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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.