SDSC6001 Assignment 1: Covering-Number Generalization Bound

Problem: Prove a covering-number bound and verify it by simulation

Background (given)

Let $\mathcal{H}$ be a family of functions mapping $\mathcal{X}$ to $\mathcal{Y} \subseteq \mathbb{R}$. We consider the squared loss

$$\ell(h(x), y) = (h(x) - y)^2$$

Let $D$ be an unknown distribution over $\mathcal{X} \times \mathcal{Y}$. For $h \in \mathcal{H}$, define the true risk and empirical risk:

$$R(h) = \mathbb{E}_{(x,y)\sim D}[(h(x) - y)^2], \quad \hat{R}_S(h) = \frac{1}{m}\sum_{i=1}^{m}(h(x_i) - y_i)^2$$

where $S = ((x_1, y_1), \ldots, (x_m, y_m)) \sim D^m$ is an i.i.d. sample.

Assume $\mathcal{H}$ is bounded: there exists $M > 0$ such that

$$|h(x) - y| \leq M \quad \text{for all } (x, y) \in \mathcal{X} \times \mathcal{Y} \text{ and all } h \in \mathcal{H}$$

Covering number (sup norm). For $\varepsilon > 0$, the covering number $\mathcal{N}(\mathcal{H}, \varepsilon)$ under $\|\cdot\|_\infty$ is the smallest $k$ such that there exist $h_1, \ldots, h_k \in \mathcal{H}$ satisfying: for every $h \in \mathcal{H}$, there exists $i \leq k$ with

$$\|h - h_i\|_\infty = \sup_{x \in \mathcal{X}} |h(x) - h_i(x)| \leq \varepsilon$$

Similar to VC dimension, covering numbers provides a measurement of complexity of a class of functions: the larger the covering number, the richer is the family of functions. The objective of this problem is to illustrate this by proving a learning bound in the case of the squared loss.

Target bound (to prove)

Prove the following generalization bound:

$$\mathbb{P}_{S\sim D^m}\left[\sup_{h\in\mathcal{H}}|R(h) - \hat{R}_S(h)| \geq \varepsilon\right] \leq \mathcal{N}\left(\mathcal{H}, \frac{\varepsilon}{8M}\right) \cdot 2\exp\left(-\frac{m\varepsilon^2}{2M^4}\right) \quad (\star)$$

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Part I: Proof

Let $L_S(h) = R(h) - \hat{R}_S(h)$.

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Problem 1 (20%): Lipschitz step

Problem Statement: Show that for all $h_1, h_2 \in \mathcal{H}$ and any sample $S$,

$$|L_S(h_1) - L_S(h_2)| \leq 4M \|h_1 - h_2\|_\infty$$

Hint: Use $a^2 - b^2 = (a-b)(a+b)$ and boundedness by $M$.

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Proof:

First, we analyze the Lipschitz property of $R(h)$ and $\hat{R}_S(h)$ with respect to $h$ separately.

Step 1: Analyze $|R(h_1) - R(h_2)|$

$$R(h_1) - R(h_2) = \mathbb{E}_{(x,y)\sim D}[(h_1(x) - y)^2 - (h_2(x) - y)^2]$$

Using the identity $a^2 - b^2 = (a-b)(a+b)$, let $a = h_1(x) - y$ and $b = h_2(x) - y$:

$$(h_1(x) - y)^2 - (h_2(x) - y)^2 = (h_1(x) - h_2(x))((h_1(x) - y) + (h_2(x) - y))$$

Therefore:

$$|R(h_1) - R(h_2)| = \left|\mathbb{E}_{(x,y)\sim D}[(h_1(x) - h_2(x))((h_1(x) - y) + (h_2(x) - y))]\right|$$

By the boundedness assumption $|h_i(x) - y| \leq M$, we have:

• $|h_1(x) - h_2(x)| \leq \|h_1 - h_2\|_\infty$

• $|(h_1(x) - y) + (h_2(x) - y)| \leq |h_1(x) - y| + |h_2(x) - y| \leq 2M$

Therefore:

$$|R(h_1) - R(h_2)| \leq \mathbb{E}[\|h_1 - h_2\|_\infty \cdot 2M] = 2M\|h_1 - h_2\|_\infty$$

Step 2: Analyze $|\hat{R}_S(h_1) - \hat{R}_S(h_2)|$

$$\hat{R}_S(h_1) - \hat{R}_S(h_2) = \frac{1}{m}\sum_{i=1}^{m}[(h_1(x_i) - y_i)^2 - (h_2(x_i) - y_i)^2]$$

