# when supermartingale is a martingale

• Jun 16th 2012, 09:43 AM
waytogo
when 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 AM
girdav
Re: 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 AM
waytogo
Re: when supermartingale is a martingale
This is what I get:

$E_s$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 AM
girdav
Re: 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 PM
waytogo
Re: 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 PM
girdav
Re: when supermartingale is a martingale
$\leq$ instead of $<$. (it hold for a non-zero measure set).