Let $(B_t:t\geq0)$ be a standard Brownian motion. I need to solve $\mathbb{P}(B_2-B_1>1|B_{0.5}=2)$. I first thought of using independent increments to get $\mathbb{P}(B_2-B_1>1)$, and then using stationarity this equals $\mathbb{P}(B_1>1)$, which can be solved with the normal cdf. However, if I first use stationarity, I obtain $\mathbb{P}(B_1>1|B_{0.5}=2)$, and then I can't use independence. Is my logic in the first part correct? Or should I approach it differently?
Stationary and independent increments in Brownian motion
brownian motionprobabilitystochastic-processes
Related Solutions
$B_1$ and $B_2$ are not independent.
Since $B_1$ and $B_2-B_1$ are independent, $$0=\mathrm{Cov}(B_1,B_2-B_1)=E[B_1 (B_2-B_1)]=E[B_1 B_2]-E[B_1^2]=E[B_1 B_2] - 1$$ So, $$\mathrm{Cov}(B_1,B_2)=E[B_1 B_2]=1$$
You can use this result and the fact that linear combinations of normal variables are normal to calculate the distribution of $B_1+B_2+B_3$.
I'm not sure what exactly you meant by "expectation over a Brownian process". Can you please give us a little background of the problem you are trying to solve? Your method of taking an expectation with respect to $N(0,\sqrt{t})$ is fine when you want the expected value of some function of $B_t$.
There is also a theory of stochastic integration with respect to stochastic processes which might be of use to you.
(i) Since $X_0 = B_0$ in distribution, we have $$\mathbb{P}(X_0 \in A) = \mathbb{P}(B_0 \in A) = \delta_0(A)$$ for any measurable set $A$. Hence, $X_0=0$ almost surely.
(ii) Take $0=t_0 < \ldots < t_n$. By assumption, the random vectors $U:= (X_{t_1},\ldots,X_{t_n})$ and $V:=(B_{t_1},\ldots,B_{t_n})$ have the same distribution. Since $(B_t)_{t \geq 0}$ is a Brownian motion, $V$ is Gaussian, and so is $U$. If we define $$f(x_1,\ldots,x_n) := \begin{pmatrix} x_1 \\ x_2-x_1 \\ \vdots \\ x_n-x_{n-1} \end{pmatrix},$$ then $$f(U) = \begin{pmatrix} X_{t_1} \\ X_{t_2} -X_{t_1} \\ \vdots \\ X_{t_n}-X_{t_{n-1}} \end{pmatrix} \quad \text{and} \quad f(V)=\begin{pmatrix} B_{t_1} \\ B_{t_2}-B_{t_1} \\ \vdots \\ B_{t_n}-B_{t_{n-1}} \end{pmatrix}$$ have the same distribution and both are Gaussian. Now recall that the entries $G_j$ of a Gaussian random vector $G=(G_1,\ldots,G_n)$ are independent iff they are uncorrelated, i.e. if the covariance matrix of $G$ is a diagonal matrix. Since the Brownian motion $(B_t)_{t \geq 0}$ has independent increments, it follows that the covariance matrix of the Gaussian random vector $f(V)$ is a diagonal matrix, and so is the covariance matrix of $f(U)$, which means that the entries $X_{t_j}-X_{t_{j-1}}$ are independent.
(iii) By assumption, $(B_s,B_{t+s})$ and $(X_s,X_{t+s})$ have the same distribution. This implies that $X_{t+s}-X_s = B_{t+s}-B_s$ in distribution, and so $X_{t+s}-X_s \sim N(0,t)$.
Best Answer
Stationarity doesn't tell you $\mathbb{P}(B_2-B_1>1|B_{0.5}=2)=\mathbb{P}(B_1>1|B_{0.5}=2)$. It only tells you $\mathbb{P}(B_2-B_1>1)=\mathbb{P}(B_1>1)$. Thus your first part is correct, but not your second approach. Think about this (counter)example: Consider $\mathbb{P}(B_{1.5}-B_1>3|B_{0.5}=2)$, if it was equal to $\mathbb{P}(B_{0.5}>3|B_{0.5}=2)$, the probability $\mathbb{P}(B_{1.5}-B_1>3|B_{0.5}=2)$ would be zero, which is definitely not true.