Noncommutative Blackwell-Ross martingale inequality

TL;DR

This paper extends classical martingale inequalities to the noncommutative setting, providing new bounds for supermartingales and deriving an Azuma-type inequality.

Contribution

It introduces a noncommutative Blackwell-Ross inequality for supermartingales, generalizing previous work to the noncommutative framework.

Findings

Established a noncommutative Blackwell-Ross inequality for supermartingales.

Derived an Azuma-type inequality from the new bounds.

Generalized classical martingale inequalities to noncommutative probability.

Abstract

We establish a noncommutative Blackwell--Ross inequality for supermartingales under a suitable condition which generalize Khan's works to the noncommutative setting. We then employ it to deduce an Azuma-type inequality.

Equations

$$\mathrm{Prob}(X_n\ge mc\mathrm{for}\mathrm{some}n\ge m)\le \exp \left(\frac{-m{c}^2}{2{\alpha }^2}\right),$$

$$\tau (e^{y_1+y_2})\le \tau (e^{y_1}{e}^{y_2})$$

$${\mathcal{E}}_{j-1}(x)={\mathcal{E}}_{\mathfrak{N}}(x)$$

$$x={\mathcal{E}}_{j-1}(x)={\mathcal{E}}_{\mathfrak{N}}(x)\in \mathfrak{N}.$$

$${\mathcal{E}}_i\circ {\mathcal{E}}_j={\mathcal{E}}_j\circ {\mathcal{E}}_i={\mathcal{E}}_{\min \{i,j\}}$$

$$\lambda e^{(1-\lambda )x}+(1-\lambda )e^{-\lambda x}\le e^{\frac{x^2}{8}}.$$

$${\mathcal{E}}_{n-1}(e^{t{x}_n})\le f(t)\le e^{-\gamma t+\lambda t^2},(\lambda >0,\gamma \ge 0,t>0)$$

$$\tau \left({\vee }_{n=m+i}^{\infty }{\text{1}}_{(a+bn,\infty )}(s_n)\right)\le A^m{e}^{\frac{-a(b+\gamma )}{\lambda }},$$

$$\tau \left({\vee }_{n=m+j}^{\infty }{\text{1}}_{(bn,\infty )}(s_n)\right)\le A_0^m{e}^{\frac{-m(b+\gamma {)}^2}{4\lambda }}$$

$$\tau (y_n)$$

$${\vee }_{n=m+i}^k{\text{1}}_{[a+bn,\infty )}(s_n)\wedge ({\wedge }_{n=m+i}^k{\text{1}}_{[0,1)}(y_n))=0.$$

$$\bigcup _{n=m+i}^k{\text{1}}_{[a+bn,\infty )}(s_n)(\mathfrak{H})\bigcap \left(\bigcap _{n=m+i}^k{\text{1}}_{[0,1)}(y_n)(\mathfrak{H})\right).$$

$$e^{⟨c(s_j-a-bj)(\xi ),\xi ⟩}\le ⟨e^{c(s_j-a-bj)}\xi ,\xi ⟩<1.$$

$$\tau ({\vee }_{n=m+i}^{\infty }{\text{1}}_{[a+bn,\infty )}(s_n))$$

$${\text{1}}_{[bn,\infty )}(s_n)\le {\text{1}}_{[\frac{m(b+\gamma )}{2}+\frac{n(b-\gamma )}{2},\infty )}(s_n)$$

$$\tau \left({\vee }_{n=m+j}^{\infty }{\text{1}}_{[\frac{m(b+\gamma )}{2}+\frac{n(b-\gamma )}{2},\infty )}(s_n)\right)\le A_0^m{e}^{\frac{-m(b+\gamma {)}^2}{4\lambda }},$$

$$\tau \left({\vee }_{n=m+j}^{\infty }{\text{1}}_{[bn,\infty )(s_n)}\right)\le \tau \left({\vee }_{n=m+j}^{\infty }{\text{1}}_{[\frac{m(b+r)}{2}+\frac{n(b-r)}{2},\infty )}\right)\le A_0^m{e}^{\frac{-m(b+\gamma {)}^2}{4\lambda }},$$

