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To prove that $(S_t-K)^{+}$ a supermartingale with $S_t$ strictly positive supermartinagale and $K$ a constant, I started like this :

$S_t$ is supermartinagale, thus : $$E[S_t|S_0] < S_0$$ $$E[S_t - K|S_0] < S_0 - K$$

at this point I'm trying to apply Jensen Inequality but seems like it doesn't lead me to the right direction. Any help?

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2 Answers 2

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(2017.09.16) Is it me or is the plague of silent revenge downvotes spreading? Yet one more...

The result cannot hold.

If it was true, for any $X$ and $Y$ integrable, the property $E(X\mid Y)\leqslant Y$ almost surely would imply that $E(X^+\mid Y^+)\leqslant Y^+$ almost surely, which in turn implies that $E(X^+\mid Y^+)=0$ almost surely on $[Y^+=0]=[Y\leqslant0]$, which is equivalent to the fact that $X\leqslant0$ almost surely on $[Y\leqslant0]$.

But this is absurd, as the case when $P(Y=0)=1$, $P(X>0)\ne0$, $E(X)\leqslant0$, shows.

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  • $\begingroup$ Sorry, I forgot to mention that $S_t$ is strictly positive. Could it be hold in that case ? I edited the post $\endgroup$
    – JasBeck
    Sep 15, 2017 at 7:30
  • $\begingroup$ Doesn't change a thing, since one uses $X=S-K$. $\endgroup$
    – Did
    Sep 15, 2017 at 7:34
  • $\begingroup$ Oh right thank you. I think I have wrong interpretation of my original problem. If you have a quick look at this paper, on page 3, last paragraph ("We emphasize the non-standard result ... "), it says that c_t is a supermartingale cause x_t is . I might be missing something. Any hints please? thank you pdfs.semanticscholar.org/ceae/… $\endgroup$
    – JasBeck
    Sep 15, 2017 at 8:02
  • $\begingroup$ Another problem? Then post another question! $\endgroup$
    – Did
    Sep 15, 2017 at 8:06
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    $\begingroup$ What the heck is the stupid downvote for?! +1 $\endgroup$
    – user9464
    Sep 25, 2017 at 14:34
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A counterexample is Brownian motion and $K=0$. Brownian motion is a martingale, hence also a supermartingale. But $E B_t^+ > 0$ for $t>0$ so $B_t^+$ is not a supermartingale.

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