When $X_n$ is a martingale we know $(X_n)^+$, i.e. the positive part of $X_n$, is also a submartingale. Is this true? It is easy to show that this is SUB-m.g. but I couldn't find a counter-example for martingale?
Thank you!
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$\begingroup$It is most certainly false.
The function $x \longmapsto x^+$ is convex so if $X_n$ is a martingale then $X_ n^+$ is a submartingale, and will not be a martingale unless $X_n \geq 0$ a.s. for all $n$ or $X_n \leq 0$ a.s. for all $n$.
Here we have used the conditional version of Jensen's inequality to show that for a convex function $\phi$ $$ E_n[\phi(X_{n+1})] \geq \phi(E_n[X_{n+1}]) = \phi(X_n) $$ almost surely (this is the submartingale property for the new process $\{\phi(X_n)\}_n$.
For an explicit example consider a symmetric random walk with $X_0 = 0$. Clearly $E[X_n^+]>0$ as $X_n^+$ is not a.s. zero, so the process $\{X_n^+\}_n$ cannot be a martingale as this would imply $E[X_n^+]=0$ for all $n$ (martingales have constant expectation).
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