This post is devoted to a sub-Gaussian tail bound and exponential square integrability for local martingales, taken from my master course on stochastic calculus.

Sub-Gaussian tail bound and exponential square integrability for local martingales. Let ${M={(M_t)}_{t\geq0}}$ be a continuous local martingale issued from the origin. Then for all ${t,K,r\geq0}$,

$\mathbb{P}\Bigr(\sup_{s\in[0,t]}|M_s|\geq r, \langle M\rangle_t\leq K\Bigr) \leq2\mathrm{e}^{-\frac{r^2}{2K}},$

and in particular, if ${\langle M\rangle_t\leq Ct}$ then

$\mathbb{P}\Bigr(\sup_{s\in[0,t]}|M_s|\geq r\Bigr) \leq2\mathrm{e}^{-\frac{r^2}{2Ct}}$

and, for all ${\alpha<\frac{1}{2Ct}}$,

$\mathbb{E}\Bigr(\mathrm{e}^{\alpha\sup_{s\in[0,t]}|M_s|^2}\Bigr)<\infty.$

The condition ${\langle M\rangle_t\leq Ct}$ is a comparison to Brownian motion for which equality holds.

Proof. For all ${\lambda,t\geq0}$, the Doléans-Dade exponential

$X^\lambda ={\Bigr(\mathrm{e}^{\lambda M_t-\frac{\lambda^2}{2}\langle M\rangle_t}\Bigr)}_{t\geq0}$

is a positive super-martingale with ${X^\lambda_0=1}$ and ${\mathbb{E}X^\lambda_t\leq1}$ for all ${t\geq0}$. For all ${t,\lambda,r,K\geq0}$, by using the maximal inequality for the super-martingale ${X^\lambda}$ in the last step,

$\begin{array}{rcl} \mathbb{P}\Bigr(\langle M\rangle_t\leq K,\sup_{0\leq s\leq t}M_s\geq r\Bigr) &\leq&\mathbb{P}\Bigr(\langle M\rangle_t\leq K,\sup_{0\leq s\leq t}X^\lambda_s\geq\mathrm{e}^{\lambda r-\frac{\lambda^2}{2}K}\Bigr) \\ &\leq&\mathbb{P}\Bigr(\sup_{0\leq s\leq t}X^\lambda_s\geq\mathrm{e}^{\lambda r-\frac{\lambda^2}{2}K}\Bigr)\\ &\leq&\mathbb{E}(X^\lambda_0)\mathrm{e}^{-\lambda r+\frac{\lambda^2}{2}K} =\mathrm{e}^{-\lambda r+\frac{\lambda^2}{2}K}. \end{array}$

Taking ${\lambda=r/K}$ gives

$\mathbb{P}\Bigr(\langle M\rangle_t\leq K,\sup_{0\leq s\leq t}M_s\geq r\Bigr) \leq\mathrm{e}^{-\frac{r^2}{2K}}.$

The same reasoning for ${-M}$ instead of ${M}$ provides (note that ${\langle -M\rangle=\langle M\rangle}$ obviously)

$\mathbb{P}\Bigr(\langle M\rangle_t\leq K,\sup_{0\leq s\leq t}(-M_s)\geq r\Bigr) \leq\mathrm{e}^{-\frac{r^2}{2K}}.$

The union bound (hence the prefactor ${2}$) gives then the first desired inequality. The exponential square integrability comes from the usual link between tail bound and integrability, namely if ${X=\sup_{s\in[0,t]}|M_s|}$, ${U(x)=\mathrm{e}^{\alpha x^2}}$, ${\alpha<\frac{1}{2Kt}}$, then, by Fubini-Tonelli,

