want to estimate the decay:
$ \mathbb{E} |e^{i\cdot B_{\frac{1}{k}}}-e^{-\frac{1}{2k}}|$, when $ k \to \infty $
where $ B_t$ is one dim standard Brownian Motion.
is it possible that it has exponential decay, say,
$ \mathbb{E} |e^{i\cdot B_{\frac{1}{k}}}-e^{-\frac{1}{2k}}|\le e^{-k} ?$
Thanks in advance!
This is only a sketch, I believe you can fill in the technical details.
It's the same as asking about $\mathbb{E} |e^{i\cdot B_{t}}-e^{-t/2}|$ as $t\to 0$. Even more tractable expression is $\mathbb{E} |e^{is X}-e^{-s^2/2}|$, where $X$ is standard normal: $$ \mathbb{E} \bigl|e^{is X}-e^{-s^2/2}\bigr| = \int_{-\infty}^{\infty} \bigl|e^{is x}-e^{-s^2/2}\bigr|e^{-x^2/2} \frac{dx}{\sqrt{2\pi}} = \\ = \int_{-\infty}^{\infty} \bigl((\cos sx -e^{-s^2/2})^2 + \sin^2 sx\bigr)^{1/2}e^{-x^2/2} \frac{dx}{\sqrt{2\pi}} $$ Note that $\cos sx - e^{-s^2/2} = O(s^2)$, $s\to 0$, (the expression $e^{-s^2/2}$ is not very relevant; $1$, $e^{s^2}$ or $\cos s^2$ would be as good), $\sin^2 sx\sim s^2x^2$, $s\to 0$. Therefore, $$ \mathbb{E} \bigl|e^{is X}-e^{-s^2/2}\bigr| \sim |s|\int_{-\infty}^\infty |x|e^{-x^2/2} \frac{dx}{\sqrt{2\pi}} = |s|\sqrt{\frac2\pi},\ s\to 0. $$