Consider two Brownian motions $(W_t)_{t\ge 0}$ with starting point $x$ and $(W'_t)_{t\ge 0}$ with starting point $y$. Define $T:=\inf\{t\ge 0:W_t=0\}$, the first time when $W_t$ is equal to $0$. Show that for every $x,y,t>0$ it holds:

$$P(W_{t\wedge T}\le y \mid \{W_0=x\})=P(|W'_t|\ge x \mid \{W'_0=y\}).$$

My attempt:

Rewrite $W'_t=B'_t+y$ for a standard Brownian motion $(B'_t)_{t\ge 0}$. Then $P(W'_0=y)=1$ and $1_{\{W'_0=y\}}=1_\Omega$, such that

\begin{align} P(|W'_t|\ge x \mid \{W'_0=y\})&=\frac{1}{P(W'_0=y)}E\Big[1_{\{|W'_t|\ge x\}}1_{\{W'_0=y\}}\Big]=P\big(|B_t'+y|\ge x\big) \end{align}

Rewrite $W_t=B_t+x$ for a standard Brownian motion $(B_t)_{t\ge 0}$. Then $P(W_0=x)=1$ and $1_{\{W_0=x\}}=1_\Omega$, such that

\begin{align} P(W_{t\wedge T}\le y \mid \{W_0=x\})&=\frac{1}{P(W_0=x)}E\Big[1_{\{W_{t\wedge T}\le y\}}1_{\{W_0=x\}}\Big]=P\big(B_{t\wedge T}+x\le y\big)\\ \end{align}

Here I get stuck. Maybe I have to use $0\le B_{t\wedge T}+x$ and the reflection principle? I would really appreciate some help! Thanks in advance!

  • $\begingroup$ What does $t \wedge T$ mean? For some $t<T$ ? I guess you assume that $x>y$... $\endgroup$ – Diger Jan 24 at 1:48
  • $\begingroup$ $t\wedge T=\min(t,T)$ $\endgroup$ – user408858 Jan 24 at 1:49
  • $\begingroup$ Never seen it, thanks. $\endgroup$ – Diger Jan 24 at 1:50
  • $\begingroup$ Where did you encounter this problem? $\endgroup$ – saz Jan 24 at 7:54
  • $\begingroup$ Probably you are telling me that you need rigorous proof, but from a heuristic perspective isn't it fairly clear that this is so, since the Normal-distribution is symmetric i.e. it doesn't matter whether you go left (from x to y) or right (from y to x); they appear with equal probability? $\endgroup$ – Diger Jan 24 at 12:55

Let's write the assertion in term of a standard Brownian motion $(B_t)_{t \geq 0}$.

If we set $T_{x} = \inf\{t \geq 0; B_t = -x\}$ then we need to show that

$$\mathbb{P}(x+ B_{t \wedge T_{x}} \leq y) = \mathbb{P}(|B_t+y| \geq x) \tag{1}$$

for any $x,y,t>0$.

Fix $x,y,t>0$. Since

$$B_{t \wedge T_x} = \begin{cases} -x & T_x \leq t, \\ B_t, & T_x>t \end{cases}$$

we have

$$\mathbb{P}(x+B_{t \wedge T_x} \leq y) = \mathbb{P}(T_x \leq t) + \mathbb{P}(x+B_t \leq y, T_x>t). \tag{2}$$

Since the reflection principle allows us to calculate explicitly the joint distribution $(B_t,\inf_{s \leq t} B_s)$ (see below), we get

\begin{align*}\mathbb{P}(x+B_t \leq y, T_x>t) &= \mathbb{P} \left( B_t \leq y-x, \inf_{s \leq t} B_s>-x \right) \\ &= - \frac{2}{\sqrt{2\pi t^3}} \int_{-x}^0 \int_{u \leq z \leq y-x} (2u-z) \exp \left(- \frac{(2u-z)^2}{2t} \right) \, dz \, du \\ &= \frac{2}{\sqrt{2\pi t}} \int_{-x}^0 \left[-\exp \left(- \frac{(2u-y+x)^2}{2t} \right) + \exp \left( -\frac{u^2}{2t} \right)\right] \, du.\end{align*}

