Proof of Running Maximum of Brownian motion has continuous distribution without using the density or the fact that it is absolutely continuous

Question

I wanted to show that the running maximum say $\max_{t\in [0,1]}W_{t}$ has continuous distribution without taking help from the fact that it is absolutely continuous and has the distribution of $|W_{1}|$. To clarify, I want to just use the definition of Brownian motion and to prove this. (I am always talking about the standard brownian motion here).

Let us restrict to the set (of probability $1$ ) where $W_{t}$ is continuous.

Then I want to show that $P(\{\max_{t\in[0,1]}W_{t}=x\})=0$ for each $x\in\Bbb{R}$ .

So let us consider an enumeration of rationals $\{r_{n}\}_{n\in\Bbb{N}}$ in $[0,1]$ and define $Q_{n}=\{r_{1},...,r_{n}\}$ .

We have that $\max_{t\in[0,1]}W_{t}=\sup_{r\in\Bbb{Q}\cap[0,1]}W_{r}$ .

Also $\sup_{r\in\Bbb{Q}\cap[0,1]}W_{r}=x\implies W_{r}\leq x\,,\forall r\in\Bbb{Q}\cap[0,1]$ and for each $m\in\Bbb{N}$ , there exists $r\in\Bbb{Q}\cap[0,1]$ such that $W_{r}\in (x,x-\frac{1}{m}]$ .

So I want to write $\displaystyle P(\sup_{r\in[0,1]\cap\Bbb{Q}} W_{r}=x)\leq P\big(\bigcap_{m\in\Bbb{N}}\bigcup_{n\in\Bbb{N}}\{\exists r\in Q_{n}\,: W_{r}\in (x-\frac{1}{m},x]\}\big)=\lim_{m\to\infty}P(\bigcup_{n\in\Bbb{N}}\{\exists r\in Q_{n}: W_{r}\in(x-\frac{1}{m},x])$

$$\leq \lim_{m\to\infty}P\bigg(\bigcup_{n\in\Bbb{N}}\big\{\exists r\in Q_{n}:W_{r}\in (x-\frac{1}{m2^{n}},x]\big\}\bigg)\leq \lim_{m\to\infty}\lim_{n\to\infty}\sum_{k=1}^{n}P(W_{r_{k}}\in(x-\frac{1}{m2^{n}},x])$$

$$\leq \lim_{m\to\infty}\lim_{n\to\infty}\sum_{k=1}^{n}\frac{n}{m2^{n}}=0$$

However I am having trouble making this precise and I obviously am making a mistake in the step of considering $\lim_{m\to\infty}\lim_{n\to\infty}\sum_{k=1}^{n}P(W_{r_{k}}\in(x-\frac{1}{m2^{n}},x])$ . Can anyone tell me how do I make it correct?. Also is there an easier way to do this? (I mean without using the fact that it has a density).

Another way I tried to think of is that if I can show that $\max_{t\in[0,1]} W_{t}=\max_{r\in\Bbb{Q}\cap[0,1]}W_{r}$ (note that I mean $\max$ and not $\sup$) , then I can do it easily as for any finitely many points $r_{1},...,r_{n}$ , we have $P(\max_{r\in Q_{n}}W_{r}=x)\leq \sum_{k=1}^{n} P(W_{r_{k}}=x)=0$ for all $n$ . So I can proceed with the countable union and conclude what I want. But this is not necessarily true for continuous functions. I mean for example , $\sin(2x)$ achieves it's maximum at $x=\frac{\pi}{4}\in[0,1]$ and not at a rational point.

EDIT:

I also tried the following but I am unsure of it. Consider the sets $A_{n}=\bigcap_{m\in\Bbb{N}}\bigg\{\exists r\in Q_{n}:W_{r}\in(x-\frac{1}{m},x]\bigg\}$ . Then $A_{n}$'s increase to $\bigcap_{m\in\Bbb{N}}\bigg\{\exists r\in \Bbb{Q}\cap[0,1]:W_{r}\in(x-\frac{1}{m},x]\bigg\}$ .