$\begin{array}{rcl} \mathbb{E}(U(X)) &=&\mathbb{E}\Bigr(\int_0^XU'(x)\mathrm{d}x\Bigr)\\ &=&\mathbb{E}\Bigr(\int_0^\infty\mathbf{1}_{x\leq X}U'(x)\mathrm{d}x\Bigr)\\ &=&\int_0^\infty U'(x)\mathbb{P}(X\geq x)\mathrm{d}x\ &\leq&\int_0^\infty2\alpha x\mathrm{e}^{\alpha x^2}\mathrm{e}^{-\frac{x^2}{2Kt}}\mathrm{d}x <\infty. \end{array}$

Doob maximal inequality for super-martingales. If ${M}$ is a continuous super-martingale, then for all ${t\geq0}$ and ${\lambda>0}$, denoting ${M^-=\max(0,-M)}$,

$\mathbb{P}\Bigr(\max_{s\in[0,t]}|M_s|\geq\lambda\Bigr) \leq\frac{\mathbb{E}(M_0)+2\mathbb{E}(M^-_t)}{\lambda}.$

In particular when ${M}$ is non-negative then ${\mathbb{E}(M^-)=0}$ and the upper bound is ${\frac{\mathbb{E}(M_0)}{\lambda}}$.

Proof. Let us define the bounded stopping time

$T=t\wedge \inf\{s\in[0,t]:M_s\geq \lambda\}.$

We have ${M_T\in\mathrm{L}^1}$ since ${|M_T|\leq\max(|M_0|,|M_t|,\lambda)}$. By the Doob stopping theorem for the sub-martingale ${-M}$ and the bounded stopping times ${0}$ and ${T}$ that satisfy ${M_0\in\mathrm{L}^1}$ and ${M_T\in\mathrm{L}^1}$, we get

$\mathbb{E}(M_0) \geq\mathbb{E}(M_T) \geq \lambda\mathbb{P}(\max_{s\in[0,t]}M_s\geq \lambda) +\mathbb{E}(M_t\mathbf{1}_{\max_{s\in[0,t]}M_s<\lambda})$

hence, recalling that ${M^-=\max(-M,0)}$,

$\lambda\mathbb{P}(\max_{s\in[0,t]}M_s\geq \lambda) \leq \mathbb{E}(M_0)+\mathbb{E}(M^-_t).$

This produces the desired inequality when ${M}$ is non-negative. For the general case, we observe that the Jensen inequality for the nondecreasing convex function ${u\in\mathbb{R}\mapsto\max(u,0)}$ and the sub-martingale ${-M}$ shows that ${M^-}$ is a sub-martingale. Thus, by the Doop maximal inequality for non-negative sub-martingales,

$\lambda\mathbb{P}(\max_{s\in[0,t]}M^-_s\geq \lambda) \leq\mathbb{E}(M^-_t).$

Finally, putting both inequalities together gives

$\lambda\mathbb{P}(\max_{s\in[0,t]}|M_s|\geq \lambda) \leq \lambda\mathbb{P}(\max_{s\in[0,t]}M_s\geq \lambda) +\lambda\mathbb{P}(\max_{s\in[0,t]}M^-_s\geq \lambda) \leq\mathbb{E}(M_0)+2\mathbb{E}(M^-_t).$

Doob maximal inequalities. Let ${M}$ be a continuous process.

1. If ${M}$ is a martingale or a non-negative sub-martingale then for all ${p\geq1}$, ${t\geq0}$, ${\lambda>0}$,

$\mathbb{P}\Bigr(\max_{s\in[0,t]}|M_s|\geq\lambda\Bigr) \leq\frac{\mathbb{E}(|M_t|^p)}{\lambda^p}.$

2. If ${M}$ is a martingale then for all ${p>1}$ and ${t\geq0}$,

$\mathbb{E}\Bigr(\max_{s\in[0,t]}|M_s|^p\Bigr) \leq\Bigr(\frac{p}{p-1}\Bigr)^p\mathbb{E}(|M_t|^p)$

in other words

$\Bigr\|\max_{s\in[0,t]}|M_s|\Bigr\|_p\leq\frac{p}{p-1}\|M_t\|_p.$

In particular if ${M_t\in\mathrm{L}^p}$ then ${M^*_t=\max_{s\in[0,t]}M_s\in\mathrm{L}^p}$.

