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Proof: the $\max(B_t,0)$ is submartingale without using convex function with Jensen's inequality

To prove it using convex function and jensen's inequality, we know the max function is convex and it can be directly from the result of jensen's inequality. However i cannot prove it using that way by my professor asking,that is, i cannot use convex function and jensen's inequality that many textbook desired.

My idea: We should prove $E[(B_t)^{+}|\mathcal{F}_{s}] \geq (B_s)^{+}$, where $B_t^{+}$ is $\max{(B_t,0)}$ i defined, $\mathcal{F}_{s}$ is filtration, $B_t$ is Brownian motion.

It is trivial that $B_s <0$. We only should prove $B_s >0$.

And we can calculate $B_t=B_t-B_s+B_s$ and maybe write the conditional expectation by integral using density function and consider the range of integral etc.

My professor give me hint that finally we should use Komatsu's inequality without proof, for $a \geq 0$: \begin{align*} \frac{2e^{-\frac{a^2}{2}}}{\sqrt{a^{2}+4}+a} \geq \int_{a}^{\infty}e^{-\frac{x^2}{2}}dx \geq \frac{2e^{-\frac{a^2}{2}}}{\sqrt{a^{2}+2}+a} \end{align*}

I totally have no idea....if any one gives some hints, it will be great.

Thank you in advance.

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  • $\begingroup$ The signs in the inequality are not correct, they should be the opposite. Regarding your post, maybe you can start with the fact that : $E[\max(B_t,0)|\mathcal{F}_s] = E[\max(B_t-B_s,-B_s)|\mathcal{F}_s] + B_s$ and remark that $B_t-B_s$ is independent of $\mathcal{F}_s$ and $B_s$ is $\mathcal{F}_s-$mesurable. Then we will have $E[\max(B_t-B_s,-B_s)|\mathcal{F}_s] = \phi(B_s) + B_s$ where $\phi(y) = E[\max(X,-y)]$ where $X \sim N(0,t-s)$. $\endgroup$ – Sesame May 21 at 18:01
  • $\begingroup$ Sorry for my careless writing. And thanks for your idea, I will try it. $\endgroup$ – Kairi May 22 at 17:21

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