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Let $X_t$ be a standard one dimensional Brownian motion. Let

$T = \inf\{t : X_t \in\{ 1,-1\} \} $ and $S = \inf\{ t : X_t \in\{ 1, -3\}\}$

a) Explain why $X_T$ and $T$ are independent.

b) Show that $X_S$ and $S$ are not independent.

For (a) it seems that independence follows somehow from the symmetry but I am not quite sure. It would have to be the case that $E[X_TT] = 0$, already know that $E[X_T] = 0$ being a standard Brownian motion. and $0 < E[T] <\infty$, I haven't shown this just assuming this based off it's similarity to Gamblers Ruin.

For (b) if it isn't independent then $E[SX_S] \neq 0 $ but I'm having trouble calculating. I tried to solve $E[SX_S \mid \mathcal{F}_{T} ] = E[SX_S - SX_T + SX_T]$ but I don't think this is what I want to do.

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Recall the following result which is a consequence of Wald's identities (see e.g. this question):

Let $(X_t)_{t \geq 0}$ be a one-dimensional Brownian motion. For $a,b>0$ define a stopping time $\tau$ by $$\tau := \inf\{t>0; X_t \in \{-a,b\}\}.$$ Then $$\mathbb{P}(X_{\tau}=-a) = \frac{b}{b+a} \qquad \mathbb{P}(X_{\tau}=b) = \frac{a}{b+a} \qquad \mathbb{E}(\tau) = ab. \tag{1}$$ In particular, $\mathbb{E}(X_{\tau})=0$.

  1. Yes, the symmetry of Brownian motion is the key to the independence of $T$ and $X_{T}$. If we set $$\tilde{T} := \inf\{t>0; -X_t \in \{-1,1\}\}$$ we clearly have $T=\tilde{T}$. As $-X$ is also a Brownian motion, we get \begin{align*} \mathbb{P}(T \leq t, X_{T}=a) &\stackrel{X\stackrel{d}{=}-X}{=} \mathbb{P}(\tilde{T} \leq t, -X_{\tilde{T}} = a) = \mathbb{P}(T \leq t, X_{T} = -a) \tag{2} \end{align*} for any $a \in \mathbb{R}$ and $t \geq 0$. Since, by $(1)$, $$\mathbb{P}(X_{T} =1) = \mathbb{P}(X_{T}=-1) = \frac{1}{2} \tag{3}$$ (in particular $X_{T}$ takes only the values $1$ and $-1$) we get \begin{align*} \mathbb{P}(T \leq t) &= \mathbb{P}(T \leq t, X_{T} = 1) + \mathbb{P}(T \leq t, X_{T}=-1) \\ &\stackrel{(1)}{=} 2 \mathbb{P}(T \leq t, X_{T}=1), \tag{4}\end{align*} i.e. $$\mathbb{P}(T \leq t, X_{T}=1) = \frac{1}{2} \mathbb{P}(T \leq t) \stackrel{(3)}{=} \mathbb{P}(X_{T}=1) \mathbb{P}(T \leq t).$$ Because of the symmetry, this also shows $$\mathbb{P}(T \leq t, X_{T}=-1)= \mathbb{P}(X_{T}=-1) \mathbb{P}(T \leq t).$$ As $X_{T}$ takes only the values $-1$ and $1$, this proves the independence of $T$ and $X_T$.

  2. It follows from the result at the beginning of my answer that $$\mathbb{E}(X_S)=0 \quad \mathbb{P}(X_S=-3) = \frac{1}{4} \quad \mathbb{P}(X_S = 1) = \frac{3}{4} \quad \mathbb{E}(S)=3. \tag{5}$$ It is not difficult to check that $M_t := X_t^3-3tX_t$ is a martingale and by the optional stopping theorem this implies $\mathbb{E}(M_{S})=0$, i.e. $$ \mathbb{E}(X_S^3) = 3 (-3) \mathbb{E}(S 1_{\{X_S=-3\}}) +3 \mathbb{E}(S 1_{\{X_S=1\}}). \tag{6}$$It follows from $(5)$ that $\mathbb{E}(X_S^3) = -6$. On the other hand we also have $$ \mathbb{E}(S 1_{\{X_S=-3\}}) + \mathbb{E}(S 1_{\{X_S=1\}}) = \mathbb{E}(S) \stackrel{(5)}{=} 3. \tag{7}$$ Solving the system of linear equations $(6)$, $(7)$ gives $$\mathbb{E}(S 1_{\{X_S=-3\}}) = \frac{15}{12} \qquad \mathbb{E}(S 1_{\{X_S=1\}}) = \frac{21}{12},$$ and so $$\mathbb{E}(S X_S) = - \frac{15}{4} + \frac{21}{12} = -2 \neq 0 = \mathbb{E}(S) \mathbb{E}(X_S)$$ which shows that $S$ and $X_S$ are not independent.

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