Comments. This inequality allows to control the tail of the supremum by the moment at the terminal time. It is a continuous time martingale version of the simpler Kolmogorov maximal inequality for sums of independent and identically distributed random variables. Note that ${q=1/(1-1/p)=p/(p-1)}$ is the Hölder conjugate of ${p}$ namely ${1/p+1/q=1}$. The inequality is often used with ${p=2}$, for which ${(p/(p-1))^p=4}$.

Proof. We can always assume that the right hand side is finite, otherwise the inequalities are trivial.

1. If ${M}$ is a martingale, then by the Jensen inequality for the convex function ${u\in\mathbb{R}\mapsto |u|^p}$, the process ${|M|^p}$ is a sub-martingale. Similarly, If ${M}$ is a non-negative sub-martingale then, since ${u\in[0,+\infty)\mapsto u^p}$ is convex and non-decreasing it follows that ${M^p=|M|^p}$ is a sub-martingale. Therefore in all cases ${{(|M_s|^p)}_{s\in[0,t]}}$ is a sub-martingale. Let us define the bounded stopping time

$T=t\wedge \inf\{s\geq0:|M_s|\geq\lambda\}.$

Note that ${|M_T|\leq\max(|M_0|,\lambda)}$ and thus ${M_T\in\mathrm{L}^1}$. The Doob stopping theorem for the sub-martingale ${|M|^p}$ and the bounded stopping times ${T}$ and ${t}$ that satisfy ${T\leq t}$ gives

$\mathbb{E}(|M_T|^p)\leq\mathbb{E}(|M_t|^p).$

On the other hand the definition of ${T}$ gives

$|M_T|^p \geq\lambda^p\mathbf{1}_{\max_{s\in[0,t]}|M_s|\geq\lambda} +|M_t|^p\mathbf{1}_{\max_{s\in[0,t]}|M_s|<\lambda}\\ \geq\lambda^p\mathbf{1}_{\max_{s\in[0,t]}|M_s|\geq\lambda}.$

It remains to combine these inequalities to get the desired result.

2. If we introduce for all ${n\geq1}$ the “localization” stopping time

$T_n=t\wedge\inf\{s\geq0:|M_s|\geq n\},$

then the desired inequality for the bounded sub-martingale ${{(|M_{s\wedge T_n}|)}_{s\in[0,t]}}$ would give

$\mathbb{E}(\max_{s\in[0,t]}|M_{s\wedge T_n}|^p) \leq\left(\frac{p}{p-1}\right)^p\mathbb{E}(|M_t|^p),$

and the desired result for ${{(M_s)}_{s\in[0,t]}}$ would then follow by monotone convergence theorem. Thus this shows that we can assume without loss of generality that ${{(|M_s|)}_{s\in[0,t]}}$ is bounded, in particular that ${\mathbb{E}(\max_{s\in[0,t]}|M_s|^p)<\infty}$. This a martingale localization argument! The previous proof gives

$\mathbb{P}(\max_{s\in[0,t]}|M_s|\geq\lambda) \leq\frac{\mathbb{E}(|M_t|\mathbf{1}_{\max_{s\in[0,t]}|M_s|\geq\lambda})}{\lambda}$

for all ${\lambda>0}$, and thus

$\int_0^\infty\lambda^{p-1} \mathbb{P}(\max_{s\in[0,t]}|M_s|\geq\lambda)\mathrm{d}\lambda \leq \int_0^\infty\lambda^{p-2} \mathbb{E}(|M_t|\mathbf{1}_{\max_{s\in[0,t]}|M_s|\geq\lambda})\mathrm{d}\lambda.$

