Probability of two times of Brownian Motion greater than 0 
For a Brownian Motion starting at $0$, I want the probability $X(2T)>0$ and $X(T)>0$. 

One idea I had was that 
\begin{align*} \mathbb{P}(X_{2T}>0,X_{T}>0) &=
\mathbb{P}(X_{2T}>0,X_{T}>0 \mid X_{2T}>X_{T})\mathbb{P}(X_{2T}>X_{T}) \\ &\quad + 
\mathbb{P}(X_{2T}>0,X_{T}>0\mid X_{T}<X_{2T})\mathbb{P}(X_{2T}<X_{T}) \end{align*}
This is $$\mathbb{P}(X_{T} > 0)\mathbb{P}(N(0,t) > 0) + \mathbb{P}(X_{2T} > 0)\mathbb{P}(N(0,t) < 0) = \frac12.$$ 
I was also thinking that every permutation of $0, X_T, X_{2T}$ are all equally likely due to symmety around $0$, and the permutations are $0, X_T, X_{2T}$ and $0, X_{2T}, X_{T}$, so it's $\frac{2}{3!} = \frac13$. 
I'm not sure if either are correct, if any. 
 A: No, it's not that simple. First of all, we would expect that the probability $\mathbb{P}(X_T>0,X_{2T}>0)$ depends on $T$. If $T$ is large, then the gap between the two "observations" at time $t=T$ and $t=2T$ is large, and so we don't expect that the value at time $t=T$ tells us much about the value at time $t=2T$. In contrast, if $T$ is small, then strict positivity of $X_T$ implies that also $X_{2T}$ is strictly positive with "high" probability (because the time difference $(2T)-T$ between the observations is small).

Fix $T>0$. The restarted process $$B_t := X_{T+t}-X_T, \qquad t \geq 0, $$ is a Brownian motion which is independent of $(X_t)_{t \leq T}$. Moreover, $$\mathbb{P}(X_T>0,X_{2T}>0) = \mathbb{P}(X_T>0, B_T>-X_T).$$ Using the independence of $(X_t)_{t \leq T}$ and $(B_t)_{t \geq 0}$, it follows from the tower property of conditional expectation that
$$\mathbb{P}(X_T>0,X_{2T}>0)  = \mathbb{E} \big[ \mathbb{E}(1_{\{X_T>0\}} 1_{\{B_T>-X_T\}} \mid X_T) \big]= \mathbb{E}(1_{\{X_T>0\}} f(X_T))$$
where
$$f(x) := \mathbb{P}(B_T>-x), \qquad x \in \mathbb{R}.$$
If we denote by $\Phi$ the cdf of the standard Gaussian distribution, then it follows from $B_T \sim N(0,T)$ that
$$f(x) = \mathbb{P}(\sqrt{T}B_1>-x) = 1- \mathbb{P}\left(B_1 \leq - \frac{x}{\sqrt{T}}\right) =1- \Phi \left( - \frac{x}{\sqrt{T}} \right).$$
Hence,
$$\mathbb{P}(X_T>0,X_{2T}>0) = \underbrace{\mathbb{P}(X_T>0)}_{=1/2} - \mathbb{E} \left( 1_{\{X_T>0\}} \Phi \left(- \frac{X_T}{\sqrt{T}} \right) \right).$$
Writing $\phi$ for the pdf of the standard Gaussian distribution, we get
$$\mathbb{P}(X_T>0,X_{2T}>0) = \frac{1}{2} - \int_0^{\infty} \Phi \left(- \frac{x}{\sqrt{T}} \right) \phi(x) \, dx.$$
The latter integral can be calculated explicitly, see here,
$$\mathbb{P}(X_T>0,X_{2T}>0) = \frac{1}{2}  - \left[ \frac{1}{4} + \frac{1}{2\pi} \arctan \left(-\frac{1}{\sqrt{T}} \right) \right].$$
Since $\arctan(x)=\arctan(-x)$ and $\arctan(1/x) = \pi/2 - \arctan(x)$ for $x>0$, we obtain
$$\mathbb{P}(X_T>0,X_{2T}>0) = \frac{1}{2} - \frac{1}{2\pi} \arctan (\sqrt{T}).\tag{1}$$
Remark: It's easy to miss some constants/signs when doing such calculations so let us briefly check whether our final result is reasonable. Since $\arctan(x) \geq 0$ for $x \geq 0$, we find from $(1)$ that $\mathbb{P}(X_T>0,X_{2T}>0) \leq 1/2$. This is what we would expect anyway since
$$\mathbb{P}(X_T>0,X_{2T}>0) \leq \mathbb{P}(X_T>0) = \frac{1}{2}.$$
Moreover, $x \mapsto \arctan(x)$ is increasing, and so $(1)$ shows that the probability $\mathbb{P}(X_T>0,X_{2T}>0)$ is decreasing in $T$. This is also reasonable (see the intuitive consideration at the beginning of this answer).
A: This is just a complement for the answer above.
$$
\mathbb{P}\left(X_{T}>0, X_{2 T}>0\right)=\frac{1}{2}-\int_{0}^{\infty} \Phi\left(-\frac{x}{\sqrt{T}}\right) \phi(x) d x
$$
The equality above in the first answer has a small mistake. The term $\phi(x)$ is the standard normal density. However, $\int_{0}^{\infty} \Phi\left(-\frac{x}{\sqrt{T}}\right) \phi(x) d x$ is derived from $\mathbb{E}\left(1_{\left\{X_{T>0\}}\right.} \Phi\left(-\frac{X_{T}}{\sqrt{T}}\right)\right)$. We can notice that $X_T$ is a normal r.v. but not standard. Therefore the density of $X_T$ should be $$\tilde\phi(x) = \dfrac{1}{\sqrt{2\pi T}}e^{-\dfrac{x^2}{2T}}$$
The integral of above equality should be $\int_{0}^{\infty} \Phi\left(-\frac{x}{\sqrt{T}}\right) \tilde\phi(x) d x$.
If we let $m = \dfrac{x}{\sqrt{T}}$, we have $\sqrt{T}dm = dx$ and
$$\sqrt{T}\tilde\phi(m\sqrt{T})  dm = \phi(m)dm$$
Now we substitute $x = \sqrt{T}m$ into $\int_{0}^{\infty} \Phi\left(-\frac{x}{\sqrt{T}}\right) \tilde\phi(x) d x$, we have
$$\begin{aligned}\int_{0}^{\infty} \Phi\left(-\frac{x}{\sqrt{T}}\right) \tilde\phi(x) d x &= 
\int_{0}^{\infty} \Phi\left(-m\right) \tilde\phi(m\sqrt{T}) d (\sqrt{T}m)\\
& =\int_{0}^{\infty} \Phi\left(-m\right) \phi(m) d m \\
&=\frac{1}{4}+\frac{1}{2 \pi} \arctan (-1)  = 0.125
\end{aligned}
$$
Hence, the answer should be $0.5-0.125 = 0.375$.
