I would like to proof the following statement:
Let $(X_t^1,\dots, X_t^n)$, $t \geq 0$, be independent brownian movements, with $\mu = 0$ and $(\sigma_1^2,\dots, \sigma_n^2)$, where $X_t = (X_t^1,\dots, X_t^n)$, $x_0 = (X_0^1,\dots, X_0^n)$ and $||x_0||<a$. ($||x|| = \sqrt{(x_1^2 + \dots + x_n^2})$).
Proof that the mean value of
$\tau = min \{t ≥ 0 : ||X_t|| = a\}$
is given by
$E[\tau] = \frac{a^2 - ||x_0||^2}{\sigma_1^2 + \dots + \sigma_n^2}$.
I have tried to use the distribution of $\tau$, but i didnt find the solution.
Note that $(X_t^j)^2-\sigma_j^2t$ is a martingale for every $j$ with respect to its standard filtration (which is a classic result on Brownian Motion - you can prove this directly from the Markov property of Brownian Motion). Accordingly, by independence, we have that $\|X_t\|^2-\sum_{j=1}^n \sigma_j^2t$ is a martingale with respect to the filtration $\mathcal{F}_t=\sigma(X_s)_{0\leq s\leq t}$. Thus,
$$ \mathbb{E}\left(\|X_t\|^2-\sum_{j=1}^n \sigma_j^2t\right)=\|X_0\|^2 $$
Accordingly, we see, for all $K>0$, by the Optional Sampling Theorem, that
$$ \mathbb{E} \left(\|X_{\tau\wedge K}\|^2-\sum_{j=1}^n \sigma_j^2 (\tau\wedge K)\right)=\|X_0\|^2 $$
Hence, $$ \mathbb{E} \tau\wedge K=\frac{\mathbb{E}\|X_{\tau\wedge K}\|^2-\|X_0\|^2}{\sum_{j=1}^n \sigma_j^2} $$
Now, as $K\to \infty$, we have $\mathbb{E} (\tau\wedge K)\to \mathbb{E}\tau$ by montone convergence, and we have $X_{\tau\wedge K}\to X_{\tau}$ a.s. and, furthermore, $\|X_{\tau\wedge K}\|^2\leq a^2$ since $\|X_0\|^2\leq a$. Accordingly, $\mathbb{E} \|X_{\tau\wedge K}\|^2\to \mathbb{E}\|X_{\tau}\|^2=a^2$ by dominated convergence. This establishes the desired.