# 3.3. Uniform Law Under Random Entropy

$\newcommand{\argmin}{\mathop{\mathrm{argmin}}\limits}$ $\newcommand{\argmax}{\mathop{\mathrm{argmax}}\limits}$

Using all the lemmas from 3.2 to 3.5, we prove another sufficiency of the uniform law: the vanishing random entropy condition. This condition is fairly weaker than the finite bracketing entropy condition in two senses. First, instead of the (sup-normed) envelope condition, it only requires the envelope to be integrable. In $L^p(Q)$ where $Q$ is a finite measure, this is clearly implied by the envelop condition. In addition, as we already saw before, the vanishing random entropy is implied by the finite bracketing entropy condition.

In fact, van de Geer (2000) named this condition simply “the random entropy condition”, but I did a small modification to the naming. The author did not clearly mention why the condition is called “random”. I think this is because the entropy is computed with respect to the empirical measure $P_n,$ which is constructed by a random sample, hence becomes random itself.

## Lemma 3.6 ($L^2$ condition)

First let’s prove the result in the stronger condition ($L^2$ case).

(a) $\sup_{g\in\mathcal G} \|g\|_\infty < R$ for some $0\le R<\infty.$

(b) $\frac1n H_{2,P_n}(\delta,\mathcal G) \overset P \to 0$ for all $\delta>0.$

If (a) and (b) holds, then $\mathcal G$ satisfies the ULLN.

First, note that $\left|\int g~d(P_n-P)\right|$ is a non-negative martingale. Suppose that $$\sup_{g\in\mathcal G}\left|\int g~d(P_n-P)\right| \overset P \to 0,$$ then by the martingale convergence theorem, it follows that $$\sup_{g\in\mathcal G}\left|\int g~d(P_n-P)\right| \to 0 \text{ a.s.} \tag{ULLN}$$ Therefore it is enough to show convergence to zero in probability. Note in addition that for given $\delta>0,$ for $n \ge 8R^2/\delta^2,$ $$\mathbb P\left( \sup_{g\in\mathcal G} \bigg| \int g~d(P_n-P) \bigg| > \delta \right) \le 4 \mathbb P\left( \sup_{g\in\mathcal G} \bigg| \frac1n\sum_{i=1}^nW_ig(X_i) \bigg| > \frac\delta4 \right).$$ Thus it suffices to show that $$\mathbb P\left( \sup_{g\in\mathcal G} \bigg| \frac1n\sum_{i=1}^nW_ig(X_i) \bigg| > \delta \right) \overset P\to 0,~ \forall \delta>0.$$ Although we cannot apply the maximal inequality directly to the left-hand side, we can apply it to the conditional probability $$\mathbb P\left( \sup_{g\in\mathcal G} \bigg| \frac1n\sum_{i=1}^nW_ig(X_i) \bigg| > \frac\delta4 ~\Bigg\vert~ A_n \right)$$ where $$A_n:=\left\{ \mathbf X:~ R\vee R\cdot H^{1/2}_{2,P_n}(\delta/32,\mathcal G) \le \frac{\sqrt n \delta}{8C} \right\}$$ for some constant $C$. This is because each condition from (a) to (c) is satisfied as follows: \begin{aligned} \text{(a)}~ & W_i\gamma_i \in \{-|\gamma_i|, |\gamma_i|\},~ \forall i.\\ & \text{Hoeffding's inequality implies that}\\ &~~~~\mathbb P\left(| W^\intercal \gamma | \ge a\right) \le 2\exp\left( -\frac{a^2}{2\|\gamma\|^2} \right),~ \forall a>0. \\ &\text{Hence, let } C_1=2, C_2=\sqrt2. \\ \text{(b)}~ & \sup_{g\in\mathcal G} \|g\|_{2,P} \le \sup_{g\in\mathcal G} \|g\|_{\infty} < R. \\ \text{(c)}~ & \text{Let } \varepsilon=\frac\delta8. \text{ Then on } A_n, \\ &~~~~ C\left(\int_{\delta/32}^R H^{1/2}_{2,P_n}(u,\mathcal G)du \vee R\right) \\ &~~~~\le C\left( R\cdot H^{1/2}_{2,P_n}(\delta/32,\mathcal G) \vee R \right) \le \frac{\sqrt n\delta}8. \end{aligned} By the maximal inequality, \begin{aligned} &\mathbb P\left( \sup_{g\in\mathcal G} \bigg| \frac1n \sum_{i=1}^n W_ig(X_i) \bigg| > \frac\delta4 \right) \\ &\le \mathbb P\left( \sup_{g\in\mathcal G} \bigg| \frac1n \sum_{i=1}^n W_ig(X_i) \bigg| > \frac\delta4,~ A_n \right) + \mathbb P(A_n^c) \\ &= \mathbb E_\mathbf{X} \mathbb P\left( \sup_{g\in\mathcal G} \bigg| \frac1n \sum_{i=1}^n W_ig(X_i) \bigg| > \frac\delta4, ~A_n ~\Bigg|~\mathbf X \right) + \mathbb P(A_n^c) \\ &\le C\exp\left( -\frac{n\delta^2}{8^2C^2R^2} \right) + \mathbb P\left( \frac1nH_{2,P_n}(\delta/32,\mathcal G) > \frac\delta{8CR} \right) \\ &\to 0. \end{aligned}

