Hello, guys!

Can anybody give a hint on how to prove that (on continuous time setting with t in [0,T]) if M is a supermartingale such that E[M(T)] = M(0), then M is in fact a martingale.

Would appreciate any help. Thanks!

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- Jun 16th 2012, 09:43 AMwaytogowhen supermartingale is a martingale
Hello, guys!

Can anybody give a hint on how to prove that (on continuous time setting with t in [0,T]) if M is a supermartingale such that E[M(T)] = M(0), then M is in fact a martingale.

Would appreciate any help. Thanks! - Jun 16th 2012, 10:30 AMgirdavRe: when supermartingale is a martingale
In the definition of super martingale, look what happen if you assume that the inequality is strict (on a non-zero measure set) and you take expectation.

- Jun 16th 2012, 11:42 AMwaytogoRe: when supermartingale is a martingale
Tried to use your advice.

This is what I get:

$\displaystyle E_s\[M(T)\] < M(s) \Rightarrow E\[E_s\[M(T)\] \] < E\[M(s)\] \Rightarrow E\[M(T)\] < E\[M(s)\] \Rightarrow M(0) < E\[M(s)\]$

So if this is correct last inequality is a contradiction to M being supermartingale. So strict inequallity does not hold, but equality must hold.

Am I right now? - Jun 16th 2012, 11:46 AMgirdavRe: when supermartingale is a martingale
Yes, it's correct. Maybe in the first inequality you have to precise that it's true on a set of positive measure, and that the inequality is large outside the set (but not necessarily strict).

- Jun 16th 2012, 12:32 PMwaytogoRe: when supermartingale is a martingale
You mean I have to add that this holds for non-zero measure sets? And what does "inequality is large" mean?

- Jun 16th 2012, 12:34 PMgirdavRe: when supermartingale is a martingale
$\displaystyle \leq $ instead of $\displaystyle <$. (it hold for

*a*non-zero measure set).