$${\mathcal{E}}_{n-1}(x_n)\le -\gamma .$$

$$\tau \left({\vee }_{n=m+i}^{\infty }{\text{1}}_{(a+bn,\infty )}(s_n)\right)\le A^m{e}^{\frac{-8a(b+\gamma )}{(\alpha +\beta {)}^2}},$$

$$\tau \left({\vee }_{n=m+j}^{\infty }{\text{1}}_{(bn,\infty )}(s_n)\right)\le A_0^m{e}^{\frac{-2m(b+\gamma {)}^2}{(\alpha +\beta {)}^2}}$$

$$\tau \left({\vee }_{n=m+i}^{\infty }{\text{1}}_{[a+bn,\infty )}(x_n)\right)\le A^m\exp \left\{\frac{-8ab}{(\alpha +\beta {)}^2}\right\},$$

$$A=\frac{\beta }{\alpha +\beta }\exp \left\{\frac{-8b(b+\alpha )}{(\alpha +\beta {)}^2}\right\}+\frac{\alpha }{\alpha +\beta }\exp \left\{\frac{-8b(b-\beta )}{(\alpha +\beta {)}^2}\right\}\le 1.$$

$$\tau ({\wedge }_{m=1}^{\infty }{\vee }_{n=m}^{\infty }{\text{1}}_{[1,\infty )}(x_n))=\underset{m\to \infty }{\lim }\tau ({\vee }_{n=m}^{\infty }{\text{1}}_{[1,\infty )}(x_n))=0$$

$$e^{st}\le e^{t\beta }\frac{s+\alpha }{\alpha +\beta }+e^{-t\alpha }\frac{\beta -s}{\alpha +\beta }.$$

$$e^{td{x}_j}\le e^{t\beta }\frac{d{x}_j+\alpha }{\alpha +\beta }+e^{-t\alpha }\frac{\beta -d{x}_j}{\alpha +\beta }.$$

$${\mathcal{E}}_{j-1}(e^{td{x}_j})\le e^{t\beta }\frac{\alpha }{\alpha +\beta }+e^{-t\alpha }\frac{\beta }{\alpha +\beta }\le e^{\frac{t^2(\alpha +\beta {)}^2}{8}},$$

$$\tau \left({\vee }_{n=m+i}^{\infty }{\text{1}}_{(cn,\infty )}(x_n)\right)\le B^m\exp \left\{\frac{-2m{c}^2}{(\alpha +\beta {)}^2}\right\},$$

$$B=\frac{\beta }{\alpha +\beta }\exp \left\{\frac{-2c(c+2\alpha )}{(\alpha +\beta {)}^2}\right\}+\frac{\alpha }{\alpha +\beta }\exp \left\{\frac{-2c(c-2\beta )}{(\alpha +\beta {)}^2}\right\}\le 1.$$

$$\tau \left({\vee }_{n=m+i}^{\infty }{1}_{[\frac{mc}{2}+\frac{nc}{2},\infty )}(x_n)\right)\le B^m\exp \left\{\frac{-8m{c}^2}{4(\alpha +\beta {)}^2}\right\}$$

$$1_{[nc,\infty )}(x_n)\le 1_{[\frac{mc}{2}+\frac{nc}{2},\infty )}(x_n)$$

Full text

1. Introduction

Blackwell [2] showed that if $\{X_{n},X_{0}=0,n\geq 0\}$ is a martingale such that $|X_{n}-X_{n-1}|\leq\alpha$ for all $n$, then for each positive number $c$,

[TABLE]

which gives a generalization of a result of Hoeffding [5]. Ross [8] extended Blakwell’s result to the case where the bound on the martingale difference is not symmetric. Indeed, Ross employed a supermartingale argument to show that the same is true when $-\alpha\leq X_{n}-X_{n-1}\leq\beta$, where $\alpha,\beta>0$. Khan [7] generalized Blackwell–Ross inequality for martingales (supermartingales) under a subnormal structure on the conditional moment generating function $\varphi_{n}(\theta)=\mathbb{E}(\exp(\theta X_{n})|\mathcal{F}_{n-1})$ subject to some mild conditions.