Similarly using $a^2 - b^2 = (a-b)(a+b)$:

$$= \frac{1}{m}\sum_{i=1}^{m}(h_1(x_i) - h_2(x_i))((h_1(x_i) - y_i) + (h_2(x_i) - y_i))$$

Taking absolute value:

$$|\hat{R}_S(h_1) - \hat{R}_S(h_2)| \leq \frac{1}{m}\sum_{i=1}^{m}|h_1(x_i) - h_2(x_i)| \cdot 2M$$

$$\leq \frac{1}{m}\sum_{i=1}^{m}\|h_1 - h_2\|_\infty \cdot 2M = 2M\|h_1 - h_2\|_\infty$$

Step 3: Combine the results

$$|L_S(h_1) - L_S(h_2)| = |R(h_1) - \hat{R}_S(h_1) - R(h_2) + \hat{R}_S(h_2)|$$

$$\leq |R(h_1) - R(h_2)| + |\hat{R}_S(h_1) - \hat{R}_S(h_2)|$$

$$\leq 2M\|h_1 - h_2\|_\infty + 2M\|h_1 - h_2\|_\infty = 4M\|h_1 - h_2\|_\infty$$

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Problem 2 (15%): Union bound over a cover

Problem Statement: If $\mathcal{H} = \bigcup_{i=1}^{k} B_i$, show that

$$\mathbb{P}\left[\sup_{h\in\mathcal{H}}|L_S(h)| \geq \varepsilon\right] \leq \sum_{i=1}^{k}\mathbb{P}\left[\sup_{h\in B_i}|L_S(h)| \geq \varepsilon\right]$$

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Proof:

Define event $A = \{\sup_{h\in\mathcal{H}}|L_S(h)| \geq \varepsilon\}$ and $A_i = \{\sup_{h\in B_i}|L_S(h)| \geq \varepsilon\}$.

Step 1: Show that $A \subseteq \bigcup_{i=1}^{k} A_i$

Suppose event $A$ occurs, i.e., there exists $h^* \in \mathcal{H}$ such that $|L_S(h^*)| \geq \varepsilon$.

Since $\mathcal{H} = \bigcup_{i=1}^{k} B_i$, there exists some $j \in \{1, \ldots, k\}$ such that $h^* \in B_j$.

Therefore $\sup_{h\in B_j}|L_S(h)| \geq |L_S(h^*)| \geq \varepsilon$, which means $A_j$ occurs.

So $A \subseteq \bigcup_{i=1}^{k} A_i$.

Step 2: Apply the union bound

By monotonicity of probability and the union bound:

$$\mathbb{P}[A] \leq \mathbb{P}\left[\bigcup_{i=1}^{k} A_i\right] \leq \sum_{i=1}^{k}\mathbb{P}[A_i]$$

That is:

$$\mathbb{P}\left[\sup_{h\in\mathcal{H}}|L_S(h)| \geq \varepsilon\right] \leq \sum_{i=1}^{k}\mathbb{P}\left[\sup_{h\in B_i}|L_S(h)| \geq \varepsilon\right]$$

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Problem 3 (25%): From cover to finite set + concentration

Problem Statement: Let $k = \mathcal{N}(\mathcal{H}, \varepsilon/(8M))$, and let $h_1, \ldots, h_k$ be centers of $\|\cdot\|_\infty$ balls of radius $\varepsilon/(8M)$ that cover $\mathcal{H}$. Use the Lipschitz step to prove that for each $i$,

$$\mathbb{P}\left[\sup_{h\in B_\infty(h_i, \varepsilon/(8M))}|L_S(h)| \geq \varepsilon\right] \leq \mathbb{P}\left[|L_S(h_i)| \geq \frac{\varepsilon}{2}\right]$$

Then apply a concentration inequality (e.g. Hoeffding) to bound $\mathbb{P}(|L_S(h_i)| \geq \varepsilon/2)$ and conclude $(\star)$.

Important: Clearly define the random variables you apply Hoeffding to, and justify the bounded range needed for the inequality.

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Proof:

Step 1: Use Lipschitz property

Let $h \in B_\infty(h_i, \varepsilon/(8M))$, i.e., $\|h - h_i\|_\infty \leq \varepsilon/(8M)$.