Using the linearity of the integral to split the integral on the right-hand side into two parts and performing a simple change of variables we get

$$\mathbb{P}(x+B_t \leq y, T_x>t) = \sqrt{\frac{2}{\pi t}} \int_{\mathbb{R}} \left(- \frac{1}{2} \cdot 1_{\{|u-y| \leq x\}} + 1_{[-x,0]}(u) \right) \exp \left(- \frac{u^2}{2t} \right) \, du.$$

On the other hand, the reflection principle also gives

$$\mathbb{P}(T_x \leq t) =2 \mathbb{P}(B_t' \leq -x) =\sqrt{\frac{2}{\pi t}} \int_{-\infty}^{-x} \exp \left( - \frac{u^2}{2t} \right) \, du \tag{3}$$

Adding both expressions and using that

$$\sqrt{\frac{2}{\pi t}} \int_{-\infty}^0 \exp \left(- \frac{u^2}{2t} \right) \, du = \frac{1}{\sqrt{2\pi t}} \int_{\mathbb{R}} \exp \left(- \frac{u^2}{2t} \right) \, du = 1,$$

it follows from $(2)$ that

$$\begin{align*} \mathbb{P}(x+B_{t \wedge T_x} \leq y) = 1- \frac{1}{\sqrt{2\pi t}} \int_{|z-y| \leq x} \exp \left( - \frac{u^2}{2t} \right) \, du&= \mathbb{P}(|B_t+y|>x) \\ &= \mathbb{P}(|B_t+y| \geq x) \end{align*}$$

which proves the assertion.

Some remarks on the reflection principle: The reflection principle states (...or implies, depending on the formulation...) that the running supremum $M_t := \sup_{s \leq t} B_s$ satisfies $$M_t \stackrel{d}{=} |B_t|.$$ Applying this result to the Brownian motion $(-B_t)_{t \geq 0}$ it follows immediately that $$- \inf_{s \leq t} B_s \sim |B_t|.$$ In particular, $$\mathbb{P}(T_x \leq t) = \mathbb{P} \left( \min_{s \leq t} B_s \leq x \right) = \mathbb{P}(|B_t| \leq -x) = 2 \mathbb{P}(B_t \leq x)$$ for the stopping time $$T_x := \inf\{t \geq 0; B_t = -x\}$$ from the first part of this answer. Combining the reflection principle with the strong Markov property of Brownian motion, it is possible to compute the joint density of $(B_t,M_t)$:

Let $(B_t)_{t \geq 0}$ be a Brownian motion. The joint distribution $(B_t,\sup_{s \leq t} B_s)$ equals $$\mathbb{P} \left[ B_t \in dx, \sup_{s \leq t} B_s \in dy \right] = \frac{2 (2y-x)}{\sqrt{2\pi t^3}} \exp \left(- \frac{(2y-x)^2}{2t} \right) 1_{[-\infty,y]}(x) \, dx \, dy. $$

For a proof see e.g. René Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Exercise 6.8 (there are full solutions available on the web), or this question (and differentiate in order to obtain the density function).

As before, we can use the symmetry of Brownian motion and apply the result to $(-B_t)_{t \geq 0}$ in order to get the joint distribution of $(B_t,\inf_{s \leq t} B_s)$.

  • $\begingroup$ I have problems understanding the reflection principle. I do not see how $P(T_x\le t)=2P(B_t'\le-x)$ does hold. Also I do not understand how to calculate the joint distribution, since there is an infimum involved. I only can rewrite $P(\inf B_s>-x)=P(\sup B_s<x)=1-2P(B_t>x)$ with the reflection principle? But how do I conclude from there on? Thank you already for your very good answer! $\endgroup$ – user408858 Jan 24 at 19:59
  • $\begingroup$ @user408858 Which version of the reflection principle do you know...? $\endgroup$ – saz Jan 24 at 20:18
  • $\begingroup$ I know, that $B_t'=B_{t\wedge T}-(B_t-B_{t\wedge T})$ is also a brownian motion. I was also looking up the article on wikipedia where it is stated, that $P(\sup_{s\in[0,t]} B_s\ge a)=2P(B_t\ge a)$, see en.wikipedia.org/wiki/Reflection_principle_(Wiener_process) $\endgroup$ – user408858 Jan 24 at 20:25
  • $\begingroup$ @user408858 I see... well, I will add some comments on that tomorrow... $\endgroup$ – saz Jan 24 at 20:27
  • $\begingroup$ Well, I think I already can answer my first question: $P(T_x\le t)=P(\inf B_s\le -x)=2P(x\le B_t)$ $\endgroup$ – user408858 Jan 24 at 20:50

