# Inequality with local martingale (find sharp estimate for $\mathbb{P}(M_t\geq \alpha)$)

I am considering a local martingale $$M_t$$ with continuous sample paths (i.e. $$M_t\in\mathcal{M}_{C,\,\text{loc}}$$) and with quadratic variation $$\langle M\rangle_t$$ (and also $$M_0=\langle M\rangle_0$$ a.s.). I am required to find a sharp estimate for $$\mathbb{P}(M_t\geq \alpha)$$ for $$\alpha\in\mathbb{R}$$. I am told to consider two points (with $$\beta>0$$):

i) Prove that if $$\langle M\rangle_t\leq t$$, then $$\mathbb{E}\left(\mathrm{e}^{\beta M_t-\frac12\beta^2t}\right)\leq1$$; and

ii) Prove that $$\mathbb{P}\left(\mathrm{e}^{\beta M_t}>\mathrm{e}^{\alpha\beta}\right)\leq \mathrm{e}^{\frac12\beta^2t-\alpha\beta}$$. (Typo corrected, thanks @UBM !)

Point i) suggests that somewhere in the proof, I must claim that non-negative local martingales are supermartingales, but I am not sure how to go about using this claim. Point ii) suggests to me that I should be doing some form of Markov inequality with i) (although I can't really see it) and minimising the right hand side with respect to $$\beta$$, is this correct? Any guidance is greatly appreciated.

• could it be a typo and it is $- \frac{1}{2} \beta^2 \langle M \rangle_t$ instead of $- \frac{1}{2} \beta^2 t$?
– UBM
Dec 22, 2020 at 16:31
• @UBM ah I think I’m meant to achieve that assuming $\langle M\rangle_t\leq t$, I’ll edit that. Dec 22, 2020 at 17:38
• Could it be a typo and be $\mathbb{P}\left(\mathrm{e}^{\beta M_t}>\mathrm{e}^{\alpha\beta}\right)\leq \mathrm{e}^{\frac12\beta^2t -\alpha\beta}$ instead of $\mathbb{P}\left(\mathrm{e}^{\beta M_t}>\mathrm{e}^{\alpha\beta}\right)\leq \mathrm{e}^{\frac12\beta^2t}-\alpha\beta$?
– UBM
Dec 23, 2020 at 15:43

The Doleans-Dade exponential of the process $$\beta M^t, 0 \leq t \leq T$$ is
$$\mathscr E (M)_t:=\exp \left(\beta M_t -\frac{1}{2} \beta^2\langle M \rangle_t \right), 0 \leq t \leq T$$ which solves the SDE $$\mathscr E (M)_t = 1 + \int_0^t \mathscr E (M)_s dM_s.$$ Thus, since $$M$$ is a continuous local martingale, so is $$\mathscr E (M).$$ Also note that is an exponential so it is a non-negative process and therefore a supermartingale. Since $$E\mathscr E (M)_0 = 1,$$ we must have $$E\mathscr E (M)_t \leq 1 \quad \text{ for all } t \in [0,T]. \tag{1}$$ So part $$(i)$$ follows from (1) and the hypothesis $$\langle M \rangle_t \leq t.$$
For part $$(ii),$$ first note that part $$(i)$$ is equivalent to $$E[e^{\beta M_t}] \leq e^{\frac{1}{2}\beta^2 t} \tag{2}.$$
Thus, using the Markov inequality and (2) we have $$P(e^{\beta M_t} > e^{\alpha \beta}) \leq \frac{E[e^{\beta M_t}]}{e^{\alpha \beta}} \leq e^{\frac{1}{2} \beta^2 t} e^{-\alpha \beta}.$$