Now the Fubini-Tonelli theorem gives

$\int_0^\infty\lambda^{p-1}\mathbb{P}(\max_{s\in[0,t]}|M_s|\geq\lambda)\mathrm{d}\lambda =\mathbb{E}\int_0^{\max_{s\in[0,t]}|M_s|}\lambda^{p-1}\mathrm{d}\lambda =\frac{1}{p}\mathbb{E}(\max_{s\in[0,t]}|M_s|^p).$

and similarly (here we need ${p>1}$)

$\int_0^\infty\lambda^{p-2}\mathbb{E}(|M_t|\mathbf{1}_{\max_{s\in[0,t]}|M_s|\geq\lambda)})\mathrm{d}\lambda =\frac{1}{p-1}\mathbb{E}(|M_t|\max_{s\in[0,t]}|M_s|^{p-1}).$

Combining all this gives

$\mathbb{E}(\max_{s\in[0,t]}|M_s|^p) \leq\frac{p}{p-1} \mathbb{E}(M_t\max_{s\in[0,t]}|M_s|^{p-1}).$

But since the Hölder inequality gives

$\mathbb{E}(|M_t|\max_{s\in[0,t]}|M_s|^{p-1}) \leq\mathbb{E}(|M_t|^p)^{1/p}\mathbb{E}(\max_{s\in[0,t]}|M_s|^p)^{\frac{p-1}{p}},$

we obtain

$\mathbb{E}(\max_{s\in[0,t]}|M_s|^p) \leq\frac{p}{p-1}\mathbb{E}(|M_t|^p)^{1/p}\mathbb{E}(\max_{s\in[0,t]}|M_s|^p)^{\frac{p-1}{p}}.$

Consequently, since ${\mathbb{E}(\max_{s\in[0,t]}|M_s|^p)<\infty}$, we obtain the desired inequality.

Doob stopping theorem for sub-martingales. If ${M}$ is a continuous sub-martingale and ${S}$ and ${T}$ are bounded stopping times such that ${S\leq T,\quad M_S\in\mathrm{L}^1,\quad M_T\in\mathrm{L}^1}$, then

$\mathbb{E}(M_S)\leq\mathbb{E}(M_T).$

Proof. We proceed as in the proof of the Doob stopping theorem for martingales, by assuming first that ${S}$ and ${T}$ take their values in the finite set ${\{t_1,\ldots,t_n\}}$ where ${t_1<\cdots<t_n}$. In this case ${M_T}$ and ${M_S}$ are in ${\mathrm{L}^1}$ automatically. The inequality ${S\leq T}$ gives ${\mathbf{1}_{S\geq t}\leq\mathbf{1}_{T\geq t}}$ for all ${t}$. Using this fact and the sub-martingale property of ${M}$, we get

$\begin{array}{rcl} \mathbb{E}(M_S) &=&\mathbb{E}(M_0) +\mathbb{E}\Big(\sum_{k=1}^n\overbrace{\mathbb{E}(M_{t_k}-M_{t_{k-1}}\mid\mathcal{F}_{t_{k-1}})}^{\geq0}\mathbf{1}_{S\geq t_k}\Bigr)\\ &\leq&\mathbb{E}(M_0) +\mathbb{E}\Big(\sum_{k=1}^n\mathbb{E}(M_{t_k}-M_{t_{k-1}}\mid\mathcal{F}_{t_{k-1}})\mathbf{1}_{T\geq t_k}\Bigr)\\ &=&\mathbb{E}(M_T). \end{array}$

More generally, when ${S}$ and ${T}$ are arbitrary bounded stopping times satisfying ${S\leq T}$, we proceed by approximation as in the proof of the Doob stopping for martingales, using again the sub-martingale nature of ${M}$ to get uniform integrability.