## Theorem 3.7 (main theorem; $L^1$ condition)

Now we prove the main theorem using this result.

(a) $\sup_{g\in\mathcal G} \|g\|_1 < \infty.$

(b) $\frac1n H_{1,P_n}(\delta,\mathcal G) \overset P \to 0$ for all $\delta>0.$

If (a) and (b) holds, then $\mathcal G$ satisfies the ULLN.

To utilize the lemma 3.6, we define a truncated version of $\mathcal G.$ For given $R>0,$ let $$\mathcal G_R = \left\{ g\mathbf 1_{(G \le R)}:~ g\in\mathcal G \right\}.$$ Then the $L^2$ norm is bounded from above by the $L^1$ norm $$\int_{G\le R} (g_1-g_2)^2 dP_n \le 2R\int |g_1-g_2|dP_n$$ so we have the vanishing $L^2$ entropy. $$\frac1n H_{2,P_n}(\delta,\mathcal G_R) \overset P\to 0,~ \forall \delta>0.$$ By the lemma 3.6, $\mathcal G_R$ satisfies the ULLN for all $R>0.$

Now what is left is proving that the ULLN still holds even if $R\to\infty.$ \begin{aligned} &\sup_{g\in\mathcal G} \bigg| \int g~d(P_n-P) \bigg| \\ &\le \color{darkblue}{ \sup_{g\in\mathcal G} \bigg| \int_{G\le R} g~d(P_n-P) \bigg| } + \color{darkgreen}{ \int_{G>R} G~dP_n } + \color{darkred}{ \int_{G>R} G~dP }. \\ \end{aligned} For given $\varepsilon>0,$ take $R>0$ large enough so that $$\color{darkred}{ \int_{G>R} G~dP } \le \varepsilon.$$ Then take $n\in\mathbb N$ large enough so that $$\color{darkgreen}{ \int_{G>R} G~dP_n } \le \varepsilon \text{ a.s.} ~\text{ and } \color{darkblue}{ \sup_{g\in\mathcal G} \bigg| \int_{G\le R} g~d(P_n-P) \bigg| } \le \varepsilon \text{ a.s.}$$ Then the desired result follows.

References

• van de Geer. 2000. Empirical Processes in M-estimation. Cambridge University Press.
• Theory of Statistics II (Fall, 2020) @ Seoul National University, Republic of Korea (instructor: Prof. Jaeyong Lee).