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I am trying to solve the following question: denote $(X_t,Y_t)$ a Brownian motion in $\mathbb{R}^n\times\mathbb{R}$ starting from $(0,a)$ with $a>0$. Let $T_a=\inf\{t:Y_t=0 \}$. Prove that the characteristic function of $X_{T_a}$ is $\exp(-a \Vert u\Vert)$. All I can prove is that by independence, $$\mathbb{E}[e^{ i\langle u,X_{T_a}\rangle}]=\mathbb{E}[e^{ i\sum_j u_jX^j_{T_a}}]=\mathbb{E}[\prod_je^{ i u_jX^j_{T_a}}]=\prod_j\mathbb{E}[e^{ i u_jX^j_{T_a}}]=\prod_je^{-a|u_j|}=e^{-a\sum_j|u_j|}$$ And obviously, $$e^{-a\sum_j|u_j|}\neq e^{-a \Vert u\Vert}$$ What am I doing wrong ?

Onstack01
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    The coordinate processes $(X_t^j){t \geq 0}$ are independent but the random variables $X{\tau_a}^j$, $j=1,\ldots,n$ are not independent... they all depend on $\tau_a$ (and hence on $(Y_t)_{t \geq 0}$). This means that $\mathbb{E}(\prod \ldots) \neq \prod \mathbb{E}(\ldots)$. – saz Feb 23 '20 at 08:09
  • I am not sure I understand: for all deterministic $t$, the $X^i_t$ are independent. But if $T_a$ is made random, the $X_{T_a}$ become dependent on each other ? We could clearly have a covariance matrix with zeros everywhere but non zero values only in the line and column corresponding to $Y$ so that the fact they are all depending on $Y$ does not mean they are not independent between themselves. – Onstack01 Feb 23 '20 at 13:21
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  • If, say $(U_t){t \geq 0}$ and $(V_t){t \geq 0}$ are two independent processes and $\tau$ is some stopping time, then $U_{\tau}$ and $V_{\tau}$ need not be independent. That's why yor calculation doesn't work. 2) Note that independence can be only characterized via covariances if we are dealing if Gaussian random variables/vectors. If $(U_t){t \geq 0}$ is Gaussian and $\tau$ some stopping time, then $U{\tau}$ is, in general, not Gaussian.
    • – saz Feb 23 '20 at 14:06
  • This is really interesting. I had never realized your last point. Thank you. – Onstack01 Feb 23 '20 at 14:25