Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $p$ be a real number greater than $1$. It is well known (see Hall and Heyde's Martingale limit theory and its applications, Theorem 2.10) that there exists a constant $C_p$ such that if $(X_i)_{i=1}^n$ is a real valued martingale difference with respect to the filtration $(\mathcal F_i)_{i=1}^n$ (that is, $(S_j:=\sum_{i=1}^jX_i)_{j=1}^n$ is a martingale with respect to this filtration), then $$\frac 1{C_p}\mathbb E\left(\sum_{i=1}^nX_i^2\right)^{p/2}\leqslant \mathbb E\left|\sum_{i=1}^nX_i\right|^p\leqslant C_p\mathbb E\left(\sum_{i=1}^nX_i^2\right)^{p/2}.$$ Hence the $\mathbb L^p$ norm of the partial sum is controlled by those of the quadratic variation.

Now define for a real valued random variable $X$: $$\lVert X\rVert_{p,\infty}:=\left(\sup_{t\geqslant 0}t^p\mu\{|X|\geqslant t\}\right)^{1/p}.$$ This is equivalent to a norm (namely $N(X):=\sup_{\mu(A)>0}\mu(A)^{-1+1/p}\int_A|X|\mathrm d\mu$).

I would like to know whether there is a similar inequality to Rosenthal's one, that is, a control of $N(S_n)$ in terms of those of $N\left(\sqrt{\sum_{i=1}^nX_i^2}\right)$ plus maybe an other term. This seems to be a natural question which has probably been investigated, but I didn't manage to find a reference.

There are weak-$\mathbb L^p$ versions of Rosenthal's inequality for independent random variables, but I would like to see a reference dealing with an extension to martingale differences, namely:

Let $(\Omega,\mathcal F,\mu)$ be a probability space $p\gt 2$. Is there a constant $C_p$ such that if $n$ is an integer and $(X_j)_{1\leqslant j\leqslant n}$ is a martingale difference with respect to the filtration $(\mathcal F_j)_{1\leqslant j\leqslant n}$ with $\lVert X_j\rVert_{p,\infty}\lt\infty$, then $$C_p^{-1}\left\lVert \sqrt{\sum_{j=1}^nX_j^2}\right\rVert_{p,\infty}\leqslant \left\lVert \sum_{j=1}^nX_j\right\rVert_{p,\infty}\leqslant C_p\left\lVert \sqrt{\sum_{j=1}^nX_j^2}\right\rVert_{p,\infty}~? $$

share|improve this question

3 Answers 3

up vote 2 down vote accepted

The result you want is mentioned in Remark 6 in

Johnson, W. B.(1-TXAM); Schechtman, G.(IL-WEIZ) Martingale inequalities in rearrangement invariant function spaces. Israel J. Math. 64 (1988), no. 3, 267–275 (1989).

share|improve this answer
    
Thank you very much! I accept this answer since there is a reference (I guessed it was already done somewhere). This question made me learn a lot of things in functional analysis, and indeed, the theory of rearrangement invariant function spaces is very practical for this kind of problem. –  Davide Giraudo Jan 30 at 19:38

In any Banach lattice, a sufficient condition for the equivalence of the norm of a sum of elements and the norm of the square function is that the sequence be unconditional and the lattice has non trivial cotype. This is proved in many places, in particular in the introductory book of Albiac and Kalton and in volume 2 of Lindenstrauss and Tzafriri. In a rearrangement invariant function space, martingale difference sequences are unconditional provided the upper and lower Boyd indices are in the range $(0,\infty)$ because you can then do an interpolation argument (vol. 2 of [LT]). IIRC, for martingale difference sequences you don't also need non trivial cotype because you get an upper estimate for a linear combination of martingale differences in terms of the square function both in the space and in the dual space. You have some duality because the filtration the martingale determines gives you commuting contractive projections that converge to the identity operator.

This is either an outline of a proof or utter nonsense.

share|improve this answer
    
Thanks for your answer. I will look at the references you mention. I'm not used to work with Banach lattices and unconditional sequences, so I will learn a lot of things and will maybe able to come out with a proof. I remember the proof of Hall and Heyde does not use such functional analytic tools (it uses stopping times, estimations of moments, etc...). –  Davide Giraudo Jan 25 at 21:57
2  
If you are functional analyst who is interested in the $L_p$ spaces, you must study probability. If you are a probabilist who is interested in limit theorems, you do yourself a disservice if you do not study the connections to functional analysis. As for myself, I do not know where Banach space theory ends and probability theory begins. –  Bill Johnson Jan 25 at 22:49
1  
I agree and retagged the question. There are indeed very interesting connection with functional analysis. –  Davide Giraudo Jan 27 at 16:41
    
Bill: I find it hard to imagine a functional analyst not interested in the $L_p$ spaces, and impossible to imagine a probabilist who is not interested in limit theorems! (Maybe that was really part of your point.) It's worth noting, though, that connections between probability and functional analysis reach well beyond $L_p$ spaces and limit theorems. –  Mark Meckes Feb 5 at 10:46