But $\displaystyle P(A_{n})=P\bigg(\bigcap_{m\in\Bbb{N}}\bigg\{\bigcup_{r\in Q_{n}}W_{r}\in (x-\frac{1}{m},x]\bigg\}\bigg)=\lim_{m\to\infty}P\bigg(\bigcup_{r\in Q_{n}}\bigg\{W_{r}\in(x-\frac{1}{m},x]\bigg\}\bigg)$

(We could switch the intersection over $m$ to limit because for each fixed $n$, the sets $\{\exists r\in Q_{n}:W_{r}\in(x-\frac{1}{m},x]\}$ is a decreasing sequence)

$$\leq \lim_{m\to\infty} \sum_{k=1}^{n}\frac{1}{m}\frac{1}{\sqrt{r_{k}}}=0$$ (By using a simple bound for standard gaussian that $P(X\in(a,b))\leq (b-a)$ for $X\sim N(0,1))$

Also to note that the maximum is always $\geq 0$ a.s. . So we can ignore $r=0$ and consider $r\in\Bbb{Q}\cap(0,1]$ as $W_{0}=0$ a.s.

So $P(A_{n})=0\,,\forall n\in\Bbb{N}$ and hence $P(\bigcup_{n\in\Bbb{N}}A_{n})=0$ which is what is required.

Answer 1

We will prove (with our hands tied behind our back, as requested) that for each $x>0$, we have $P(\{\max_{t\in[0,1]}W_{t}=x\})=0$ (The case $x=0$ follows from time-inversion, for instance.)

It suffices to show that for each $\epsilon \in (0,1)$, the event $A^{\epsilon}_x:= \{\max_{t\in[\epsilon,1]}W_{t}=x\}$ has $P(A^{\epsilon}_x)=0$, since for $x>0$, $$\{\max_{t\in[0,1]}W_{t}=x\}=\cup_{m=2}^\infty A^{1/m}_x\,.$$ Let $H^{\epsilon}_z:=\{\max_{t\in[0,1-\epsilon]}W_{t}=z\}$ and let $\mu$ be the distribution of $W_\epsilon$. Then by independence of increments of Brownian motion, we have $$P(A^{\epsilon}_x)= \int_{\mathbb R} P(H^{\epsilon}_{x-y})\, d\mu(y) \,. \tag{*} $$

Since $P(H_{x-y}^\epsilon)$ could be positive for at most countably many $y$, and $\mu$ is a continuous measure, the integral in $(*)$ must vanish.

Answer 2

We recall the following facts.

• For any Brownian motion $W_t$, $\limsup_{t \to \infty} W_t = +\infty$ (this follows from showing that $\limsup_{n \in \mathbb N, n \to \infty} W_n = +\infty$, a proof of which can be found here).

• Time inversion : if $W_t$ is a standard BM then $B_t = tW_{\frac 1t}$ is also a standard BM.

• Time reversal : Notice the difference with the previous part. If $W_t$ is a Brownian motion and $T \in (0,\infty)$ is fixed, then $X_s = W_{T} - W_{T-s}$ is a Brownian motion for time $s \in [0,T]$.

• Sign inversion : If $B_t$ is a Brownian motion, so is $-B_t$.

• The strong Markov property : if $B$ is a Brownian motion and $\tau$ is a stopping time, then conditioned on $\{\tau<\infty\}$, the process $W_{t} = B_{\tau+t} - B_t$ is a Brownian motion started at the point $B_{\tau}$ and independent of the filtration $\mathcal F_{\tau}$ (generated by events that occurred "before" $\tau$) given the value of $X_{\tau}$.

Let us now use time inversion to make a local statement about the Brownian motion. As $\limsup_{t \to \infty} W_t = +\infty$ and $B_t = t W_{\frac 1t}$ is also a standard BM, $\limsup_{t \to \infty} tW_{\frac 1t} = +\infty$. Let $s = \frac 1t$, then $\limsup_{s \to 0^+} \frac{W_s}{s} = +\infty$. (all these statements occur with probability $1$, of course).

