \mathbf{d}} \ell(y_i, f(\mathbf{x}_i)).$$

This estimator is unbiased and can be used for finding a good enough approximation of $f_*$. This results into the empirical risk minimization principle: $$f_*^{\mathbf{d}} = \arg \min_{f \in \mathcal{F}} \hat{R}(f, \mathbf{d})$$

???

What does unbiased mean?

=> The expected empirical risk estimate (over d) is the expected risk.


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Most machine learning algorithms, including neural networks, implement empirical risk minimization.

Under regularity assumptions, empirical risk minimizers converge:

$$\lim_{N \to \infty} f_*^{\mathbf{d}} = f_*$$

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This is why tuning the parameters of the model to make it work on the training data is a reasonable thing to do.


Polynomial regression

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Consider the joint probability distribution $p_{X,Y}$ induced by the data generating process $$(x,y) \sim p_{X,Y} \Leftrightarrow x \sim U[-10;10], \epsilon \sim \mathcal{N}(0, \sigma^2), y = g(x) + \epsilon$$ where $x \in \mathbb{R}$, $y\in\mathbb{R}$ and $g$ is an unknown polynomial of degree 3.


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Our goal is to find a function $f$ that makes good predictions on average over $p_{X,Y}$.

Consider the hypothesis space $f \in \mathcal{F}$ of polynomials of degree 3 defined through their parameters $\mathbf{w} \in \mathbb{R}^4$ such that $$\hat{y} \triangleq f(x; \mathbf{w}) = \sum_{d=0}^3 w_d x^d$$


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For this regression problem, we use the squared error loss $$\ell(y, f(x;\mathbf{w})) = (y - f(x;\mathbf{w}))^2$$ to measure how wrong the predictions are.

Therefore, our goal is to find the best value $\mathbf{w}_*$ such that $$\begin{aligned} \mathbf{w}_* &= \arg\min_\mathbf{w} R(\mathbf{w}) \\ &= \arg\min_\mathbf{w} \mathbb{E}_{(x,y)\sim p_{X,Y}}\left[ (y-f(x;\mathbf{w}))^2 \right] \end{aligned}$$


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Given a large enough training set $\mathbf{d} = \{(x_i, y_i) | i=1,\ldots,N\}$, the empirical risk minimization principle tells us that a good estimate $\mathbf{w}_*^{\mathbf{d}}$ of $\mathbf{w}_*$ can be found by minimizing the empirical risk: $$\begin{aligned} \mathbf{w}_*^{\mathbf{d}} &= \arg\min_\mathbf{w} \hat{R}(\mathbf{w},\mathbf{d}) \\ &= \arg\min_\mathbf{w} \frac{1}{N} \sum_{(x_i, y_i) \in \mathbf{d}} (y_i - f(x_i;\mathbf{w}))^2 \\ &= \arg\min_\mathbf{w} \frac{1}{N} \sum_{(x_i, y_i) \in \mathbf{d}} (y_i - \sum_{d=0}^3 w_d x_i^d)^2 \\ &= \arg\min_\mathbf{w} \frac{1}{N} \left\lVert \underbrace{\begin{pmatrix} y_1 \\ y_2 \\ \ldots \\ y_N \end{pmatrix}}_{\mathbf{y}} - \underbrace{\begin{pmatrix} x_1^0 \ldots x_1^3 \\ x_2^0 \ldots x_2^3 \\ \ldots \\ x_N^0 \ldots x_N^3 \end{pmatrix}}_{\mathbf{X}} \begin{pmatrix} w_0 \\ w_1 \\ w_2 \\ w_3 \end{pmatrix} \right\rVert^2 \end{aligned}$$


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This is ordinary least squares regression, for which the solution is derived as $$\mathbf{w}_*^{\mathbf{d}} = (\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T\mathbf{y}.$$

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The expected risk minimizer $\mathbf{w}_*$ within our hypothesis space is $g$ itself.

Therefore, on this toy problem, we can verify that $f(x;\mathbf{w}_*^{\mathbf{d}}) \to f(x;\mathbf{w}_*) = g(x)$ as $N \to \infty$.


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Under-fitting and over-fitting

What if we consider a hypothesis space $\mathcal{F}$ in which candidate functions $f$ are either too "simple" or too "complex" with respect to the true data generating process?