In this paper, we adopt the classical ideas in probability theory and the Golden–Thompson inequality to establish a Blackwell–Ross martingale inequality under a non-symmetric bound on the martingales differences in the framework of noncommutative probability spaces.

A von Neumann algebra $\mathfrak{M}$ on a Hilbert space $\mathfrak{H}$ with unit element $1$ equipped with a normal faithful tracial state $\tau:\mathfrak{M}\to\mathbb{C}$ is called a noncommutative probability space. We denote by $\leq$ the usual order on the self-adjoint part $\mathfrak{M}^{sa}$ of $\mathfrak{M}$. For each self-adjoint operator $x\in\mathfrak{M}$, there exists a unique spectral measure $E$ as a $\sigma$-additive mapping with respect to the strong operator topology from the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$ of $\mathbb{R}$ into the set of all orthogonal projections such that for every Borel function $f:\sigma(x)\to\mathbb{C}$ the operator $f(x)$ is defined by $f(x)=\int f(\lambda)dE(\lambda)$, in particular, $\textbf{1}_{B}(x)=\int_{B}dE(\lambda)=E(B)$.

The celebrated Golden–Thompson inequality [9] states that for any self-adjoint elements $y_{1},y_{2}$ in a noncommutative probability space $\mathfrak{M}$, the inequality

[TABLE]

holds; see also [12] for some Golden–Thompson type inequalities.

For $p\geq 1$, the noncommutative $L^{P}$-space $L^{p}(\mathfrak{M})$ is defined as the completion of $\mathfrak{M}$ with respect to the $L^{p}$-norm $\|x\|_{p}:=\left(\tau(|x|^{p})\right)^{1/p}$. The commutative cases of discussed spaces are usual $L^{p}$-spaces. For further information we refer the reader to [3] and references therein.

Let $\mathfrak{N}$ be a von Neumann subalgebra of $\mathfrak{M}$. Then there exists a normal positive contractive projection $\mathcal{E}_{\mathfrak{N}}:\mathfrak{M}\to\mathfrak{N}$ satisfying the following properties:

(i) $\mathcal{E}_{\mathfrak{N}}(axb)=a\mathcal{E}_{\mathfrak{N}}(x)b$ for any $x\in\mathfrak{M}$ and $a,b\in\mathfrak{N}$;

(ii) $\tau\circ\mathcal{E}_{\mathfrak{N}}=\tau$.

Moreover, $\mathcal{E}_{\mathfrak{N}}$ is the unique mapping satisfying (i) and (ii). The mapping $\mathcal{E}_{\mathfrak{N}}$ is called the conditional expectation of $\mathfrak{M}$ with respect to $\mathfrak{N}$.

Let $\mathfrak{N}\subseteq\mathfrak{A}_{j}\,\,(1\leq j\leq n)$ be von Neumann subalgebras of $\mathfrak{M}$. We say that the $\mathfrak{A}_{j}$ are order independent over $\mathfrak{N}$ if for every $2\leq j\leq n$, the equality

[TABLE]

holds for all $x\in\mathfrak{A}_{j}$, where $\mathcal{E}_{j-1}$ is the conditional expectation of $\mathfrak{M}$ with respect to the von Neumann subalgebra generated by $\mathfrak{A}_{1},\ldots,\mathfrak{A}_{j-1}$; cf. [6]. Note that this notion of independence implies that $\mathfrak{N}$ should be the intersection of all $\mathfrak{A}_{j}$. In fact, if $x\in\mathfrak{A}_{j-1}\cap\mathfrak{A}_{j}$, then

[TABLE]

A filtration of $\mathfrak{M}$ is an increasing sequence $(\mathfrak{M}_{j},\mathcal{E}_{j})_{0\leq j\leq n}$ of von Neumann subalgebras of $\mathfrak{M}$ together with the conditional expectations $\mathcal{E}_{j}$ of $\mathfrak{M}$ with respect to $\mathfrak{M}_{j}$ such that $\bigcup_{j}\mathfrak{M}_{j}$ is $w^{*}$–dense in $\mathfrak{M}$. It follows from $\mathfrak{M}_{j}\subseteq\mathfrak{M}_{j+1}$ that

