X(t) follows a Brownian Motion. What is P(X(T) and X(2T) >0)?

Question

Let $X(t)$ follows a Brownian Motion. What is $P(X(T)>0 \cap X(2T)>0)$?

Running MC simulations I found out $\frac{3}{8}$ but I cannot prove it mathematically.

Answer 1

HINT Let's denote the event $$ \lim_{h \to 0^+} \left\{ x-h < X(T) < x+h \right\} = A_x. $$ This is intuitively just the continuous analog of the discrete event $\{X(T) = x\}$.

Also denote the pdf and cdf of $X(T)$ as $f_T$ and $F_T$, respectively.

Note that $$ \begin{split} p &= \mathbb{P}[X(T)>0 \text{ and } X(2T)>0] \\ &= \int_0^{\infty} \mathbb{P}\left[X(T)>0 \text{ and } X(2T)>0 \left| A_x \right.\right] \mathbb{P}[A_x] dx \\ &= \int_0^{\infty} \mathbb{P}\left[X(2T)>0 \left| A_x \right.\right] f_T(x) dx\\ &= \int_0^{\infty} \mathbb{P}\left[X(T+T) - X(T) > -x \left| A_x \right.\right] f_T(x) dx \quad \text{independent increments implies}\\ &= \int_0^{\infty} \mathbb{P}\left[X(T) > -x \right] f_T(x) dx\\ &= \int_0^{\infty} (1-F_T(-x)) f_T(x) dx\\ \end{split} $$ Can you take it from here?

UPDATE

As you correctly noticed, using $F_T(-x) = 1-F_T(x)$, you get $$ p = \int_0^\infty F_T(x) f_T(x) dx $$ and the integral can be substituted using $u = F_T(x)$. What do you get?