If $(B_t)_{t\ge 0}$ is a Brownian motion and $\tau$ is a stopping time, then the stopped process $(B_{\min(\tau,t)})_{t\ge 0}$ is integrable

Question

Let $B=(B_t)_{t\ge 0}$ be a Brownian motion on a probability space $(\Omega,\mathcal A,\operatorname{P})$. By definition $B_t$ is normally distributed with mean $0$ and variance $t$.

Now, let $\mathbb F=(\mathcal F_t)_{t\ge 0}$ be a filtration on $(\Omega,\mathcal A)$ and $\tau$ be a $\mathbb F$-stopping time. Is the stopped process $B^\tau:=(B_{\min(\tau,t)})_{t\ge 0}$ integrable, i.e. does it hold $$B_{\min(\tau,t)}\in\mathcal{L}^1(\operatorname{P})\;\;\;\text{for all }t\ge 0\;?\tag{1}$$

I assume, that $(1)$ can be easily shown and most likely $B_{\min(\tau,t)}$ is normally distributed, too. However, how can we prove that?

Maybe we need boundedness of $\tau$. Then, there exists a $T>0$ such that $$B_{\min(\tau,t)}=B_T\;\;\;\text{for all }t>T$$ and $B_{\min(\tau,t)}$ would be normally distributed with mean $0$ and variance $T$ for all $t>T$.

But that is not exactly the desired statement $(1)$.

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Remark: $\;\;\;$ $B$ is called Brownian motion $:\Leftrightarrow$ $B$ is a real-valued stochastic process on $(\Omega,\mathcal A,\operatorname{P})$ with

• $B_0=0$ almost surely
• $B$ has independent and stationary increments
• $B_t\;\tilde\;\mathcal{N}_{0,\;t}$
• $B$ is almost surely continuous

Please note, that out there exists a definition of a so-called $\mathbb{G}$-Brownian motion, where $\mathbb{G}$ is a given filtration on $(\Omega,\mathcal{A})$. One can show, that the Brownian motion defined above is a $\mathbb{G}$-Brownian motion, if $G$ satisfies the usual conditions.

Answer 1

Doob says: $B_{\tau \wedge t} = E[B_{t \wedge t}\mid \mathcal{F}_\tau]$.

Jensen says: $$E[\lvert B_{\tau \wedge t} \rvert] = E[\lvert E[B_{t \wedge t}\mid \mathcal{F}_\tau]\rvert] \leq E[ E[\lvert B_{t \wedge t}\rvert\mid \mathcal{F}_\tau]] = E[\lvert B_t \rvert] < \infty$$

Answer 2

There is an answer to this once one knows the density of the signed running minimum/maximum of a reflected Brownian motion, $$M_t^{\pm} = \sup_{0 < s < t}\pm B_s \lor 0.$$ Clearly $$|B_{\tau \land t}| \leq M_t^+ + M_t^-$$ almost surely. In fact, for each $t,$ $M_t^\pm \overset{d}{=} |B_t|.$ (See Karatzas and Shreve, for instance). Then $$E|B_{\tau \land t}| \leq EM_t^+ + EM_t^- < \infty.$$