Let $X$ be a Brownian Motion with drift $\mu>0$ and $X^*$ its running maximum. Is $X^*$ a local martingale? If I construct the sequence of stopping times $\tau_k = \inf \{s >\tau_{k-1}, X(s)>X(\tau_{k-1}) \}$ and let $Y(t) = X(t) \mathbb{1} _{t \in [\tau_{k-1},\tau_{k})}$ I am almost there because $Y(t)=X^*(t)$. Is there a better (more direct or more elegant) proof or I am wrong?
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$X^{\ast}$ is not a martingale (except for the trivial case where the quadratic variation of $X-\mu t$ vanishes). This is because $X^{\ast}$ is an increasing process. – Vasily Melnikov Oct 01 '23 at 15:14
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Thanks Vasily. I am little confused: not every martingale is a local martingale (while every martingale is a local martingale) so it is not enough. I am more convinced that my statement is wrong and my argument is shiberish (using a an indirect argument with stop process I reach a contradiction) – Santiago Oct 03 '23 at 06:05
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nonnegative local martingales are supermartingales, showing that $X^{\ast}$ is a supermartingale if it is a local martingale. But in general the only continuous local martingales with finite variation are constant, which already shows $X^{\ast}$ must be constant to be a local martingale. – Vasily Melnikov Oct 03 '23 at 18:27