[TABLE]

for all $i,j\geq 0$. Generally, a sequence $(x_{j})_{j\geq 0}$ in $L^{1}(\mathfrak{M})$ is called a martingale (supermartingale, resp.) with respect to the filtration $(\mathfrak{M}_{j})_{0\leq j\leq n}$ if $x_{j}\in L^{1}(\mathfrak{M}_{j})$ and $\mathcal{E}_{j}(x_{j+1})=x_{j}$ ($\mathcal{E}_{j}(x_{j+1})\leq x_{j}$, resp.) for every $j\geq 0$. It follows from (1.2) that $\mathcal{E}_{j}(x_{i})=x_{j}$ for all $i\geq j$. Put $dx_{j}=x_{j}-x_{j-1}\,\,(j\geq 0)$ with the convention that $x_{-1}=0$. Then $dx=(dx_{j})_{j\geq 0}$ is called the martingale difference of $(x_{j})$. The reader is referred to [13] for more information.

2. Main Results

In this section, we provide a noncommutative Blackwell–Ross inequality. To this end, we will need the following lemma which was proved by Alon, et al. [1].

Lemma 2.1**.**

For $0\leq\lambda\leq 1$

[TABLE]

We are inspired by some ideas in the commutative case, e.g. [7], to provide our main result.

Theorem 2.2**.**

Let $\left\{s_{n}=\sum_{i=1}^{n}x_{n},~{}~{}n\geq 0\right\}$ be a self-adjoint supermartingale in $\mathfrak{M}$ with respect to a filtration $(\mathfrak{M}_{n},\mathcal{E}_{n})_{n\geq 0}$ such that

[TABLE]

where $f(t)$ is a continuous positive function on $[0,\infty)$. Then for positive numbers $a$ and $b$, there exists $i\geq 1$ such that for any positive integer $m$

[TABLE]

where $A=e^{-bt_{0}}f(t_{0})\leq 1$ and $t_{0}=\frac{b+\gamma}{\lambda}$. Moreover,

[TABLE]

for some $j\geq 1$, where $A_{0}=e^{-\frac{1}{2}(b-\gamma)t_{0}}f(t_{0})$ and $t_{0}=\frac{b+\gamma}{2\lambda}$.

Proof.

Let $y_{n}=\exp(ts_{n}-at-bnt)$, $t>0$. We show that the sequence $(y_{n})_{n\geq 0}$ satisfies the following inequality at $t=t_{0}=\frac{b+\gamma}{\lambda}$. To this end, note that

[TABLE]

if $A=e^{-bt}f(t)$ and $t=t_{0}=\frac{b+\gamma}{\lambda}$, in which the first and second inequalities follows from (2.1).

We have $\vee_{n=m+i}^{k}\textbf{1}_{[a+bn,\infty)}(s_{n})\preceq\vee_{n=m+i}^{k}\textbf{1}_{[1,\infty)}(y_{n})$, for every $k\geq m+i$, in which $m,i$ are positive integers, since

[TABLE]

To show this assume that $\xi$ is an unit element in

[TABLE]

Therefore $\langle s_{j}\xi,\xi\rangle\geq a+bj$ and $\langle e^{c(s_{j}-a-bj)}\xi,\xi\rangle<1$, for some $m+i\leq j\leq k$. By the operator version of the classical Jensen’s inequality for the convex function $t\mapsto e^{c(t-a-bj)}$, we get

[TABLE]

Consequently, $\langle c(s_{j}-a-bj)(\xi),\xi\rangle<0$ and hence $\langle s_{j}\xi,\xi\rangle<a+bj$ which gives rise to a contradiction. Choose $i\in\mathbb{N}$ such that $\frac{A^{i}}{1-A}\leq 1$. Hence

[TABLE]

for any positive integer $m$, and this ensures (2.2).

