I need to proof a corollary of Hoeffding's inequality, and since I'm not used to doing proofs I really don't know where to begin.

Hoeffding's inequality: Let $X_1,...,X_n$ be independent real-valued random variables, such that for each $i \in \{1,...,n\}$ there exists $a_i\leq b_i$, such that $X_i \in [a_i, b_i]$. Then for every $\varepsilon >0$: \begin{equation} P(\sum_{i=1}^nX_i - E[\sum_{i=1}^nX_i] \geq \varepsilon) \leq e^{-2 \varepsilon^2 / \sum_{i=1}^n (b_i-a_i)^2} \end{equation}

If we assume that $X_i$'s are identically distributed and belong to the $[0,1]$ interval we obtain the following corollary.

Corollary: Let $X_1,...,X_n$ be independent random variables, such that $X_i\in [0,1]$ and $E[X_i]=\mu$ for all $i$, then for every $\varepsilon >0$:

\begin{equation} P(\frac{1}{n}\sum_{i=1}^nX_i - \mu \geq \varepsilon) \leq e^{-2n\varepsilon^2} \end{equation}

So I need to prove that the corollary follows from Hoeffding's inequality. Could anyone please share some reference of proof or just prove it here?

  • 2
    $\begingroup$ There are some typos in the formula... $\endgroup$
    – utobi
    Sep 17, 2022 at 12:52
  • 1
    $\begingroup$ I suspect that when the typos are cleaned up there will be nothing to prove, because it should be just a matter of plugging suitable values of $(a_i,b_i)$ into the inequality. $\endgroup$
    – whuber
    Sep 17, 2022 at 12:56
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    $\begingroup$ Re the edit: apply the inequality to the case $n\varepsilon$ where $a_i=0$ and $b_i=1$ for all $i.$ $\endgroup$
    – whuber
    Sep 17, 2022 at 13:12
  • $\begingroup$ Thanks, whuber - that seems to do it! $\endgroup$
    – random1234
    Sep 17, 2022 at 13:22

1 Answer 1


With $$ \begin{align} & \mathbb{E}\left(\sum_{i=1}^nX_i\right) = n \cdot \mu, \\ & a_i=0, \\ & b_i=1, \\ & \tilde{\varepsilon} \mathrel{:=} n\cdot\varepsilon, \end{align} $$ we have $$ \mathbb{P}\left(\sum_{i=1}^nX_i-\mathbb{E}\left(\sum_{i=1}^nX_i\right)\geq \tilde{\varepsilon}\right) \leq \exp\left(-\frac{2n^2\varepsilon^2}{\sum_{i=1}^n1^2}\right) = \exp\left(-2n\varepsilon^2\right) $$ and $$ \mathbb{P}\left(\sum_{i=1}^nX_i-\mathbb{E}\left(\sum_{i=1}^nX_i\right)\geq \tilde{\varepsilon}\right) = \mathbb{P}\left(\frac{1}{n} \cdot \sum_{i=1}^nX_i - \mu \geq \varepsilon\right). $$


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