Doob stopping theorem for martingales. If ${M}$ is a continuous martingale and ${T:\Omega\rightarrow[0,+\infty]}$ is a stopping time then ${{(M_{t\wedge T})}_{t\geq0}}$ is a martingale: for all ${t\geq0}$ and ${s\in[0,t]}$, we have

$M_{t\wedge T}\in\mathrm{L}^1 \quad\text{and}\quad \mathbb{E}(M_{t\wedge T}\mid\mathcal{F}_s)=M_{s\wedge T}.$

Moreover, if ${T}$ is bounded, or if ${T}$ is almost surely finite and ${{(M_{t\wedge T})}_{t\geq0}}$ is uniformly integrable (for instance dominated by an integrable random variable), then

$M_T\in\mathrm{L}^1 \quad\text{and}\quad \mathbb{E}(M_T)=\mathbb{E}(M_0).$

Comments. The most important is that ${{(M_{t\wedge T})}_{t\geq0}}$ is a martingale. We have always ${\lim_{t\rightarrow\infty}M_{T\wedge t}\mathbf{1}_{T<\infty}=M_T\mathbf{1}_{T<\infty}}$ almost surely. When ${T<\infty}$ almost surely we could use what we know on ${M}$ and ${T}$ to deduce by monotone or dominated convergence that this holds in ${\mathrm{L}^1}$, giving ${\mathbb{E}(M_T)=\mathbb{E}(\lim_{t\rightarrow\infty}M_{t\wedge T})=\lim_{t\rightarrow\infty}\mathbb{E}(M_{t\wedge T})=\mathbb{E}(M_0)}$.

The theorem states that this is automatically the case when ${T}$ is bounded or when ${M^T}$ is uniformly integrable. Furthermore, if ${M^T}$ is uniformly integrable then it can be shown that ${M_\infty}$ exists, giving a sense to ${M_T}$ even on ${\{T=\infty\}}$, and then ${\mathbb{E}(M_T)=\mathbb{E}(M_0)}$.

Proof. Let assume first that ${T}$ takes a finite number of values ${t_1<\cdots<t_n}$. Let us show that ${M_T\in\mathrm{L}^1}$ and ${\mathbb{E}(M_T)=\mathbb{E}(M_0)}$. We have ${M_T=\sum_{k=1}^nM_{t_k}\mathbf{1}_{T=t_k}\in\mathrm{L}^1}$, and moreover, using

$\{T\geq t_k\}=(\cup_{i=1}^{k-1}\{T=t_i\})^c\in\mathcal{F}_{t_{k-1}},$

and the martingale property ${\mathbb{E}(M_{t_k}-M_{t_{k-1}}\mid\mathcal{F}_{t_{k-1}})=0}$, for all ${k}$, we get

$\mathbb{E}(M_T) =\mathbb{E}(M_0) +\mathbb{E}\Big(\sum_{k=1}^n\mathbb{E}(M_{t_k}-M_{t_{k-1}}\mid\mathcal{F}_{t_{k-1}})\mathbf{1}_{T\geq t_k}\Bigr) =\mathbb{E}(M_0).$

Suppose now that ${T}$ takes an infinite number of values but is bounded by some constant ${C}$. For all ${n\geq0}$, we approximate ${T}$ by the piecewise constant random variable (discretization of ${[0,C]}$).

$T_n=C\mathbf{1}_{T=C}+\sum_{k=1}^{n}t_{k}\mathbf{1}_{t_{k-1}\leq T<t_{k}} \quad\text{where}\quad t_k=t_{n,k}=C\frac{k}{n}.$

This is a stopping time since for all ${t\geq0}$, ${\{T_n\leq t\}=\{T_n\leq\lfloor t\rfloor\}\in\mathcal{F}_{\lfloor t\rfloor}}$, which reduces the problem to a discrete ${t}$, and then for all integer ${m\geq0}$, we have that ${\{T_n=m\} =\varnothing\in\mathcal{F}_0}$ if ${m\not\in\{t_{k}:1\leq k\leq n\}}$, while ${\{T_n=m\}=\{T=C\}\in\mathcal{F}_C}$ if ${m=C}$, and

$\{T_n=m\} =\{T<t_{k-1}\}^c\cap\{T<t_{k}\}\in\mathcal{F}_{t_{k}}$

if ${m=t_{k}}$, ${1\leq k\leq n}$, where we used the fact that for all ${t\geq0}$,

