# Brownian Motion and Reflection Principle

I am currently studying the Brownian motion and I am stuck with a problem related to the reflection principle.

What I am trying to calculate is the probability that a standard Brownian Motion $X_t$ returns to zero given that it starts in $X_{t_a} = a$ and ends in $X_{t_b} = b$. ($t_a,t_b,a,b > 0$)

That is:

$$P [X_t = 0 \hspace{1 mm}for \hspace{1 mm}some \hspace{1 mm} t∈[t_a,t_b]|X_{t_a} = a, X_{t_b} = b], \quad t_a,t_b,a,b > 0.$$

Any answer or comment is greatly appreciated, thanks!

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Your question is quite unclear, in particular its formulation in English and the formula do not match. Are you assuming that $X_0=a$ and asking for $P(T_0\lt T_b)$ where $T_x$ is the first hitting time of $x\ne a$? – Did Aug 30 '12 at 12:18
As did said, the question is unclear: incorrectly formulated I suppose. Is what you want to ask for $P[X_t=0\text{ for some } t\in[\alpha,\beta]|X_\alpha=a, X_\beta=b]$ for some $0<\alpha<\beta$ and $a,b>0$? You should also outline what you have tried and where you got stuck: e.g. how well do you understand the reflection principle. – Einar Rødland Aug 30 '12 at 14:15
The reflection principle lets you calculate $\mathbb P (\tau_0 < t_b, X_{t_b} > x)$ and related quantities from which the conditional distribution can be derived. There is a more direct argument based on change of measure for brownian bridge which can be found in D Siegmund's Sequential Analysis – mike Aug 30 '12 at 14:30
Thank you all and apologize for my mistake. Einar, you are right. What I’m trying to calculate is P[Xt=0 for some t∈[α,β]|Xα=a,Xβ=b] for some 0<α<β and a,b>0. I understand the basics of the reflection principle (ie: the distribution of Mt and the distribution of the stopping time given the Xt distribution). However I’m having problems with the starting point at Xta = a and ending point at Xtb = b. I also know that the solution is exp(-2ab/t), but I’m stuck with the proof. Thanks a lot!! – RCA Aug 30 '12 at 16:45
question edited – RCA Aug 30 '12 at 17:27

## 1 Answer

1. foregoing consideration: Since $$\mathbb{P}[\exists s \in [t_a,t_b]: X_s = 0|X_{t_a}=a, X_{t_b} = b] = \mathbb{P}[\exists s \in [t_a,t_b]: X_s = -a | X_{t_a}=0, X_{t_b}=b-a] = \mathbb{P}[\exists s \in [0,t_b-t_a]: B_s = -a|B_{t_b}=b-a]$$ we can consider instead the problem $$\mathbb{P}[\exists s \in [0,t]: B_s = -a|B_t=c]$$ where $(B_t)_{t \geq 0}$ is an arbitrary Brownian Motion, $c:=b-a$, $t:=t_b-t_a$. Moreover, $$\mathbb{P}[\exists s \in [0,t]: B_s = -a|B_t=c] = \mathbb{P} \left[ \inf_{s \leq t} B_s \leq -a |B_t = c \right] = \mathbb{P} \left[\sup_{s \leq t} B_s \geq a|B_t = -c \right]$$ using that $(B_t)_{t \geq 0}$ has continuous sample paths and $B \sim -B$.
2. calculation of the joint distribution of $(M_t,B_t)$ where $M_t := \sup_{s \leq t} B_s$: Let $y \geq x$ and $\tau_y := \inf \{t \geq 0; B_t = y\}$. Then $$\mathbb{P}[M_t \geq y,B_t \leq x] = \mathbb{P}[\tau_y \leq t, B_t \leq x] = \mathbb{E}(1_{[\tau_y \leq t]} \cdot 1_{[B_t \leq x]}) \\ = \mathbb{E}(1_{[\tau_y \leq t]} \cdot \mathbb{E}(1_{[B_t \leq x]}|\mathcal{F}_{\tau_y^+}))$$ using the tower property. Now we have $$\mathbb{E}(1_{[B_t \leq x]}|\mathcal{F}_{\tau_y^+})(w) = \mathbb{E}^{B_{\tau_y}(w)}(1_{(-\infty,x]}(B_{t-\tau_y(w)}) = \mathbb{P}^y[B_{t-\tau_y(w)} \leq x] \\ = \mathbb{P}[B_{t-\tau_y(w)} \leq x-y]\stackrel{B \sim -B}{=} \mathbb{P}[B_{t-\tau_y(w)} \geq y-x] \\ = \mathbb{P}^y[B_{t-\tau_y(w)} \geq 2y-x] = \mathbb{E}(1_{[B_t \geq 2y-x]}|\mathcal{F}_{\tau_y^+})(w)$$ for all $w \in [\tau_y \leq t]$ (In this part we used the reflection principle! See "René L. Schilling/Lothar Partzsch: Brownian Motion", Theorem 6.11.) Thus (using again the tower property) $$\mathbb{P}[M_t \geq y,B_t \leq x] = \mathbb{P}[M_t \geq y, B_t \geq 2y-x] \stackrel{y \geq x}{=} \mathbb{P}[B_t \geq 2y-x] \\ = \int_{2y-x}^\infty \frac{1}{\sqrt{2\pi \cdot t}} \cdot \exp \left(-\frac{z^2}{2t} \right) \, dz$$ By differentiation we obtain $$(M_t,B_t) \sim \underbrace{\frac{1}{\sqrt{2\pi \cdot t^3}} \cdot \exp \left(- \frac{(2y-x)^2}{2t} \right) \cdot 2 (2y-x)}_{=:f(y,x)} \, dy \, dx$$
3. Known formula: $$\mathbb{E}(h(Y,X)|X=x) = \int h(Y,x) \cdot \frac{f_{Y,X}(Y,x)}{f_X(x)} \, d\mathbb{P}$$ where $(Y,X) \sim f_{Y,X}(y,x) \, dy \, dx$, $X \sim f_X(x) \,dx$. Apply this to $Y:=M_t$, $X:=B_t$, $h(y,x) := 1_{(a,\infty]}(y)$. We know (from the 2nd step) that $$f_{Y,X}(y,x) = \frac{1}{\sqrt{2\pi \cdot t^3}} \cdot \exp \left(- \frac{(2y-x)^2}{2t} \right) \cdot 2 (2y-x)$$ By straight-forward calculations we get $$\mathbb{E}(1_{[a,\infty)}(M_t)|B_t = x) = \exp \left(\frac{2}{t} \cdot a \cdot (x-a) \right)$$

We conclude:

$$\mathbb{P}[\exists s \in [0,t]: B_s = -a|B_t=c] \stackrel{1}{=} \mathbb{P} \left[\sup_{s \leq t} B_s \geq a|B_t = -c \right] = \mathbb{E}(1_{[a,\infty)}(M_t)|B_t = -c) \\ \stackrel{3}{=} \exp \left(\frac{2}{t} \cdot a \cdot (-c-a) \right) \stackrel{c=b-a}{=} \exp \left(\frac{2}{t} \cdot a \cdot b \right)$$

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