Say we have a black-box access to a distribution $\mathcal{D}$ with finite support $\{1,2,...,n\}$ with probability mass function $i \mapsto p_i$. How many samples of $\mathcal{D}$ are needed to learn the mode of $\mathcal{D}$, i.e. the $j$ such that $p_j=\max\{p_i\}$, (with error probability $\delta$)? Is there any known lower bound? Any idea, comment, or reference will be very helpful.

  • I know that $\Theta(\frac{1}{d_H^2} \log\frac{1}{\delta})$ samples are required regarding Hellinger distance of $\mathcal{D}=(p,1-p)$ and $\mathcal{D}'=(1-p,p)$ in the case of $n=2$.
  • I proved an upper bound $O(\frac{1}{\varepsilon^2} \log \frac{n}{\delta})$ with standard Hoeffding bound, where $\varepsilon$ is the difference between the largest and the second largest among $p_i$'s.

Update: I proved a lower bound $\Omega(\frac{1}{\gamma}\log\frac{1}{\delta})$ by reducing the problem to some $n=2$ case, where $\gamma=\min_i\left(1-2\sqrt{\frac{pp_i}{(p+p_i)^2}}\right)$ with $p$ being the mode. But, I still want to reduce the gap between upper and lower bounds.

A followup question: Say we have a black-box access to a distribution $\mathcal{D}$ with finite support $A\cup B:=\{a_1,...,a_n\}\cup\{b_1,...,b_m\}$ with probability mass function $a_i \mapsto p_i$ and $b_i \mapsto q_i$. How many samples of $\mathcal{D}$ are required to make sure the most frequent sample is in $A$ (with error probability $\delta$)? For sanity, it is guaranteed that $A$ contains the mode of $\mathcal{D}$.

  • Following Clement's answer, I could prove an upper bound of $O(\frac{1}{\varepsilon^2}\log\frac{1}{\delta})$, where $\varepsilon=\max\{p_i\}-\max\{q_i\}$. However, a nice lower bound eluded me, mainly because reducing to a Bernoulli case does not seem to work in this case.

1 Answer 1


From the DKW inequality, it follows one can learn an arbitrary distribution over $\mathbb{R}$ (and in particular over $\{1,2,\dots,n\}$ to Kolmogorov distance* $\varepsilon$ with probability at least $1-\delta$ using $$ O\!\left(\frac{1}{\varepsilon^2}\log\frac{1}{\delta}\right)$$ samples; which is optimal in view of the complexity of learning even a single Bernoulli (biased coin): embed the biased coin (with bias either $1/2+\varepsilon$ or $1/2-\varepsilon$) in the middle of the domain, where all other elements have mass 0: figuring out which element is the mode corresponds to figuring out the bias.

This implies the same upper (and lower) bounds for your question, where $\varepsilon$ is defined as the gap (assuming you have a priori knowledge on that gap; this is not quite apparent from your post. If not, doing something adaptive and slightly more involved might be required).

$*$ Kolmgorov distance is $\ell_\infty$ distance between CDFs. This guarantee is stronger (and implies) learning the pdf in $\ell_\infty$. For more, see. e.g., this short note of mine, specifically Section 4 (not new results, just exposition of "folklore" facts)

  • $\begingroup$ isn't $\Omega\left(\frac{1}{\epsilon^2}\log\frac{1}{\delta}\right)$ a lower bound for any $\mathcal{D}$? $\endgroup$ Commented May 20, 2022 at 5:27
  • $\begingroup$ @mathworker21 yes, you can start with any distribution and embed the corresponding biased coin problem (with a suitable $\varepsilon$) in there. Is that what you mean? $\endgroup$
    – Clement C.
    Commented May 20, 2022 at 5:37
  • $\begingroup$ By "what you mean", do you mean "what you had in mind for a proof"? I meant what I asked, namely if $\Omega(\epsilon^{-2}\log\delta^{-1})$ is always a lower bound; it seems you agree that is always a lower bound. $\endgroup$ Commented May 20, 2022 at 6:02
  • $\begingroup$ @ClementC. Wow, thank you for your answer! And also writing up folklores. It really helps non-specialists like me. Actually, this question was already inspired by your note on distinguishing discrete distributions:) $\endgroup$
    – actcon
    Commented May 20, 2022 at 12:47
  • 2
    $\begingroup$ Erm, why the downvote? Is there something wrong with this answer? $\endgroup$
    – Clement C.
    Commented May 22, 2022 at 1:11

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.