# Brownian motion independence from stopping time

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.

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.