I'm trying to prove that For a non-negative supermartingale $M$ it holds that for all $\lambda>0$ we have $$\lambda P\{\sup_{n}M_{n}\geq\lambda\}\leq E(M_{0})$$

My idea was to use Markov's inequality which states that $$\lambda P(M_{n}\geq\lambda)\leq E(M_{n})$$

As it holds for all $M_{n}$ it must also hold for the supremum of $M_{n}$ and using the supermartingale property $E(M_{n})\leq E(M_{0})$ one finds the desired result.

However I'm not sure if I can just say that it also holds for the supremum, could anyone help me out with this?

Help is much appreciated.

  • 2
    $\begingroup$ "As it holds for all Mn it must also hold for the supremum" is as wrong as can be. $\endgroup$ – Did Apr 14 '13 at 11:21
  • $\begingroup$ @Did, since Markov's inequality holds for any non-negative random variable, $M_n$ is a non-negative process, and $\sup_n M_n$ is also non-negative, why cannot we conclude that $P(\sup_n M_n \geq \lambda) \leq \frac{1}{\lambda} E \, \sup_n M_n$? Thank you. $\endgroup$ – Ivan Sep 8 '13 at 7:05
  • $\begingroup$ @Ivan We certainly can, but the next sentence in the post seems to indicate that the OP's idea was to conclude that $P(\sup_nM_n>λ)⩽(1/λ)\sup_nE(M_n)$ (which does not follow). On the other hand, using the (different, correct) argument in your comment, how to reach $E(M_0)$ in the RHS? $\endgroup$ – Did Sep 8 '13 at 8:28
  • $\begingroup$ @Did, in the first sentence in your response to my comment, you meant $E(M_0)$ on the right-hand side, right? I agree that, following this logic, it is not straightforward (not possible?) to reach the desired result; I just wanted to clarify your statement. Thank you for the answer. So, it seems, Lost1's solution is the way to go in this situation. $\endgroup$ – Ivan Sep 8 '13 at 8:50
  • $\begingroup$ This question is related to math.stackexchange.com/questions/47118/doob-like-inequality, and I am currently trying to clarify it for myself. $\endgroup$ – Ivan Sep 8 '13 at 8:54


Define a stopping time $T=\inf\{k: X_k\geq \lambda \}$

Write $\mathbb{E}X_{n\wedge T}=\mathbb{E}X_T1_{\{T\leq n\}} + \mathbb{E}X_n1_{\{T> n\}} $

Use optional sampling theorm: $\mathbb{E}X_{n\wedge T}\leq\mathbb{E}X_0$ drop something you don't want because it is positive.

fill in the details yourself :)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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