Let $$ \mathcal{F}=\{F:\mathbb{R}\longrightarrow\mathbb{R}: \text{$F$ is the CDF of some probability measure on $\mathbb{R}$}\}. $$ Consider the loss function, $L:\mathcal F\times\mathcal F\to\mathbb R$, defined by $$ L(F,G) = \sup_{x\in\mathbb R}|F(x) - G(x)|. $$ (This is just the metric induced by the $L^\infty$-norm.)

Let $\Theta$ be a subset of $\mathcal{F}$. For $\theta\in\Theta$ and a random sample $X_1,\ldots,X_n$ drawn from $\theta$, let $\hat{\theta}(X_1,\ldots,X_n)$ be the associated empirical distribution function: $$ \hat\theta(X_1,\ldots,X_n)(x) = \frac 1 n\sum_{i=1}^n I(X_i\leq x) $$ Are there "large" sets $\Theta$ for which one can one explicitly compute, or bound, the risk $R$ associated to the loss function $L$ in terms of $n$ and $\Theta$, where $R$ is defined by $$ R(\theta,\hat{\theta}) = E_\theta[L(\theta,\hat\theta(X_1,\ldots,X_n))]? $$


1 Answer 1


It is easy to get an upper bound for $\Theta = \mathcal{F}$ using the Dvoretzky-Kiefer-Wolfowitz (DKW) inequality and $EL = \int_0^\infty\Pr(L>u)du$, see https://en.wikipedia.org/wiki/Dvoretzky–Kiefer–Wolfowitz_inequality#The_DKW_inequality

The DKW inequality gives us $$R(\theta,\hat\theta) = \int_0^\infty \Pr\left(L(\theta,\hat\theta)>u\right)du \leq 2\int_0^\infty e^{-2nu^2}du.$$

I'm not sure if tighter bounds exist for $\Theta\subset\mathcal{F}$.

  • 2
    $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Nov 21, 2023 at 21:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.