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Let X be a Levy process with no positive jumps and $\tau_y:=\inf\{t> 0: X_t > y\}$ then we have

$$X_{\tau_y}=y\text{ on }\{\tau_y <\infty\}.$$

Could you explain that why? and does it hold for Levy process with no negative jumps? If X be a Feller process with no positive jumps then does this hold? How about if we state with $T_y:=\inf\{t>0: X_t=y\}.$ Thank you!

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I don't think it's true unless you assume that $X_0\leq y$. Then, if you consider the process with $X_0\leq y$ and no positive jumps, the idea is that the process can reach $(y,\infty)$ only by the continuous dynamics (I cannot give a formal proof though). If the Levy process does not have negative jumps but can have positive, it is not true: take the Poisson process $N_t$ and put $y = \frac12$. For the case $$ T_y:=\inf\{t>0:X_t = y\} $$ the statement $X_{T_y} = y$ should hold for any process with trajectories continuous from the right. –  Ilya Feb 17 '12 at 14:13
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1 Answer

This has nothing to do with being a Lévy process or even with randomness.

Assume that the function $f:t\mapsto f(t)$ has no positive jump. Let $t_y=\inf\{t\gt0\mid f(t)\gt y\}$. Assume that $t_y$ is finite and that $y\gt f(0)$.

Then $f(t_y-h)\leqslant y$ for every $h\gt0$, by definition of $t_y$, hence $\limsup\limits_{s\to t,s\lt t}f(s)\leqslant y$. Since $f$ has no positive jumps, this implies that $f(t_y)\leqslant y$. On the other hand, if $f(t_y)\lt y$, then $f(t_y+h)\lt y$ for every $h\gt0$ small enough, otherwise $f$ would make a positive jump at time $t_y$. This is in contradiction with the definition of $t_y$ hence $f(t_y)=y$.

Likewise, $s_y=\inf\{t\gt0\mid f(t)=y\}$ yields $f(s_y)=y$, but $s_y\lt t_y$ is possible.

On the other hand, if $f$ has positive jumps, everything is possible, even that $f(t_y)\lt y$. If $f$ is càdlàg however, $f(t_y)\geqslant y$ but $f(t_y)\gt y$ may happen (consider the function entire part).

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A very nice and clear argument –  S.D. Feb 17 '12 at 19:49
    
@Didier Piau, thank you very much for your answer. I think that with the case $T_y=\inf\{t>0: X_t=y\}$ we need surely a condition that the point y is regular, i.e $P_{y}(T_y=0)=1$. But I do not know which the following cases are correct i) the process X have no positive jumps then $$X_{T_y}=y.$$ ii)the process X have no negative jumps then $$X_{T_y}=y.$$ Do you know about these? Thank you! –  Duy Son Son Feb 17 '12 at 20:43
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