Perhaps we may go for the proof as follows.

First, note that \begin{align} &\mathbb{P}(W_{t\wedge T}\le y|W_0=x)\\ &=\mathbb{P}(W_t\le y|W_0=x,T>t)\,\mathbb{P}(T>t)\\ &+\mathbb{P}(W_T\le y|W_0=x,T\le t)\,\mathbb{P}(T\le t)\\ &=\mathbb{P}(W_t\le y|W_0=x,T>t)\,\mathbb{P}(T>t)+\mathbb{P}(T\le t)\tag{1}, \end{align} where $\mathbb{P}(W_T\le y|W_0=x,T\le t)=1$ as $W_T=0<y$ always holds.

Second, note that \begin{align} \mathbb{P}(W_t\le y|W_0=x)&=\mathbb{P}(W_t\le y|W_0=x,T>t)\,\mathbb{P}(T>t)\\ &+\mathbb{P}(W_t\le y|W_0=x,T\le t)\,\mathbb{P}(T\le t). \end{align} Substitute this result into $(1)$, and we obtain \begin{align} &\mathbb{P}(W_{t\wedge T}\le y|W_0=x)\\ &=\mathbb{P}(W_t\le y|W_0=x)+\mathbb{P}(T\le t)\\ &-\mathbb{P}(W_t\le y|W_0=x,T\le t)\,\mathbb{P}(T\le t)\\ &=\mathbb{P}(W_t\le y|W_0=x)\\ &+\mathbb{P}(W_t>y|W_0=x,T\le t)\,\mathbb{P}(T\le t)\tag{2}. \end{align}

Next, note that $$ \mathbb{P}(W_t>y|W_0=x,T\le t)\,\mathbb{P}(T\le t)=\mathbb{P}(W_t>y,T\le t|W_0=x) $$ denotes the probability of a Brownian motion that starts from $x$, hits $0$ at time $T$, and then goes beyond $y$ at time $t$. By the reflection principle, it follows that $$ \mathbb{P}(W_t>y|W_0=x,T\le t)\,\mathbb{P}(T\le t)=\mathbb{P}(W_t<-y|W_0=x). $$ Substitute this result into $(2)$, and we obtain \begin{align} \mathbb{P}(W_{t\wedge T}\le y|W_0=x)&=\mathbb{P}(W_t\le y|W_0=x)\\ &+\mathbb{P}(W_t<-y|W_0=x).\tag{3} \end{align}

Thereafter, by $W_t'=W_t-x+y$, we have \begin{align} &\mathbb{P}(W_{t\wedge T}\le y|W_0=x)\\ &=\mathbb{P}(W_t'\le 2y-x|W_0'=y)+\mathbb{P}(W_t'<-x|W_0'=y)\\ &=\mathbb{P}(W_t'\ge x|W_0'=y)+\mathbb{P}(W_t'<-x|W_0'=y),\tag{4} \end{align} where $\mathbb{P}(W_t'\le 2y-x|W_0'=y)=\mathbb{P}(W_t'\ge x|W_0'=y)$ due to the fact that $2y-x$ and $x$ are symmetric with respect to $y$.

Finally, note that $x>0$, and $(4)$ is obviously the desired equality $$ \mathbb{P}(W_{t\wedge T}\le y|W_0=x)=\mathbb{P}(\left|W_t'\right|\ge x|W_0'=y). $$ This completes the proof.


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.