We claim now that $$\mathbb P(\limsup_{s \to 0^+} W_s \leq 0) =0$$ Indeed, if this event had non-zero probability, then $\frac 1s$ is positive for $s>0$, therefore $\{\limsup_{s \to 0^+} \frac{W_s}{s} \leq 0\}$ would also have non-zero probability, a contradiction to $\limsup_{s \to 0^+} \frac{W_s}{s} = +\infty$ a.s. This is a local statement asserting , in conjunction with the strong Markov Property, that from the right the Brownian motion ,even at a stopping time, is not going to be a local maximum.

Now, let's see how we would argue that for $x \geq 0$, the event $E_x = \{\sup_{t \in [0,1]} W_t = x\}$ has probability $0$, where $W_t$ is a standard Brownian motion. We will show that it is contained in a set of measure zero, and assuming completeness of the filtration, we will be done. Alternately, it is possible to express $E_x$ as a countable union of cylinder sets which I leave an interested reader to do.

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Let $\tau_x$ be the hitting time of $x$, $\tau_x = \inf\{t \geq 0 : W_t \geq x\}$. We claim that

$$ E_x \subset \left(\{\tau_x \in [0,1)\} \cup \{W_{\tau_x} = x\} \cup \left\{\limsup_{t \to \tau_x^+} W_t \leq x\right\}\right) \\ \bigcup \left(\{\tau_x = 1\} \cup \{B_{\tau_x} = x\}\right) $$

Indeed, suppose that $E_x$ occurs. Then the maximum of $W_t$ over $[0,1]$ is $x$. As $W_t, t \in [0,1]$ is a continuous function a.s., the maximum $x$ is attained. By definition of $\tau_x$, this implies that $\tau_x \in [0,1]$, and furthermore that $W_{\tau_x} = x$ since $W_{\tau_x}>x$ would contradict $x$ being the maximum.

Now, if $\tau_x \in [0,1)$ then $x$ is certainly a local maximum of $W_t$ "from the right" and this is reflected in the event $\limsup_{t \to \tau_x^+} W_t \leq x$. However, $\tau_x = 1$ could also happen with $B_{\tau_x} = x$, which we write as a separate event. Obviously, the "from the right" doesn't apply at $\tau_x=1$ because the supremum time range ends at $1$.

Handling the first event is quite easy. Suppose that the event $F_x = \{\tau_x \in [0,1)\} \cup \{W_{\tau_x} = x\}$ has non-zero probability. Then, conditioned on $F_x$, the process $B_t = W_{\tau_x+ t} - W_{\tau_x}, t \geq 0$ is a Brownian motion. However, because $W_{\tau_x} = x$, the equality $$ \left\{\limsup_{t \to \tau_x^+} W_t \leq x\right\} = \left\{\limsup_{s \to 0^+}B_s \leq 0\right\} $$ holds by making the change of variable $s = t-\tau_x$. This has probability zero, therefore we are done. (Of course, if $\mathbb P(F_x) = 0$ then we're done as well).

Note that $x=0$ is handled by the first part, we assume WLOG that $x>0$ in the second part if it is necessary.

What about the second part? Here, we use time reversal. If $\tau_x = 1$ then $W_t <x$ for $t<1$. As $B_{\tau_x} = x$, this means that $W_1-W_t > 0$ for $t \in (0,1)$ (on a set of non-zero probability). Let $s = 1-t$, then we get $W_1 - W_{1-s}>0$ for $s \in (0,1)$. This is a new Brownian motion say $B_s = W_{1} - W_{1-s}$, and we have $B_s > 0$ for $s \in (0,1)$.

That's clearly an event which should have probability $0$, which is easy to see because it implies that $\liminf_{s \to 0^+} B_s \geq 0$, and by taking the negative sign, $\limsup_{s \to 0^+} X_s \leq 0$ where $X_s = -B_s$ is also a Brownian motion on $(0,1)$. That has probability $0$ by what we know, concluding the argument.

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Perhaps this argument can be simplified in places, but it definitely avoids the joint relation of the process and running maximum which is required to make the conclusion that $|W_1|$ has the same distribution as the running maximum.