We want to prove that $$\mathbb P(\sup_{0\leq s\leq t}W_s\geq a)=2\mathbb P(W_t\geq a).$$

Here is the proof.

My mistakes are in the proof of $$\mathbb P(\sup_{0\leq s\leq t}W_s\geq a, W_t<a)=\frac{1}{2}\mathbb P(W_t\geq a).$$

So let $\tau_a=\inf\{t\geq 0\mid W_t=a\}$. A famous result says that $(X_t:=W_{t+\tau_a}-W_{\tau_a})_t$ is a Brownian motion independent of $\mathcal F_{\tau_a}$. Then, they say :

\begin{align*} \mathbb P\left(\sup_{0\leq s\leq t}W_s\geq a,W_t<a\right) =&\mathbb P\left(\sup_{0\leq s\leq t}W_s\geq a, X(t-\tau_a)<0\right) \\ &=\mathbb E[\mathbb E[\boldsymbol 1_{\{\sup_{0\leq s\leq t}W_s\geq a\}}\boldsymbol 1_{\{X(t-\tau_a)<0\}}\mid \mathcal F_{\tau_a}]]\\ &=\mathbb E[\boldsymbol 1_{\{\sup_{0\leq s\leq t}W_s\geq a\}}\mathbb E[\boldsymbol 1_{\{X(t-\tau_a)<0\}}\mid \mathcal F_{\tau_a}]]. \end{align*}

Q1) Why $\boldsymbol 1_{\sup_{0\leq s\leq t}W_s\geq a}$ is $\mathcal F_{\tau_a}-$measurable ? I agree that it's $\mathcal F_t-$measurable, but since a priori $\mathcal F_t$ is not in $\mathcal F_{\tau_a}$, I don't understand why we can take it out of the expectation.

Q2) (with several under questions)

  • After they say that $\mathbb E[\boldsymbol 1_{\{X(t-\tau_a)\}}\mid \mathcal F_{\tau_a}]=\frac{1}{2}$, and also, I don't understand why.

  • I agree that $(X_t)$ is a brownian motion, and so, indeed, $\mathbb P\{X_t\leq 0\}=\frac{1}{2}$, but here we have $X(t-\tau_a)$ (which is not clear what is this process... why should it be a Brownian motion ?) By the way, I have the impression that $t-\tau_a$ can be negative, and so $X_{t-\tau_a}$ could be not well defined. Indeed, it's weird to say that $t>\tau_a$ since $\tau_a$ is random, so we don't know pointwise what is it.


1 Answer 1


Since $$\left\{ \sup_{0 \leq s \leq t} W_s \geq a \right\} = \{\tau_a \leq t\} \tag{1}$$ and $\tau_a$ is $\mathcal{F}_{\tau_a}$-measurable, it follows that $$\left\{\sup_{0 \leq s \leq t} W_s \geq a \right\} \in \mathcal{F}_{\tau_a}.$$ This answers your first question. Re the 2nd one: You are right that one has to be careful with these computations. I would argue as follows: Since $(X_t)_{t \geq 0}$ is independent from $\mathcal{F}_{\tau_a}$ and $\tau_a$ is $\mathcal{F}_{\tau_a}$-measurable, it holds that $$\mathbb{E}(1_{\{\tau_a \leq t\}} 1_{\{X(t-\tau_a)<0\}} \mid \mathcal{F}_{\tau_a}) = g(\tau_a)$$ where $$g(s) := \mathbb{E}(1_{\{s \leq t\}} 1_{\{X(t-s)<0\}}),$$see the proposition in this answer. Clearly, $$g(s) = 1_{\{s \leq t\}} \mathbb{P}(X(t-s)<0) = \frac{1}{2} 1_{\{s \leq t\}},$$

and so

$$\mathbb{E}(1_{\{\tau_a \leq t\}} 1_{\{X(t-\tau_a)<0\}} \mid \mathcal{F}_{\tau_a})= \frac{1}{2} 1_{\{\tau_a \leq t\}}.$$

Taking expectation and using $(1)$ this gives

$$\mathbb{P}(\tau_a \leq t, X(t-\tau_a) <0) = \frac{1}{2} \mathbb{P}(\sup_{s \leq t} W_s \geq a).$$

Combining this with the calculations in your question, the assertion follows.

  • $\begingroup$ Just a small question, for $\mathbb E[\boldsymbol 1_{\tau_a\leq t}\boldsymbol 1_{X(t-\tau_a)}\mid \mathcal F_{\tau_a}]=g(\tau_a)$, can I argue as follow (even if the proposition of your link works, it's just by curiosity) : $$\mathbb E[\boldsymbol 1_{\{\tau_a\leq t\}}\boldsymbol 1_{\{X(t-\tau_a)<0\}}\mid \mathcal F_{\tau_a}]=\boldsymbol 1_{\tau_a\leq t}\mathbb E[\boldsymbol 1_{\{X(t-\tau_a)<0\}}\mid \mathcal F_{\tau_a}]=\boldsymbol 1_{\tau_a\leq t}\mathbb E[\boldsymbol 1_{\{X(t-\tau_a)<0\}}]=g(\tau_a)$$ where $g(s)=\boldsymbol 1_{s\leq t}\mathbb E[\boldsymbol 1_{X(t-s)}<0]$. $\endgroup$
    – John
    Aug 30, 2019 at 9:51
  • 1
    $\begingroup$ @John It's not a good idea to pull the indicator function outside the (conditional) expectation. The problem is that $X(t-\tau_a)$ is not well-defined if $\tau_a>t$. As long as the indicator function is inside the expectation, there is no problem because we only consider $\omega$ with $\tau_{a}(\omega) \leq t$ (... because the indicator function is zero for all other $\omega$) $\endgroup$
    – saz
    Aug 30, 2019 at 9:57
  • $\begingroup$ Good point. Thank you :) $\endgroup$
    – John
    Aug 30, 2019 at 10:09
  • $\begingroup$ You say that $\mathbb E[\boldsymbol 1_{T-\tau_a}<0\mid \mathcal F_a]$ is well define only if $\tau_a\leq T$. But here it's rather $$\boldsymbol 1_{\tau_a\leq T}\mathbb E[\boldsymbol 1_{\{X(T-\tau_a)\}}\mid \mathcal \tau_a].$$ So, it's rather cleat that $\tau_a\leq T$ (because if $\tau_a>T$, then the previous expression is $0$). So, the comment of john on August30'19 works, no ? (I don't get the problem). $\endgroup$
    – Walace
    May 9, 2020 at 14:06
  • $\begingroup$ @Walace No, it doesn't work that way. Note that the expression equals $$1_{\tau_a \leq T}(\omega) \mathbb{E}[ 1_{X(T-\tau_a)}(\omega') \mid \tau_a](\omega),$$ i.e. the conditional expectation is taking with respect to $\omega'$ (not with respect to $\omega$).... and $X(T-\tau_a)(\omega')$ is not well defined for $\omega'$ with $\tau_a(\omega')>T$. $\endgroup$
    – saz
    May 10, 2020 at 10:56

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