I am studying Larry Wasserman's lecture notes on Statistics which uses Casella and Berger as its primary text. I am working through his lecture notes set 2 and got stuck in the derivation of lemma used in Hoeffding's inequality (pp.2-3). I am reproducing the proof in the notes below and after the proof I will point out where I am stuck.


Suppose that $\mathbb{E}(X) = 0$ and that $ a \le X \le b$. Then $\mathbb{E}(e^{tX}) \le e^{t^2 (b-a)^2/8}$.


Since $a \le X \le b$, we can write $X$ as a convex combination of $a$ and $b$, namely $X = \alpha b + (1 - \alpha) a$ where $\alpha = \frac{X-a}{b-a}$. By convexity of the function $y \to e^{ty}$ we have

$e^{tX} \le \alpha e^{tb} + (1 - \alpha) e^{ta} = \frac{X-a}{b-a} e^{tb} + \frac{b-X}{b-a} e^{ta}$

Take expectations of both sides and use the fact $\mathbb{E}(X) = 0$ to get

$\mathbb{E}(e^{tX}) \le \frac{-a}{b-a} e^{tb} + \frac{b}{b-a} e^{ta} = e^{g(u)}$

where $u = t(b-a)$, $g(u) = -\gamma u + \log(1-\gamma + \gamma e^{u})$ and $\gamma = -a /(b-a)$. Note that $g(0) = g^{'}(0) = 0$. Also $g^{''}(u) \le 1/4$ for all $u > 0 $.

By Taylor's theorem, there is a $\varepsilon \in (0, u)$ such that $g(u) = g(0) + u g^{'}(0) + \frac{u^2}{2} g^{''}(\varepsilon) = \frac{u^2}{2} g^{''}(\varepsilon) \le \frac{u^2}{8} = \frac{t^2(b-a)^2}{8}$

Hence $\mathbb{E}(e^{tX}) \le e^{g(u)} \le e^{\frac{t^2(b-a)^2}{8}}$.

I could follow the proof till

$\mathbb{E}(e^{tX}) \le \frac{-a}{b-a} e^{tb} + \frac{b}{b-a} e^{ta} = e^{g(u)}$ but I am unable to figure out how to derive $u, g(u), \gamma$.

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    $\begingroup$ It is interesting that the maximum value of $\text{var}(X)$ is $\sigma_{\max}^2 = (b-a)^2/4$ and thus the result is effectively $$E[e^{tX}] \leq e^{\sigma_{\max}^2t^2/2}$$ which looks far too familiar to be arising out of sheer coincidence. I suspect there may be another, possibly easier, way to derive the result via a probabilistic argument. $\endgroup$ – Dilip Sarwate Jan 14 '12 at 1:42
  • $\begingroup$ @DilipSarwate My understanding is that the maximum variance occurs for an uniform random variable $ X \sim \mathcal{U}(a,b)$. The variance of $X$ is $\mathsf{Var}(X) = \frac{(b-a)^2}{12}$. Can you please explain how you got $\frac{(b-a)^2}{4}$? $\endgroup$ – Anand Jan 14 '12 at 3:59
  • $\begingroup$ By concentrating the mass on the endpoints... $\endgroup$ – Elvis Jan 14 '12 at 7:39
  • $\begingroup$ @DilipSarwate I added a few comments in the proof, that may clarify a liitle bit why the worst case is the max variance. $\endgroup$ – Elvis Jan 14 '12 at 10:43
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    $\begingroup$ @DilipSarwate - See lemma 1 and exercise 1 here: terrytao.wordpress.com/2010/01/03/… . It seems there is a simpler derivation relying on Jensen's inequality and taylor's expansion. Yet the details of this are unclear to me. Perhaps someone can make sense of it. (derivation of (9) to (10) and exercise 1) $\endgroup$ – Leo Aug 13 '13 at 20:21

I’m not sure I understood your question correctly. I’ll try to answer: try to write $$-\frac{a}{b-a} e^{tb} + \frac{b}{b-a} e^{ta}$$ as a function of $u = t(b-a)$ : this is natural as you want a bound in $e^{u^2 \over 8}$.

Helped by the experience, you will know that it is better to chose to write it in the form $e^{g(u)}$. Then $$e^{g(u)} = -\frac{a}{b-a} e^{tb} + \frac{b}{b-a} e^{ta}$$ leads to $$\begin{align*} g(u) &= \log\left( -\frac{a}{b-a} e^{tb} + \frac{b}{b-a} e^{ta} \right)\\ &= \log\left( e^{ta} \left( -\frac{a}{b-a} e^{t(b-a)} + \frac{b}{b-a} \right)\right)\\ &= ta + \log\left( \gamma e^u + (1-\gamma) \right)\\ &= -\gamma u + \log\left( \gamma e^u + (1-\gamma) \right),\\ \end{align*}$$ with $\gamma = - {a \over b-a}$.