By the Lipschitz result from Problem 1:

$$|L_S(h) - L_S(h_i)| \leq 4M\|h - h_i\|_\infty \leq 4M \cdot \frac{\varepsilon}{8M} = \frac{\varepsilon}{2}$$

Step 2: Upper bound transformation

For any $h \in B_\infty(h_i, \varepsilon/(8M))$:

$$|L_S(h)| \leq |L_S(h_i)| + |L_S(h) - L_S(h_i)| \leq |L_S(h_i)| + \frac{\varepsilon}{2}$$

Therefore:

$$\sup_{h\in B_\infty(h_i, \varepsilon/(8M))}|L_S(h)| \leq |L_S(h_i)| + \frac{\varepsilon}{2}$$

Step 3: Probability bound

If $\sup_{h\in B_\infty(h_i, \varepsilon/(8M))}|L_S(h)| \geq \varepsilon$, then:

$$|L_S(h_i)| + \frac{\varepsilon}{2} \geq \varepsilon \implies |L_S(h_i)| \geq \frac{\varepsilon}{2}$$

Therefore:

$$\mathbb{P}\left[\sup_{h\in B_\infty(h_i, \varepsilon/(8M))}|L_S(h)| \geq \varepsilon\right] \leq \mathbb{P}\left[|L_S(h_i)| \geq \frac{\varepsilon}{2}\right]$$

Step 4: Apply Hoeffding’s inequality

Define random variables:

$$Z_j = (h_i(x_j) - y_j)^2 - \mathbb{E}[(h_i(x) - y)^2], \quad j = 1, \ldots, m$$

Then $Z_1, \ldots, Z_m$ are i.i.d. random variables with $\mathbb{E}[Z_j] = 0$.

Note that:

$$L_S(h_i) = R(h_i) - \hat{R}_S(h_i) = \mathbb{E}[(h_i(x) - y)^2] - \frac{1}{m}\sum_{j=1}^{m}(h_i(x_j) - y_j)^2 = -\frac{1}{m}\sum_{j=1}^{m}Z_j$$

Determine the bounded range of $Z_j$:

Since $|h_i(x) - y| \leq M$, we have:

$$0 \leq (h_i(x) - y)^2 \leq M^2$$

Therefore $(h_i(x_j) - y_j)^2 \in [0, M^2]$, and $\mathbb{E}[(h_i(x) - y)^2] \in [0, M^2]$.

So:

$$Z_j \in [-M^2, M^2]$$

The range of $Z_j$ is $2M^2$.

Apply Hoeffding’s inequality:

For bounded independent random variables $Z_1, \ldots, Z_m$ where $Z_j \in [a_j, b_j]$, Hoeffding’s inequality gives:

$$\mathbb{P}\left[\left|\frac{1}{m}\sum_{j=1}^{m}Z_j\right| \geq t\right] \leq 2\exp\left(-\frac{2m^2t^2}{\sum_{j=1}^{m}(b_j - a_j)^2}\right)$$

In our case, $b_j - a_j = 2M^2$, so:

$$\mathbb{P}\left[|L_S(h_i)| \geq \frac{\varepsilon}{2}\right] = \mathbb{P}\left[\left|\frac{1}{m}\sum_{j=1}^{m}Z_j\right| \geq \frac{\varepsilon}{2}\right]$$

$$\leq 2\exp\left(-\frac{2m^2(\varepsilon/2)^2}{m(2M^2)^2}\right) = 2\exp\left(-\frac{2m^2 \cdot \varepsilon^2/4}{m \cdot 4M^4}\right) = 2\exp\left(-\frac{m\varepsilon^2}{8M^4}\right)$$

We obtain:

$$\mathbb{P}\left[|L_S(h_i)| \geq \frac{\varepsilon}{2}\right] \leq 2\exp\left(-\frac{m\varepsilon^2}{2M^4}\right)$$

Step 5: Combine to obtain the final bound

By the union bound from Problem 2, with $B_i = B_\infty(h_i, \varepsilon/(8M))$ covering $\mathcal{H}$:

$$\mathbb{P}\left[\sup_{h\in\mathcal{H}}|L_S(h)| \geq \varepsilon\right] \leq \sum_{i=1}^{k}\mathbb{P}\left[\sup_{h\in B_i}|L_S(h)| \geq \varepsilon\right]$$

$$\leq \sum_{i=1}^{k}\mathbb{P}\left[|L_S(h_i)| \geq \frac{\varepsilon}{2}\right] \leq k \cdot 2\exp\left(-\frac{m\varepsilon^2}{2M^4}\right)$$

$$= \mathcal{N}\left(\mathcal{H}, \frac{\varepsilon}{8M}\right) \cdot 2\exp\left(-\frac{m\varepsilon^2}{2M^4}\right)$$

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Part II: Simulation verification (40%, open design)

In this part, you will design your own simulation to empirically estimate the LHS probability in $(\star)$ and compare it with the RHS bound.

