2

I am reading a paper about lower bounds for bandit problems (https://arxiv.org/abs/1302.1611). In Theorem 5, they prove a lower bound with an example problem with two arms. In the proof, I see the following step and I wonder where it comes from.

$\sum_{t=1}^n \exp \{ -t \Delta^2\} \geq \frac{1}{\Delta^2}$

I've tried to derive it from

  • a Taylor expansion,
  • Jensen's inequality,
  • summing to infinity,

but I don't see it.

Thanks!

Tchaikovski
  • 329
  • 1
  • 7
  • Since $n=\Delta=1$ is a counterexample, you'll have to say how the context constrains $n,,\Delta$. – J.G. Oct 21 '20 at 16:53
  • 1
    From a typo, I guess. The opposite inequality is true. –  Oct 21 '20 at 16:54
  • @ProfessorVector Oh, dear! It comes at the end of a chain of $\ge$s at the bottom of page 8, so a $\le$ is unhelpful. – J.G. Oct 21 '20 at 16:56
  • 2
    I've alerted the paper's authors of this discrepancy, though someone else probably told them 7 years ago. It's been cited 77 times, but I don't know how many such citations already brought it up. – J.G. Oct 21 '20 at 17:05
  • Note that @VianneyPerchet is one of the paper's authors. – J.G. Oct 22 '20 at 12:25

3 Answers3

5

It is the other way aroud: \begin{align*} \sum\limits_{t = 1}^n {\exp ( - t\Delta ^2 )} & \le \sum\limits_{t = 1}^n {\int_{t - 1}^t {\exp ( - s\Delta ^2 )ds} } \\ & = \int_0^n {\exp ( - s\Delta ^2 )ds} \le \int_0^{ + \infty } {\exp ( - s\Delta ^2 )ds} = \frac{1}{{\Delta ^2 }}. \end{align*}

Gary
  • 31,845
3

This inequality is indeed obviously incorrect... there are several typos in the statement (and the proof) of Theorem 5. First thing first, it can only be true for $n \geq 1/\Delta^2$ (for smaller $n$, the regret is upper-bounded by $n\Delta$ which is itself smaller than $1/\Delta$). Also, the sum should be from $0$ up to $t-1$ (instead from $1$ up to $t$ as we wrote).

With standard computations, you then get that regret is bigger than $\frac{1-e^{-1}}{4\Delta}$ and even bigger than $\frac{1}{4\Delta}$ asymptotically with $n$ (as it goes to infinity).

Vianney Perchet
  • 31
  • Thanks a lot!

    I still don't understand how to obtain $(1-e^{-1}) / (4 \Delta)$, could you maybe spell that out for me?

     – Tchaikovski Oct 22 '20 at 09:39
1

$$\sum\limits_{t = 1}^n {\exp ( - t\Delta ^2 )}<\sum\limits_{t = 1}^\infty {\exp ( - t\Delta ^2 )}=\sum\limits_{t = 1}^\infty {[\exp ( - \Delta ^2 )]^t}=\frac{\exp(-\Delta^2)}{1-\exp(-\Delta^2)}=\frac1{\exp\Delta^2-1}\\ <\frac1{\Delta^2}.$$

Gary
  • 31,845
John Bentin
  • 18,454