I am currently reading the paper Functional Classification in Hilbert Spaces by Biau, Bunea and Wegkamp, and there is one step in the proof of Theorem 1 that is not clear to me. I give below a simplified version of the steps of the proof that I do not understand, but the original calculations can be found in Section V-B, page 7 of the original paper.

To briefly introduce the notations, we are dealing with a binary classification problem. The dataset is made of $n$ i.i.d. pairs $(X_i,Y_i) \in \mathcal X \times \left\{0,1\right\}$ and we denote by $\hat\phi_n$ the classifier which belongs to a certain hypothesis class and is a function of the data. Lastly, $L$ and $\hat L$ are respectively the true and empirical losses of the classifier $\hat\phi_n$, i.e. $$L:= \mathbb P\left(\hat\phi_n(X)\ne Y\ |\ (X_i,Y_i),1\le i\le n\right) \quad \text{ and }\quad \hat L :=\frac{1}{n}\sum_{i=1}^n\mathbf 1\left\{\hat\phi_n(X_i)\ne Y_i\right\} $$ The inequality I'm having trouble with is the following : $$\begin{align} \mathbb P\left(L-\hat L\ge \varepsilon\right) &= \mathbb E \mathbb P\left(L-\hat L\ge \varepsilon\ |\ (X_i,Y_i),1\le i\le n\right) \\ &\le \exp\left(-2n\varepsilon^2\right) \end{align}$$ The first line is clearly true by the law of total expectation, and I understand that the second line is a direct application of Hoeffding's inequality since, conditional on the data, $n\hat L$ is a sum of i.i.d Bernoulli variables of parameter $L$. My issue is that I don't know how to formalize Hoeffding's inequality in this case where the probability is conditional on a sigma algebra, since it then becomes a random variable. Should I maybe show that Hoeffding's inequality holds for the conditional probability conditional on any event in $\sigma\left((X_i,Y_i),1\le i \le n\right) $ ?

In summary, my question is the following : How to formally derive Hoeffding's inequality for conditional probability with respect to a sigma algebra ?

Many thanks in advance for the help.

Update : First off, I noticed I got confused with the notations and $\hat\phi_n$ is, in fact, dependent on the data (in the paper at least), I fixed the mistake.

Regarding my question, I attempted to prove Hoeffding's inequality for conditional probability by following the proof of the original inequality and adapting it. So let $X_i$ be independent r.v.'s respectively bounded in $[a_i,b_i]$, $S_n = \sum_{i=1}^n X_i$ and $\mathcal F$ be a sigma-algebra :
By conditional Markov inequality, we have $$\begin{align} \mathbb P\left(S_n-\mathbb E S_n \ge \varepsilon \ \vert\ \mathcal F\right)&\le e^{-s\varepsilon}\mathbb E \left[e^{s(S_n-\mathbb E S_n)} \ \vert\ \mathcal F\right]\\ &=e^{-s\varepsilon}\mathbb E \left[\exp\left(s\left(\sum_{i=1}^n X_i - \mathbb E X_i\right)\right) \ \vert\ \mathcal F\right] \\ &= e^{-s\varepsilon} \prod_{i=1}^n\mathbb E \left[e^{s\left( X_i - \mathbb E X_i\right)} \ \vert\ \mathcal F\right] \text{ (by independence)}\\ \end{align}$$ To conclude, I then have to prove/apply a conditional version of Hoeffding's lemma, but sadly that doesn't seem to work since $\mathbb E\left[X_i - \mathbb E[X_i]\ \vert \ \mathcal F\right]\ne 0$ in general. I however believe it's still possible to upper bound this conditional expectation and get a similar, but weaker version of the lemma for conditional probability, but that would yield a different result (I haven't done the calculation yet).

Am I on the right track ?

  • $\begingroup$ Am I correct that you are using $L$ to refer to $L(k,d)$, as defined in the beginning of section V-B? I ask because their definition of $L(k,d)$ is also conditional on the data $(X_i,Y_i)_{i=1}^n$. They say $$ L(k,d)=\Bbb P\{\phi_{\ell,k,d}({\bf X}^{(d)}\neq Y)\mid ({\bf X}_i,Y_i),1\le i\le n\} $$ I think this is an important detail you have left out. $\endgroup$ Jan 24, 2022 at 21:17
  • $\begingroup$ @MikeEarnest yes you are correct, I initially left it out thinking it didn't make much difference but I added it in now ! $\endgroup$ Jan 25, 2022 at 9:06

1 Answer 1


After looking at the problem again, I figured out what was wrong in my "conditional Hoeffding inequality" proof attempt : In the paper's setting, $L$ is not equal to $\mathbb E[\hat L]$ but rather $L = \mathbb E[\hat L\color{red}{\ \vert \mathcal F}] $ (by definition of conditional probability).
Therefore, the "true" conditional Hoeffding inequality I want to prove is in fact (with the same notations) : $$\mathbb P\left(S_n-\mathbb E [S_n\color{red}{\ \vert \mathcal F}] \ge \varepsilon \ \vert \mathcal F\right)\le \exp\left(\frac{-2\varepsilon}{\sum_{i=1}^n(b_i-a_i)^2}\right)$$ If I proceed similarly as in my initial attempt, I have $$\begin{align}\mathbb P\left(S_n-\mathbb E [S_n\color{red}{\ \vert \mathcal F}] \ge \varepsilon \ \vert \mathcal F\right)&\le e^{-s\varepsilon} \prod_{i=1}^n\mathbb E \left[e^{s\left( X_i - \mathbb E [X_i\color{red}{\ \vert \mathcal F}]\right)}\ \vert \mathcal F\right] \end{align} \tag1$$ And all I need to conclude is, once again, a conditional version of Hoeffding's lemma. But thankfully $\mathbb E\left[X_i - \mathbb E[X_i\ \vert \mathcal F]\ \vert \ \mathcal F\right]=0$ and by monotonicity of $\mathbb E[\cdot\ \vert \mathcal F]$ and convexity of $\exp$, one can follow the steps of the original proof to conclude that $$\mathbb E \left[e^{s\left( X_i - \mathbb E [X_i\color{red}{\ \vert \mathcal F}]\right)}\ \vert \mathcal F\right] \le \exp\left(\frac{s^2(b_i-a_i)^2}{8}\right)\ \ \forall\ 1\le i\le n $$

Finally, as in the original proof, plugging this inequality in $(1)$ and letting $s =\frac{4\varepsilon}{\sum_{i=1}^n(b_i-a_i)^2}$ gives the desired result.

  • $\begingroup$ Don’t you need the $X_i$ to be conditionally independent given $\mathcal{F}$ instead of being independent? $\endgroup$ Aug 17, 2022 at 12:56
  • 1
    $\begingroup$ Yes, you are right, simple independence is not sufficient, and conditional independence is required for this proof to work. In the paper in question, it seems that the authors don't explicitly mention it either, but I guess that is what should be implicitly understood whenever they say "independent". $\endgroup$ Aug 18, 2022 at 9:19

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