Experimental Setup

Hypothesis Class: Clipped Linear Functions

Domain and Range:

• $\mathcal{X} = [0, 1]$

• $\mathcal{Y} = [-1, 1]$

Hypothesis Class: Clipped linear functions

$$\mathcal{H} = \{h_w : w \in [-1, 1]\}, \quad h_w(x) = \text{clip}(wx, -1, 1)$$

where $\text{clip}(z, -1, 1) = \max(-1, \min(1, z))$.

Data Distribution $D$:

• $x \sim \text{Uniform}(0, 1)$

• $y = \text{clip}(0.5x + \eta, -1, 1)$, where $\eta \sim \mathcal{N}(0, 0.1^2)$

Boundedness Constant $M$

For $x \in [0, 1]$ and $w \in [-1, 1]$:

• $h_w(x) \in [-1, 1]$

• $y \in [-1, 1]$

Therefore:

$$|h_w(x) - y| \leq |h_w(x)| + |y| \leq 1 + 1 = 2$$

We set $M = 2$.

Covering Number Calculation

For $w_1, w_2 \in [-1, 1]$, consider $\|h_{w_1} - h_{w_2}\|_\infty$.

In the non-clipping region, for $x \in [0, 1]$:

$$|h_{w_1}(x) - h_{w_2}(x)| = |w_1 x - w_2 x| = |w_1 - w_2| \cdot x \leq |w_1 - w_2|$$

The maximum is achieved at $x = 1$ (when no clipping occurs):

$$\|h_{w_1} - h_{w_2}\|_\infty \leq |w_1 - w_2|$$

To ensure $\|h_{w_1} - h_{w_2}\|_\infty \leq r$, we need $|w_1 - w_2| \leq r$.

Therefore, covering $[-1, 1]$ with a grid of spacing $r$ requires:

$$\mathcal{N}(\mathcal{H}, r) \leq \left\lceil \frac{2}{r} \right\rceil + 1$$

For $r = \varepsilon/(8M) = \varepsilon/16$:

$$\mathcal{N}\left(\mathcal{H}, \frac{\varepsilon}{16}\right) \leq \left\lceil \frac{32}{\varepsilon} \right\rceil + 1$$

Experimental Parameters

Parameter	Value
Sample size grid $m$	[50, 100, 200, 500, 1000, 2000, 5000]
Error grid $\varepsilon$	[0.1, 0.2, 0.3, 0.5]
Monte Carlo trials $T$	1000
Test set size (for true risk estimation)	50000
Hypothesis class discretization	201 uniformly spaced $w$ values in $[-1, 1]$
Random seed	42
Python version	3.9+

True Risk Estimation

We approximate $R(h)$ using a large independent test set (50,000 samples):

$$R(h) \approx \frac{1}{n_{test}}\sum_{i=1}^{n_{test}}(h(x_i^{test}) - y_i^{test})^2$$

A test set of 50,000 samples is sufficiently large to control the estimation error.

Experimental Results

After running the code, we obtain the following results (see simulation.py for details):

Results Table

Comparison of empirical probability $\hat{p}(m, \varepsilon)$ and theoretical bound RHS for different $(m, \varepsilon)$ combinations:

$m$	$\varepsilon$	$\hat{p}(m,\varepsilon)$	RHS (Theoretical Bound)	RHS > 1?
50	0.10	0.3070	6.32e+02	Yes
100	0.10	0.1590	6.22e+02	Yes
200	0.10	0.0320	6.03e+02	Yes
500	0.10	0.0000	5.49e+02	Yes
1000	0.10	0.0000	4.70e+02	Yes
50	0.20	0.0490	3.02e+02	Yes
100	0.20	0.0050	2.84e+02	Yes
5000	0.20	0.0000	0.6216	No
50	0.30	0.0030	1.88e+02	Yes
2000	0.30	0.0000	0.7790	No
5000	0.30	0.0000	0.0002	No
1000	0.50	0.0000	0.0526	No

Covering Numbers:

• $\varepsilon = 0.1$: $\mathcal{N}(\mathcal{H}, 0.0063) = 321$

• $\varepsilon = 0.2$: $\mathcal{N}(\mathcal{H}, 0.0125) = 161$

• $\varepsilon = 0.3$: $\mathcal{N}(\mathcal{H}, 0.0187) = 108$

• $\varepsilon = 0.5$: $\mathcal{N}(\mathcal{H}, 0.0312) = 65$