To prove (2.3), let $g(\alpha,n)=m(b-\alpha)+\alpha n$, $n\geq m$, $\alpha\leq b$ and note that $bn\geq g(\alpha,n)$ for every $n\geq m$. A minimization consideration leads to the choice of $\alpha=\alpha_{0}=\frac{b-\gamma}{2}$. Thus

[TABLE]

for any $n\geq m$. From (2.2) we infer that

[TABLE]

for some $j\geq 1$, where $A_{0}=e^{\frac{-1}{2}(b-\gamma)t_{0}}f(t_{0})$ and $t_{0}=\frac{b+\gamma}{2\lambda}$. Hence

[TABLE]

which implies (2.3). ∎

Note that, in view of the Jensen inequality for conditional expectations in the above Theorem, we lead to the following inequality:

[TABLE]

This special case have investigated by Khan. Similar to arguments in [7], we may conclude that if $\left\{s_{n}=\sum_{i=1}^{n}x_{n},~{}~{}n\geq 0\right\}$ is a self-adjoint supermartingale with respect to a filtration $(\mathfrak{M}_{n},\mathcal{E}_{n})_{n\geq 0}$ such that $-\alpha\leq x_{n}\leq\beta$ and $\mathcal{E}_{n-1}(x_{n})\leq-\gamma$  $(\alpha>\lambda\geq 0,\beta>0)$ for all $n$, then for positive numbers $a$ and $b$, there exists $i\geq 1$ such that for any positive integer $m$

[TABLE]

in which $A=e^{-bt_{0}}f(t_{0})$ and $t_{0}=\frac{8(b+\gamma)}{(\alpha+\beta)^{2}}$, where $f(t)=e^{-\gamma t}\left(pe^{(\alpha+\beta)tq}+qe^{-(\alpha+\beta)tp}\right)$ and $p=\frac{\alpha-\gamma}{\alpha+\beta}$ and $\frac{\beta+\gamma}{\alpha+\beta}$. Similarly,

[TABLE]

for some $j\geq 1$, where $A_{0}=e^{-\frac{1}{2}(b-\gamma)t_{0}}f(t_{0})$ and $t_{0}=\frac{4(b+\gamma)}{(\alpha+\beta)^{2}}$.

Corollary 2.3**.**

(Noncommutative Blackwell–Ross inequality) Let $x=(x_{j})_{0\leq j\leq n}$ be a self-adjoint martingale in $\mathfrak{M}$ with respect to a filtration $(\mathfrak{M}_{j},\mathcal{E}_{j})_{0\leq j\leq n}$ with $x_{0}=0$ and $dx_{j}=x_{j}-x_{j-1}$ be its associated martingale difference. Assume that $-\alpha\leq dx_{j}\leq\beta$ for some positive constants $\alpha,\beta\,\,(1\leq j\leq n)$. Then for any positive values $a,b$, there exists $i\geq 1$ such that for any positive integer $m$

[TABLE]

where

[TABLE]

Moreover,

[TABLE]

Proof.

Note that $x_{n}=\sum_{k=1}^{n}dx_{k}$ for all $n$. Let $t>0$. The function $s\mapsto e^{ts}$ is convex, therefore for any $-\alpha\leq s\leq\beta$,

[TABLE]

Since $-\alpha\leq dx_{j}\leq\beta$, by the functional calculus, we have

[TABLE]

Since $\mathcal{E}_{j-1}$ is a positive map and $\mathcal{E}_{j-1}(dx_{j})=0$, we reach

[TABLE]

where the second inequality is deduced from Lemma 2.1 with $\lambda=\frac{\alpha}{\alpha+\beta}$ and $x=c(\alpha+\beta)$. Hence the desired result can be deduced from Theorem 2.2 with $f(t)=e^{t\beta}\frac{\alpha}{\alpha+\beta}+e^{-t\alpha}\frac{\beta}{\alpha+\beta}$, $\gamma=0$ and $\lambda=\frac{(\alpha+\beta)^{2}}{8}$. ∎

The authors of [10, 11] proved a noncommutative Azuma-type inequality for noncommutative martingales in noncommutative probability spaces, and as applications, the authors obtained a noncommutative Heoffding inequality. In the next corollary we give a noncommutative Blackwell inequality from which we deduce an extension of commutative Azuma-type inequality. One may regard the following conclusion as a stronger result than the noncommutative Azuma-type inequality.