Is that the kind of thing you were asking for?

Edit: a few comments on the proof

  1. The first trick deserves to be looked at carefully: if $\phi$ is a convex function, and $a\le X\le b$ is a centered random variable, then $$\mathbb{E}(\phi(X)) \le -{a\over b-a} \phi(b) + {b \over b-a} \phi(a) = \mathbb{E}(\phi(X_0)),$$ where $X_0$ is the discrete variable defined by $$\begin{align*} \mathbb P(X_0=a) &= {b \over b-a}\\ \mathbb P(X_0=b) &= -{a \over b-a}.\end{align*}$$ As a consequence, you get that $X_0$ is the centered variable with support in $[a,b]$ which has the highest variance: $$\mathsf{Var} (X) = \mathbb{E}(X^2) \le \mathbb{E}(X_0^2) = {ba^2 - ab^2 \over b-a } = - ab.$$ Note that if we fix a support width $(b-a)$, this is less than $(b-a)^2\over 4$ as Dilip says in the comments, this is because $(b-a)^2 + 4ab \ge 0$; the bound is attained for $a=-b$.
  2. Now turn to our problem. Why is it possible to get a bound depending only on $u = t(b-a)$? Intuitively, it is just a matter of rescaling of $X$: if you have a bound $\mathbb{E}\left( e^{tX} \right) \le s(t)$ for the case $b-a=1$, then the general bound can be obtained by taking $s( t(b-a) )$. Now think to the set of centered variables with support of width 1: there isn’t so much freedom, so a bound like $s(t)$ should exist.

    An other approach is to say simply that by the above lemma on $\mathbb{E}(\phi(X))$, then more generally $\mathbb{E}(\phi(tX)) \le \mathbb{E}(\phi(tX_0))$, which depends only on $u$ and $\gamma$: if you fix $u = u_0 = t_0 (b_0 - a_0)$ and $\gamma = \gamma_0 = - {a_0 \over b_0-a_0}$, and let $t, a, b$ vary, there is only one degree of freedom, and $t = {t_0 \over \alpha}$, $a = \alpha a_0$, $b = \alpha a_0$. We get $$-{a\over b-a} \phi(tb) + {b \over b-a} \phi(ta) = -{a_0\over b_0-a_0} \phi(tb_0) + {b_0 \over b_0-a_0} \phi(a_0).$$ You just have to find a bound involving only $u$.

  3. Now we are convinced that it can be done, it must be much easier! You don’t necessarily think to $g$ to begin with. The point is that you must write everything as a function of $u$ and $\gamma$.

    First note that $\gamma = -{a\over b-a}$, $1 -\gamma = {b\over b-a}$, $at = -\gamma u$ and $bt = (1-\gamma)u$. Then $$\begin{align*} \mathbb{E}(\phi(tX)) \le &-{a\over b-a} \phi(tb) + {b \over b-a} \phi(ta) \\ = & \gamma \phi((1-\gamma)u) + (1-\gamma) \phi(-\gamma u) \end{align*}$$

    Now we are in the particular case $\phi=\exp$... I think you can finish.

I hope I clarified it a little bit.

  • $\begingroup$ thats exactly what I was looking for. Thanks a lot. $\endgroup$ – Anand Jan 13 '12 at 23:32
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    $\begingroup$ @Anand I know it’s a hard to follow advice, however I think you shouldn’t start by focusing on technical details but rather try to get why such a bound can exist... then the proof should appear easier. I tried to show you the why in the second part, added this morning (you need to sleep on a question like this – at least I need to). I think it’s terrible how this kind of intuitions doesn’t appear in most textbooks... even if you get the technical part, as long as you don’t have the ideas everything looks magic. Thank you and CrossV for giving me the opportunity to think to this in details! $\endgroup$ – Elvis Jan 14 '12 at 13:20
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    $\begingroup$ Wow! +1 for the edit. Thanks. But wouldn't it be nice if it were possible to get something like $$E[e^{tX}] \leq e^{E[t^2X^2/2]} = e^{(t^2/2)E[X^2]} = e^{(t^2/2)\text{var}(X)} \leq e^{t^2\sigma_{\max}^2/2}?$$ $\endgroup$ – Dilip Sarwate Jan 14 '12 at 13:40
  • $\begingroup$ @Elvis Thanks for the advise and for taking the time to write down the intuitive part. I need to spend some time to understand this! $\endgroup$ – Anand Jan 14 '12 at 14:30
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    $\begingroup$ @Elvis Taking about intuition, I want to clarify my understanding. To get sharper bounds one needs higher moments. Markov uses the first moment, Chebyshev the second moment and the Hoeffding uses mgf. Is this correct? If someone can expand and clarify this part it would be great. $\endgroup$ – Anand Jan 14 '12 at 14:34

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