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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.

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    $\begingroup$ "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
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Hint:

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 :)

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