$\{T=t\}=\{T\leq t\}\cap\{T<t\}^c=\{T\leq t\}\cap\cap_{r=1}^\infty\{T>t-1/r\}\in\mathcal{F}_t.$

Since ${T_n}$ takes a finite number of values, the previous step gives ${\mathbb{E}(M_{T_n})=\mathbb{E}(M_0)}$. On the other hand, almost surely, ${T_n\rightarrow T}$ as ${n\rightarrow\infty}$. Since ${M}$ is continuous, it follows that almost surely ${M_{T_n}\rightarrow M_T}$ as ${n\rightarrow\infty}$. Let us show now that ${{(M_{T_n})}_{n\geq1}}$ is uniformly integrable. Since for all ${n\geq0}$, ${T_n}$ takes its values in a finite set ${t_1<\cdots<t_{m_n}\leq C}$, the martingale property and the Jensen inequality give, for all ${R>0}$,

$\begin{array}{rcl} \mathbb{E}(|M_C|\mathbf{1}_{|M_{T_n}|\geq R}) &=&\sum_k\mathbb{E}(|M_C|\mathbf{1}_{|M_{t_k}|\geq R,T_n=t_k})\\ &=&\sum_k\mathbb{E}(\mathbb{E}(|M_C|\mid\mathcal{F}_{t_k})\mathbf{1}_{|M_{t_k}|\geq R,T_n=t_k})\\ &\geq&\sum_k\mathbb{E}(|\mathbb{E}(M_C\mid\mathcal{F}_{t_k})|\mathbf{1}_{|M_{t_k}|\geq R,T_n=t_k})\\ &=&\sum_k\mathbb{E}(|M_{t_k}|\mathbf{1}_{|M_{t_k}|\geq R,T_n=t_k})\\ &=&\mathbb{E}(|M_{T_n}|\mathbf{1}_{|M_{T_n}|\geq R}). \end{array}$

Now ${M}$ is continuous and thus locally bounded, and ${M_C\in\mathrm{L}^1}$, thus, by dominated convergence,

$\sup_n\mathbb{E}(|M_{T_n}|\mathbf{1}_{|M_{T_n}|>R}) \leq\mathbb{E}(|M_C|\mathbf{1}_{\sup_{s\in[0,C]}|M_s|\geq R}) \underset{R\rightarrow\infty}{\longrightarrow}0.$

Therefore ${{(M_{T_n})}_{n\geq0}}$ is uniformly integrable. As a consequence

$\overset{\mathrm{a.s.}}{\lim_{n\rightarrow\infty}}M_{T_n}=M_T\in\mathrm{L}^1 \quad\text{and}\quad \mathbb{E}(M_T)=\lim_{n\rightarrow\infty}\mathbb{E}(M_{T_n})=\mathbb{E}(M_0).$

Let us suppose now that ${T}$ is an arbitrary stopping time. For all ${0\leq s\leq t}$ and ${A\in\mathcal{F}_s}$, the random variable ${S=s\mathbf{1}_A+t\mathbf{1}_{A^c}}$ is a (finite) stopping time, and what precedes for the finite stopping time ${t\wedge T\wedge S}$ gives ${M_{t\wedge T\wedge S}\in\mathrm{L}^1}$ and ${\mathbb{E}(M_{t\wedge T\wedge S})=\mathbb{E}(M_0)}$. Now, using the definition of ${S}$, we have

$\mathbb{E}(M_0) =\mathbb{E}(M_{t\wedge T\wedge S}) =\mathbb{E}(\mathbf{1}_AM_{s\wedge T}) +\mathbb{E}(\mathbf{1}_{A^c}M_{t\wedge T}) =\mathbb{E}(\mathbf{1}_A(M_{s\wedge T}-M_{t\wedge T}))+\mathbb{E}(M_{t\wedge T}).$