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Which model would you choose?

.grid[ .kol-1-3[

$f_1(x) = w_0 + w_1 x$

] .kol-1-3[

$f_2(x) = \sum_{j=0}^3 w_j x^j$

] .kol-1-3[

$f_3(x) = \sum_{j=0}^{10^4} w_j x^j$

] ]

???

In this course, we will argue for $f_3$.

Large parameter spaces are not a problem, as long as the capacity of the hypothesis space is controlled. For example, by using stochastic gradient descent, we can optimize $f_3$ without overfitting.


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.center[$\mathcal{F}$ = polynomials of degree 1]


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.center[$\mathcal{F}$ = polynomials of degree 2]


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.center[$\mathcal{F}$ = polynomials of degree 3]


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.center[$\mathcal{F}$ = polynomials of degree 4]


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.center[$\mathcal{F}$ = polynomials of degree 5]


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.center[$\mathcal{F}$ = polynomials of degree 10]


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Degree $d$ of the polynomial VS. error.

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Why shouldn't we pick the largest $d$?


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Let $\mathcal{Y}^{\mathcal X}$ be the set of all functions $f : \mathcal{X} \to \mathcal{Y}$.

We define the Bayes risk as the minimal expected risk over all possible functions, $$R_B = \min_{f \in \mathcal{Y}^{\mathcal X}} R(f),$$ and call the Bayes optimal model the model $f_B$ that achieves this minimum.

No model $f$ can perform better than $f_B$.


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The capacity of an hypothesis space induced by a learning algorithm intuitively represents the ability to find a good model $f \in \mathcal{F}$ for any function, regardless of its complexity.

In practice, capacity can be controlled through hyper-parameters of the learning algorithm. For example:

  • The degree of the family of polynomials;
  • The number of layers in a neural network;
  • The number of training iterations;
  • Regularization terms.

???

We talk about the capacity of the hypothesis space induced by the learning algorithm (parametric model + optimization algorithm). This is different from the capacity of the model itself.


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  • If the capacity of $\mathcal{F}$ is too low, then $f_B \notin \mathcal{F}$ and $R(f) - R_B$ is large for any $f \in \mathcal{F}$, including $f_*$ and $f_*^{\mathbf{d}}$. Such models $f$ are said to underfit the data.
  • If the capacity of $\mathcal{F}$ is too high, then $f_B \in \mathcal{F}$ or $R(f_*) - R_B$ is small.
    However, because of the high capacity of the hypothesis space, the empirical risk minimizer $f_*^{\mathbf{d}}$ could fit the training data arbitrarily well such that $$R(f_*^{\mathbf{d}}) \geq R_B \geq \hat{R}(f_*^{\mathbf{d}}, \mathbf{d}) \geq 0.$$ In this situation, $f_*^{\mathbf{d}}$ becomes too specialized with respect to the true data generating process and a large reduction of the empirical risk (often) comes at the price of an increase of the expected risk of the empirical risk minimizer $R(f_*^{\mathbf{d}})$. In this situation, $f_*^{\mathbf{d}}$ is said to overfit the data.

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Therefore, our goal is to adjust the capacity of the hypothesis space such that the expected risk of the empirical risk minimizer gets as low as possible.

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???

Comment that for deep networks, training error may goes to 0 while the generalization error may not necessarily go up!


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When overfitting, $$R(f_*^{\mathbf{d}}) \geq R_B \geq \hat{R}(f_*^{\mathbf{d}}, \mathbf{d}) \geq 0.$$

This indicates that the empirical risk $\hat{R}(f_*^{\mathbf{d}}, \mathbf{d})$ is a poor estimator of the expected risk $R(f_*^{\mathbf{d}})$.

Nevertheless, an unbiased estimate of the expected risk can be obtained by evaluating $f_*^{\mathbf{d}}$ on data $\mathbf{d}_\text{test}$ independent from the training samples $\mathbf{d}$: $$\hat{R}(f_*^{\mathbf{d}}, \mathbf{d}_\text{test}) = \frac{1}{N} \sum_{(\mathbf{x}_i, y_i) \in \mathbf{d}_\text{test}} \ell(y_i, f_*^{\mathbf{d}}(\mathbf{x}_i))$$

This test error estimate can be used to evaluate the actual performance of the model. However, it should not be used, at the same time, for model selection.


