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I am not sure that this is a proper question. I am looking for the verification of the proof rather than for an answer.

There is one inequality I use pretty often and I would like to be sure that it is correct. Namely, if $X_n$ is a non-negative supermartingale then $$ \mathsf P\left\{\sup\limits_n X_n\geq \delta\right\}\leq\frac 1 \delta\mathsf EX_0.\quad(1) $$

First I derived it from the version of Doob's inequality presented in the book "Optimal Stopping and Free-Boundary Problems" by Peskir and Shiryaev (2006). On the other hand here Doob's inequality in probability you can see a counterexample for the inequality stated in the book. Also it that topic mpiktas find a reference to the earlier book by Shiryaev with a weaker inequality, which I believe is correct.

Anyway, I have a proof of (1) and would like you to help me verify if it is strict or not.

Let $Y$ be a process with non-negative values. Then for all $N\in \mathbb{N}$ and $\delta>0$ holds $$ \mathsf P\left\{\sup\limits_{n\leq N} Y_n\geq \delta\right\}\leq\frac 1 \delta\sup\limits_{\tau\leq N}\,\,\,\mathsf EY_\tau\quad (2) $$ where $\tau$ is a stopping time with respect to the natural filtration of $Y$.

Proof: Take $\tau^*=\inf\{ n\geq 0:Y_n\geq\delta\}$ and $\tau = \tau^*\wedge N$. Now $\tau\leq N$ a.s. and $\tau$ is a stopping time with respect to the natural filtration of $Y$. Denote $$p = \mathsf P\left\{\sup\limits_{n\leq N} Y_n\geq \delta\right\},$$ so now $\mathsf{P}\{\tau = \tau^*\} = p$.

We have $$ \mathsf EY_\tau \geq \delta p $$ since $Y$ is a non-negative process, which leads to the inequality (2). End of the proof.

We can obtain (1) from (2) since $\sup\limits_{\tau\leq N} \,\,\,\mathsf EX_\tau = \mathsf E X_0$ if $X$ is a supermartingale. Taking $N\to\infty$ we obtain (1).

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This is one of the exercises from Continuous Martingales and Brownian Motion by D. Revuz and M. Yor. More precisely, Exercise 1.25 from Chapter II, called Maximal inequality for positive supermartingales. –  Ivan Jul 27 '13 at 19:13
    
@Ivan: feel free to post it as an answer, I have already found similar proofs in other sources, but it will take me time to recall where precisely. –  Ilya Jul 29 '13 at 8:03
    
Ilya, how did you conclude from $P\{\tau = \tau^*\} = p$ that $E \, Y_\tau \geq \delta \, p$? I see that one can apply Markov's inequality to the random variable $\sup_{n \leq N} Y_n$, but then we have $\delta \, p \leq E \, \sup_{n \leq N} Y_n$. Thank you. –  Ivan Sep 8 '13 at 6:35
    
@Ivan: when we stop $Y$ at $\tau$, there are two possibilities: either we stop at $\tau^*$, or we stop at $N$. In the former case, the value would be at least $\delta$, and it happens with probability $p$. In the latter case the value would be at least zero as the process is non-negative. Since $$ \mathsf E Y_\tau = \mathsf EY_\tau\cdot 1(\tau = \tau^*) + \mathsf EY_\tau\cdot 1(\tau \neq \tau^*) $$ we obtain that $\mathsf E Y_\tau \geq \delta\cdot p + 0\cdot (1-)p$. –  Ilya Sep 9 '13 at 7:51
    
I would be glad if someone knows a reference to prove the formula (2), posted in the first topic. Thanks! Niccolo –  user103845 Oct 28 '13 at 16:25

1 Answer 1

The problem is taken from "Continuous Martingales and Brownian Motion" by D. Revuz and M. Yor, Exercise 1.15 in Chapter II. The following proof is based on this question and that one.

Problem:

If $X$ is a right-continuous positive supermartingale, prove that \begin{equation} P\left( \sup_t X_t > \lambda \right) \leq \lambda^{-1} E \, X_0. \end{equation}

Proof:

Let $T_\lambda = \inf\{ s \geq 0: X_s \geq \lambda \}$. $T_\lambda$ is a stopping time with respect to the natural filtration of $X_t$. First, we note that $$ E \, X_{T_\lambda \wedge t} = E \, \left( X_{T_\lambda} \, 1_{\{ T_\lambda \leq t \}} \right) + E \, \left( X_t \, 1_{\{ T_\lambda > t \}} \right) \geq E \, \left( X_{T_\lambda} \, 1_{\{ T_\lambda \leq t \}} \right). $$ On the other hand, since $X_t$ is a supermartingale, and $T_\lambda \min t$ is an almost-surely-bounded stopping time, by the optional stopping theorem, $$ E X_{T_\lambda \wedge t} \leq E X_0. $$ Hence, $$ E \, \left( X_{T_\lambda} \, 1_{\{ T_\lambda \leq t \}} \right) \leq E X_0. $$ Since $X_t$ is a positive process, by Markov's inequality, $$ \lambda \, P \left( X_{T_\lambda} \, 1_{\{T_\lambda \leq t\}} \geq \lambda \right) \leq E \, \left( X_{T_\lambda} \, 1_{\{T_\lambda \leq t\}} \right). $$ At the same time, $$ \left\{ X_{T_\lambda} \, 1_{\{T_\lambda \leq t\}} \geq \lambda \right\} = \left\{ T_\lambda \leq t\ \right\} = \left\{ \sup_{s \leq t} X_s \geq \lambda \right\}. $$ Consequently, the desired result follows.


I would be grateful if anybody could give me any feedback. Thank you.

Regards, Ivan

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Neither $X_{T_\lambda} > \lambda$ nor $[T_\lambda \leq t ] = [ \sup_{0 \leq s \leq t} X_s > \lambda ]$ hold in general. –  Did Aug 31 '13 at 14:44
    
@Did, I have updated the proof. May I please ask you to explain what I am missing when I claim $\{ T_\lambda \leq t \} = \{ \sup_{s \leq t} X_s \geq \lambda \}$. Thank you. –  Ivan Sep 8 '13 at 9:26
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You changed an inequality sign from $\gt$ to $\geq$. Now the events are equal (because the paths of $X$ are continuous). –  Did Sep 8 '13 at 9:37

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