Let $W = \{W_t, t \geq 0\}$ be an $m$-dimensional Brownian motion. The process $( \|W_t\|, \mathcal{F}_t )_{t ≥ 0}$ is then a Bessel process, where $\|\cdot\|$ is the Euclidean norm in $R^m$.
Question. Using Statement 1 (with the function $h(x) = e^tx$ and $t>0$), prove that the inequality is true for every $s>0$ and all $x>\sqrt{ms}$:
$$P(\sup\limits_{t∈[0,s]} ||W_t||≥x) ≤ (\frac{sd}{ex^2})^{-m/2} e^{-x^2/(2s)}$$
Statement 1: Let $h: R \rightarrow R_+ $ be a non-decreasing convex function and $(X_n, \mathcal{F}_n)_{1 \leqslant n \leqslant N}$ be a submartingale. Then for any $ u \in R$ and $t>0$
\begin{equation} P( \max\limits_{1 \leqslant n \leqslant N} X_n \geqslant u ) \leqslant \mathbb{E} h(t X_n)/h(tu). \end{equation}
Solution:
- First, I need to prove that the process is a submartingale, that is, to prove the most important condition that:
$$E(\|W_t\| \, | \, \mathcal{F}_s)\geq \| W_s \|$$
And since $\mathcal{F}_s$ is a natural filtration, I need to show that
$$E(\|W_t\| \, | \, W_s)\geq \| W_s \|$$
We know that $\|W_t\|= \sqrt{W_1^2 +...+W_t^2}$
And this is where I got stuck.
- Then I was only able to substitute $h$ and get this expression:
$P(\max\limits_{0 \leqslant t \leqslant m} \|W_t\| \geq u)\leq \frac{E(e^{t^2\|W_t\|})}{e^{t^2 u}}$
I'm having trouble calculating the expectation of such complex functions, so I'm hoping for some advice.