I think I came out with a proof of the existence of a constant $C_p$ such that if $(S_n,\mathcal F_n)$ is a martingale with differences $X_i$, then $$C_p^{-1}\left\lVert\left(\sum_{i=1}^nX_i^2\right)^{1/2}\right\rVert_{p,\infty} \leqslant \lVert S_n\rVert_{p,\infty}\leqslant C_p\left\lVert\left(\sum_{i=1}^nX_i^2\right)^{1/2}\right\rVert_{p,\infty} +C_p\left\lVert\left(\sum_{i=1}^n\mathbb E[X_i^2\mid\mathcal F_{i-1}]\right)^{1/2}\right\rVert_{p,\infty}.$$

The proof is quite technical and relies on the proof of classical Rosenthal's inequality for the $p$-th moment. Some intermediate steps gives estimates on the tails and then are converted into bounds on the moments.

The following lemma is a part of Lemma 2.3. in Hall and Heyde's book.

Lemma. Let $(S_i,\mathcal F_i)_{i\geqslant 1}$ be a non-negative sub-martingale and for $\theta>0$, define $$Y:=\max\left\{\theta\left(\sum_{i=1}^nX_i^2\right)^{1/2}, \max_{1\leqslant i\leqslant n}S_i\right\}.$$ Then for each $t>0$, $$t\mu\{Y\gt \sqrt{1+2\theta^2}t\}\leqslant 3\mathbb E[S_n\chi_{Y>t}].$$

The proof relies on the use of the non-negative submartingale $(T_i,\mathcal F_i)$ with $$T_i:=S_i\chi\left\{\theta\left(\sum_{j=1}^iX_j^2\right)^{1/2}\gt t\right\}$$ and the stopping time $\tau:=\min\{n,\inf\{i\mid\theta\left(\sum_{j=1}^iX_j^2\right)^{1/2}\gt t \}\}$.

Choosing $\theta:=2$, noticing that $2\left(\sum_{i=1}^nX_i^2\right)^{1/2}\leqslant Y$ and the link between $\lVert X\rVert_{p,\infty}$ and $\sup t^p\mu\{X\geqslant t\}$, we obtain the inequality $$\left\lVert\left(\sum_{i=1}^nX_i^2\right)^{1/2}\right\rVert_{p,\infty} \leqslant \frac{3^{p+1}}2\lVert S_n\rVert_{p,\infty}.$$ Using the martingales $(\mathbb E[S_n^+\mid\mathcal F_i])_i$ and $(\mathbb E[S_n^-\mid\mathcal F_i])_i$ we can extend to any martingale $(S_n)$, namely $$\left\lVert\left(\sum_{i=1}^nX_i^2\right)^{1/2}\right\rVert_{p,\infty} \leqslant 3^{p+1}\lVert S_n\rVert_{p,\infty}.$$

Here is the weak moment version of Lemma 2.3 of Hall and Heyde

Lemma. If $X$ and $Y$ are two non-negative random variables, $\beta>1$, $\delta\gt 0$ and $\varepsilon\gt 0$ are such that for each positive $t$, $$\mu\{X\geqslant\beta t,Y\leqslant \delta t\}\leqslant \varepsilon\mu\{X\geqslant t\}$$ and $\beta^p\lt 1$, then $$\lVert X\rVert_{p,\infty}\leqslant \frac{\beta}{(1-\beta^p\varepsilon)^{1/p}}\delta^{-1}\lVert Y\rVert_{p,\infty}.$$

To conclude the proof, we need to apply this lemma to $X:=\max_{1\leqslant i\leqslant n}|S_i|$ and $Y:=\max\left\{\left(\sum_{i=1}^n\mathbb E[X_i^2\mid\mathcal F_{i-1}]\right)^{1/2}, \max_{1\leqslant i\leqslant n}|X_i|\right\}$ with $\varepsilon:=\frac{\delta^2}{(\beta-\delta-1)^2}$, $\beta >1$ and $\delta$ small enough. For a fixed $\lambda$, define $$E_k:=\left\{\lambda<\max_{1\leqslant i \leqslant k-1}|S_i|\leqslant \beta\lambda;\max_{1\leqslant i\leqslant k-1}|X_i|\leqslant \delta\lambda;\sum_{i=1}^k \mathbb E[X_i^2\mid\mathcal F_{i-1}]\leqslant \delta^2\lambda^2\right\}$$ and $T_i:=\sum_{k=1}^i\chi_{E_k}X_k$. From the inclusion $$\{X>\beta\lambda;Y\leqslant \delta\lambda\}\subset \{\max_{1\leqslant i\leqslant n} |T_i|>(\beta-\delta-1)\lambda\},$$ the fact that $(T_n,\mathcal F_n)$ is a martingale and Doob's maximal inequality $$\mu\{X>\beta\lambda;Y\leqslant \delta\lambda\}\leqslant (\beta-\delta-1)^{-2}\lambda^{-2} \mathbb E(T_n^2).$$ We conclude by the estimate $$\mathbb E(T_n^2)\leqslant \delta^2\lambda^2\mu\left\{\max_{1\leqslant i\leqslant n}|S_i|\geqslant \lambda\right\}.$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.