Corollary 2.4**.**

Let $x=(x_{j})_{0\leq j\leq n}$ be a self-adjoint martingale in $\mathfrak{M}$ with respect to a filtration $(\mathfrak{M}_{j},\mathcal{E}_{j})_{0\leq j\leq n}$ and $dx_{j}=x_{j}-x_{j-1}$ be its associated martingale difference. Assume that $-\alpha\leq dx_{j}\leq\beta$ for some nonnegative constants $\alpha,\beta>0\,\,(1\leq j\leq n)$. Then for any positive value $c$, there exists $i\geq 1$ such that for any positive integer $m$

[TABLE]

where

[TABLE]

Proof.

For $a=\frac{mc}{2}$ and $b=\frac{c}{2}$, it follows form Corollary 2.3 that

[TABLE]

for some $i\geq 1$. Moreover, we have

[TABLE]

for every $n\in\mathbb{N}$. Hence the result is deduced from (2.4) and (2.5). ∎

Corollary 2.5** (Azuma-type inequality).**

Let $Z_{n}$,   $n\geq 0$ be a martingale sequence of bounded random variables with respect to a filtration $(\mathcal{F}_{n},\mathbb{E}_{n})_{n\geq 1}$ on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ with $Z_{0}=0$. If $-\alpha\leq dZ_{n}\leq\alpha$ for all $n$, then for each $c>0$ there exists $i\geq 1$ such that for any positive integer $m$

[TABLE]

Proof.

It immediately follows from Corollary 2.4. ∎

Now we can state a version of classical Blackwell-Ross supermartingale inequality as follows; cf. [7].

Corollary 2.6**.**

Let $\left\{S_{n}=\sum_{i=1}^{n}X_{n},~{}~{}n\geq 0\right\}$ be a supermartingale of bounded random variables with respect to a filtration $(\mathcal{F}_{n},\mathbb{E}_{n})_{n=1}^{N}$ on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ such that

[TABLE]

where $f(t)$ is a positive continuous function. Then for positive numbers $a$ and $b$, there exists $i\geq 1$ such that for any positive integer $m$,

[TABLE]

where $A=e^{-bt_{0}}f(t_{0})<1$ and $t_{0}=\frac{b+\gamma}{\lambda}$. Moreover,

[TABLE]

for some $j\geq 1$, where $A_{0}=e^{-\frac{1}{2}(b-\gamma)t_{0}}f(t_{0})$ and $t_{0}=\frac{b+\gamma}{2\lambda}$.

Corollary 2.7**.**

Let $\mathfrak{N}\subseteq\mathfrak{A}_{j}(\subseteq\mathfrak{M})$ be order independent over $\mathfrak{N}$. Let $x_{j}\in\mathfrak{A}_{j}$ be self-adjoint such that $\mathcal{E}_{\mathfrak{N}}(x_{j})\leq 0$ and

[TABLE]

where $f(t)$ is a continuous positive function on $[0,\infty)$ such that $f(0)=1$. Then for positive numbers $a$ and $b$, there exists $i\geq 1$ such that for any positive integer $m$

[TABLE]

Proof.

Let $\mathfrak{M}_{0}=\mathfrak{N}$ and $\mathcal{E}_{0}=\mathcal{E}_{\mathfrak{N}}$. For every $1\leq j\leq n$, let $\mathfrak{M}_{j}$ be the von Neumann subalgebra generated by $\mathfrak{A}_{1},\ldots,\mathfrak{A}_{j-1}$ and $\mathcal{E}_{j}$ be the corresponding conditional expectation. Put $s_{0}:=0$ and $s_{j}:=\sum_{k=1}^{j}x_{k}$ for $1\leq j\leq n$. Then

[TABLE]

It follows that $(s_{j})_{0\leq j\leq n}$ is a supermartingale with respect to the filtration $(\mathfrak{M}_{j},\mathcal{E}_{j})_{0\leq j\leq n}$. Hence, the result follows via $\left\{s_{n}=\sum_{i=1}^{n}x_{n},~{}~{}n\geq 0\right\}$ in Theorem 2.2. ∎

Acknowledgement. The authors are grateful to Prof. Fedor Sukochev for his useful comments.