Since ${\mathbb{E}(M_{t\wedge T})=\mathbb{E}(M_0)}$, we get the martingale property for ${{(M_{t\wedge T})}_{t\geq0}}$, namely

$\mathbb{E}((M_{t\wedge T}-M_{s\wedge T})\mathbf{1}_A)=0.$

Finally, suppose that ${T<\infty}$ almost surely and ${{(M_{t\wedge T})}_{t\geq0}}$ is uniformly integrable. The random variable ${M_T}$ is well defined and ${\lim_{t\rightarrow\infty}M_{t\wedge T}=M_T}$ almost surely because ${M}$ is continuous. Furthermore, since ${{(M_{t\wedge T})}_{t\geq0}}$ is uniformly integrable, it follows that ${M_T\in\mathrm{L}^1}$ and ${\lim_{t\rightarrow\infty}M_{t\wedge T}=M_T}$ in ${\mathrm{L}^1}$. In particular ${\mathbb{E}(M_0)\underset{\forall t}{=}\mathbb{E}(M_{t\wedge T})=\lim_{t\rightarrow\infty}\mathbb{E}(M_{t\wedge T})=\mathbb{E}(M_T)}$. Further reading in the same spirit.

Last Updated on 2020-12-03

1. Jean-René Chazottes 2020-11-26

Dear Djalil,
Thank you for this post! I would like to make a historical comment on the so-called Chernoff’s bound or Chernoff bounding method. According to Chernoff himself, this idea is due to Donald Rubin (https://en.wikipedia.org/wiki/Donald_Rubin):
“ He produced a rather simple argument, using the Markov inequality, for the upper bound. Since that seemed to be a minor lemma in the ensuing paper I published (Chernoff, 1952), I neglected to give him credit. I now consider it a serious error in judgment […].”
Once again, the Arnold Principle applies: If a notion bears a personal name, then this name is not the name of the discoverer.

The quotation above is taken from page 35 of the book:
Chernoff, Herman (2014). “A career in statistics” (PDF). In Lin, Xihong; Genest, Christian; Banks, David L.; Molenberghs, Geert; Scott, David W.; Wang, Jane-Ling (eds.). Past, Present, and Future of Statistics

2. Djalil Chafaï 2020-11-26

Thank you very much Jean-René ! In fact, it is also in the article A conversation with Herman Chernoff, by John Bather, Statistical Science 11(4) (1996) 335-350. An instance of Stigler’s law of eponymy.

Bather: Let us go back to the 1950s and talk about some of the research you did in those early years at Stanford. Tell me about the result that is always known as Chernoff’s lemma, but that you attribute to Herman Rubin.

Chernoff: In the work that I did on my visit to Stanford, there were two papers. One was called the “Measure of asymptotic efficiency” and the other “Locally optimal designs,” and both of these came out of problems that had important relevance to design of experiments. The first one was, as far as I know, the first application of large deviation theory to statistical problems. I had two simple hypotheses and I noticed that for discriminating between the two, in the range outside the 5% significance level, we were in the case of large deviations. Being ignorant at the time of the beautiful work by Cramer, I derived a slightly overlapping, but not nearly as elegant, result which showed that asymptotically the probability of falling in the tails approaches zero at an exponential rate, under mild assumptions. Rubin claimed that part of my derivation, giving the lower bounds, could be obtained much more easily. After working so hard, I doubted it very much. He showed me the Chebyshev type of proof that gives rise to what’s now called the Chernoff bound, but it is certainly Rubin’s. When I wrote up the technical report, I mentioned his assistance but when I submitted the paper for publication, I left it out because it was so trivial and it never occurred to me that this would be one of the things that would lead to my fame in electrical engineering circles. That inequality turned out to be a very important result as far as information theory is concerned, and so the lower bound has been called the Chernoff bound ever since. I am very unhappy about the fact that I did not properly credit Rubin at that time because I thought it was a rather trivial lemma, but many things are only trivial once you know them.

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