Consider a collection of $n$ i.i.d. Bernoulli random variables $\{ X_i \}_{i=1}^{n}$ with $\mathbb{E}[X_i] = \mu$.

Then, if $\hat{\mu}$ is the mean of the $n$ random variables, i.e. if, $$\hat{\mu} = \frac{1}{n} \sum_{i=1}^{n} X_i,$$ then, by Hoeffding's inequality we have the following concentration result,

$$\mathbb{P} (\hat{\mu} - \mu \geq \Delta) \leq \exp(-2n\Delta^{2}).$$

If instead of a concentration result on the mean $\hat{\mu}$, we were interested in a concentration result on its inverse, $\frac{1}{\hat{\mu}}$, what approach can we take?

That is, how can we obtain a Hoeffding like exponential decay bound on the probability,

$$\mathbb{P} \left( \frac{1}{\hat{\mu}} \geq a \right),$$

for some suitably chosen form of constant $a$.

Since $\frac{1}{\hat{\mu}}$ does not involve a mean over the direct samples of any random variable, I am not sure how to proceed.


1 Answer 1


It is possible to come up with something like this. Here is an example.

The first thing to note is that,


Where $S_n = X_1 + \dots + X_n$. Now under the assumption that the $X_i \sim Bern(p_i)$ we have the following Multiplicative Chernoff bound,

$$\Pr[S_n\geq(1+\delta)\mu]\leq 2\exp(-\mu\delta^2/3)$$

where $\mu = \mathbb{E} [X]$ and $\delta \in (0,1)$. Then notice that $f(x)=1/x$ is one-to-one and is decreasing on $\mathbb{R}_+$. Note that $S_n \in [0,\infty)$. Let us assume that $S_n \neq 0$. Then we have, $$\Pr[S_n\geq(1+\delta)\mu]=\Pr[S_n^{-1}\leq\frac{1}{(1+\delta)\mu}]=\Pr[nS_n^{-1}\leq\frac{n}{(1+\delta)\mu}]=\Pr[\hat{\mu}^{-1}\leq\frac{n}{(1+\delta)\mu}]$$




$$Pr[\frac{1}{\hat{\mu}}\geq\frac{n}{(1+\delta)\mu}] \geq 1-2\exp(-\mu\delta^2/3)$$

Which I guess in some sense is more of an anti-concentration result. I imagine using this approach you can come up with other bounds depending on your desired outcome.

  • $\begingroup$ Thank you very much for your answer, I had seen the multiplicative chernoff bound in a HW problem once but its applicability here slipped me. I'll keep the answer open for another week before accepting the answer in case anyone else has a different idea. $\endgroup$
    – ijuneja
    Commented Jun 16, 2021 at 7:06

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