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Degree $d$ of the polynomial VS. error.

???

What value of $d$ shall you select?

But then how good is this selected model?


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(Proper) evaluation protocol

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There may be over-fitting, but it does not bias the final performance evaluation.

.footnote[Credits: Francois Fleuret, EE559 Deep Learning, EPFL.]


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.center[This should be avoided at all costs!]

.footnote[Credits: Francois Fleuret, EE559 Deep Learning, EPFL.]


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.center[Instead, keep a separate validation set for tuning the hyper-parameters.]

.footnote[Credits: Francois Fleuret, EE559 Deep Learning, EPFL.]

???

Comment on the comparison of algorithms from one paper to the other.


Bias-variance decomposition

Consider a fixed point $x$ and the prediction $\hat{Y}=f_*^\mathbf{d}(x)$ of the empirical risk minimizer at $x$.

Then the local expected risk of $f_*^{\mathbf{d}}$ is $$\begin{aligned} R(f_*^{\mathbf{d}}|x) &= \mathbb{E}_{y \sim p_{Y|x}} \left[ (y - f_*^{\mathbf{d}}(x))^2 \right] \\ &= \mathbb{E}_{y \sim p_{Y|x}} \left[ (y - f_B(x) + f_B(x) - f_*^{\mathbf{d}}(x))^2 \right] \\ &= \mathbb{E}_{y \sim p_{Y|x}} \left[ (y - f_B(x))^2 \right] + \mathbb{E}_{y \sim p_{Y|x}} \left[ (f_B(x) - f_*^{\mathbf{d}}(x))^2 \right] \\ &= R(f_B|x) + (f_B(x) - f_*^{\mathbf{d}}(x))^2 \end{aligned}$$ where

  • $R(f_B|x)$ is the local expected risk of the Bayes model. This term cannot be reduced.
  • $(f_B(x) - f_*^{\mathbf{d}}(x))^2$ represents the discrepancy between $f_B$ and $f_*^{\mathbf{d}}$.

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If $\mathbf{d} \sim p_{X,Y}$ is itself considered as a random variable, then $f_*^\mathbf{d}$ is also a random variable, along with its predictions $\hat{Y}$.


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???

What do you observe?


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Formally, the expected local expected risk yields to: $$\begin{aligned} &\mathbb{E}_\mathbf{d} \left[ R(f_*^{\mathbf{d}}|x) \right] \\ &= \mathbb{E}_\mathbf{d} \left[ R(f_B|x) + (f_B(x) - f_*^{\mathbf{d}}(x))^2 \right] \\ &= R(f_B|x) + \mathbb{E}_\mathbf{d} \left[ (f_B(x) - f_*^{\mathbf{d}}(x))^2 \right] \\ &= \underbrace{R(f_B|x)}_{\text{noise}(x)} + \underbrace{(f_B(x) - \mathbb{E}_\mathbf{d}\left[ f_*^\mathbf{d}(x) \right] )^2}_{\text{bias}^2(x)} + \underbrace{\mathbb{E}_\mathbf{d}\left[ ( \mathbb{E}_\mathbf{d}\left[ f_*^\mathbf{d}(x) \right] - f_*^\mathbf{d}(x))^2 \right]}_{\text{var}(x)} \end{aligned}$$

This decomposition is known as the bias-variance decomposition.

  • The noise term quantifies the irreducible part of the expected risk.
  • The bias term measures the discrepancy between the average model and the Bayes model.
  • The variance term quantities the variability of the predictions.

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Bias-variance trade-off

  • Reducing the capacity makes $f_*^\mathbf{d}$ fit the data less on average, which increases the bias term.
  • Increasing the capacity makes $f_*^\mathbf{d}$ vary a lot with the training data, which increases the variance term.

.footnote[Credits: Francois Fleuret, EE559 Deep Learning, EPFL.]


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What about a neural network with .bold[millions] of parameters?


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.footnote[Credits: Belkin et al, 2018.]

???

This plot is known as the "double descent" curve. It shows that the test error can decrease as the number of parameters increases, even after the model has enough capacity to fit the training data.

The x-axis is misleading, as the number of parameters is not the same as the capacity.


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.footnote[Credits: Belkin et al, 2018.]


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The end.