Let $T_{a}$ denote the first time the Brownian motion process hits $a$. When $a>0$, then $P\{X(t)\ge a|T_{a}\le t\}=\frac{1}{2}$

Question

Let $T_{a}$ denote the first time the Brownian motion process hits $a$. When $a>0$, then $P\{X(t)\ge a|T_{a}\le t\}=\frac{1}{2}$

I cannot see how it can be true, anyone could help me? Thanks very much.

Answer 1

The idea is as follows. Consider some realisation of the process. Since $T_a\leq t$, the process must have hit $a$ by time $t$, at some $T_a$. Now imagine the process behaviour from $T_a$ to $t$. Starting with $B(T_a)$, by symmetry, it is just as likely to move below $a$ or above $a$. More precisely:

$$P\{X(t)\geq a|T_a\leq t\}=P\{X(T_a+(t-T_a))-a\geq 0|T_a\leq t\}=P\{X(T_a+(t-T_a))-X(T_a)\geq 0|T_a\leq t\}=\frac12.$$

To see why this is the case, we might think that $X(t)-a=X(t)-X(T_a)\sim\mathcal{N}(0,t-T_a)$, which is symmetric, and hence gives $\frac12$. The only problem here is that $T_a$ is random. However, $X(T_a+s)-X(T_a)$ is actually a Brownian motion for $t\geq 0$; this result follows from the property of strong independent increments or strong Markov property (see, e.g., Adventures in Stochastic Processes by S. Resnick, page 504).

Answer 2

One can also use reflection principle for Wiener process to find out the probability. Define the reflected Wiener process $Y(t) := X(t)1_{t < T_a} + (2a - X(t))1_{t\ge T_a}$. Note that $Y(t)$ is also a Wiener process and $T_a$ is also the first time $Y(t)$ hits $a$. We have $$\operatorname{Pr}\left(Y(T) \ge a \mid T_a \le T\right) = \operatorname{Pr}\left(X(T) \ge a \mid T_a \le T\right).$$ On the other hand, $$\begin{eqnarray}\operatorname{Pr}\left(Y(T) \ge a \mid T_a \le T\right) &=& \operatorname{Pr}\left(2a - X(T) \ge a \mid T_a \le T\right)\\& =& \operatorname{Pr}\left(X(T) \le a \mid T_a \le T\right) ,\end{eqnarray}$$ hence $$\operatorname{Pr}\left(X(T) \ge a \mid T_a \le T\right) = \frac{1}{2}.$$