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I’ve got another martingale inequality that I would be grateful for a kickstart on.

Suppose $X_t$ is a local martingale such that $|X_t|$ and $\langle X_t\rangle\leq c$ $\forall$ $t\geq 0$ and for some constant $c\in\mathbb R$. I need to show that $$\mathbb{E}\left(\sup_{t\geq 0}X_t\right)^4\leq361\mathbb{E}\langle X\rangle_\infty^2.$$

How should I approach this? I asked a question of similar nature previously, so I suspect I will need Cauchy-Schwarz's, Doob's, and Grönwall's inequalities here. However, in the previous question, $X_t$ was given in SDE form, so applying the inequalities made sense — here, all I know is that it is a martingale, so I'm not too sure where to start. Please guide me, thank you!

UBM
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user107224
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  • Can you attach a link to the question of similar nature previously asked? – Sarvesh Ravichandran Iyer Oct 30 '20 at 02:51
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    The inequality is a particular case of the Burkholder-Davis-Gundy inequality. – saz Oct 30 '20 at 07:04
  • @saz hi, I just took a look at the inequality — it’s not very illuminating to me, where do the constant on both sides of the equation come from? i can understand it for p=2 but for a general p (as well as this case) what would be the course of action? – user107224 Oct 30 '20 at 12:46
  • @TeresaLisbon hi, the question I asked is here: https://math.stackexchange.com/q/3854669/320332 – user107224 Oct 30 '20 at 12:48
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    @user107224 Noted. As mentioned, your inequality is a case of the BDG inequality. However, I will see if I can present a proof specific to this case. – Sarvesh Ravichandran Iyer Oct 31 '20 at 09:17
  • @user107224: Is ${X_t }$ a continuous local martingale such that $X_0=0$? – UBM Nov 09 '20 at 19:14

1 Answers1

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I am assuming that $X$ is a continuous local martingale such that $X_0=0$ and that $\langle X \rangle$ is the quadratic variation process of $X.$

Applying Ito's formula with $f(x)=x^4$ we have $$|X_t|^4 = 4 \int_0^t X_t^3 dX_t + 6 \int_0^t X_t^2 d\langle X \rangle_t. \tag{1}$$

Since $X$ is bounded, $E \int_0^t|X_s|^6ds < \infty,$ so the process $\{ \int_0^t X^3_s dX_s, t \geq 0 \}$ is a martingale and therefore, $E \int_0^t X^3_s dX_s = 0.$

Thus, taking expectations in (1) $$E|X_t|^4 \leq 6 E[\sup_{0 \leq s \leq t} |X_s|^2 \langle X \rangle_t]$$ and by the Holder inequality $$E|X_t|^4 \leq 6 \left(E\sup_{0 \leq s \leq t} |X_s|^4\right)^{\frac{1}{2}} \left(E\langle X \rangle^2_t \right)^{\frac{1}{2}}. \tag{2}$$ Now (apart from the above) by Doob's martingale inequality $$E \sup_{0 \leq s \leq t}|X_s|^4 \leq \left(\frac{4}{3}\right)^4 E|X_t|^4. \tag{3}$$ Substituting (2) in (3) $$E \sup_{0 \leq s \leq t}|X_s|^4 \leq \left(\frac{4}{3}\right)^4 6 \left(E\sup_{0 \leq s \leq t} |X_s|^4\right)^{\frac{1}{2}} \left(E\langle X \rangle^2_t \right)^{\frac{1}{2}}.$$ Thus $$E \left( \sup_{0 \leq s \leq t}X_s \right)^4 \leq E \sup_{0 \leq s \leq t}|X_s|^4 \leq \left(\frac{4}{3}\right)^8 6^2 E\langle X \rangle^2_t.$$

UBM
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  • beautiful solution! thank you! – user107224 Nov 10 '20 at 16:56
  • just a quick general question — when are the absolute values necessary? usually for ito’s formula I hardly see absolute values put into place, plus the RHS if the first line doesn’t involve any, which confuses me slightly. – user107224 Nov 10 '20 at 20:49
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    @user107224: since $X^4_t \geq 0,$ they are not necessary. – UBM Nov 10 '20 at 21:44
  • another quick general question — is the line about the integral of X^3 wrt dX an isometry argument? Is there a general statement about the expectation of (the powers of) bounded martingales? – user107224 Nov 11 '20 at 14:58
  • also for the last line, should the first $\leq$ be an = since the supremum of a set of the squares of non-negative numbers is equal to the square of the supremum? – user107224 Nov 11 '20 at 15:29
  • see the edit for the first question. It's not related with the Ito isometry, it's because the integral (as a process) is a martingale and a martingale has constant expectation. – UBM Nov 11 '20 at 16:59
  • For the second question, the $\leq$ is correct, but if you prefer to put an $"="$ is also fine. It doesn't change the result anyway. The reason why I didn't wrote $"="$ is because you have to prove prove that the supremum is always going to be positive (which is the case because $X$ is a martingale that starts at $0$. However, the way I wrote it, I don't need to say anything about that and it's still correct. – UBM Nov 11 '20 at 17:11
  • I’m starting to overthink and worry about the martingale property now. Is the lebesgue integral of X_s^6 < infty because X_s is bounded, or because the quadratic variation is bounded? Likewise, is the Ito integral of X^3 a martingale because it is L2 bounded due to bounded quadratic variation? if you could explain a bit more about it that would be awesome! – user107224 Nov 11 '20 at 20:42
  • first question: because $X$ is bounded. Second question: have a look at my answer here: https://math.stackexchange.com/questions/3473273/to-show-martingale-what-other-conditions-do-you-need-to-check-if-the-process-ha/3473689#3473689 – UBM Nov 11 '20 at 22:11