Probability associated with Brownian motion

Question

I would like a nod in the right direction with the following problem.

Let $0<t_1<t_2<\cdots<t_n<\infty$, and $\{a_i\}_1^n$ be real numbers, find a function $f(x_1,\ldots,x_n)$ so that

$$P\{B(t_1)\leq a_1, B(t_2)\leq a_2,\ldots, B(t_n)\leq a_n\} > =\int_{-\infty}^{a_n} \cdots \int_{-\infty}^{a_1} f(x_1,\ldots,x_n)\,dx_1\cdots dx_n$$

My assumption is that you try to re-write the expression so that you can use the fact that Brownian motion has independent increments, however I am getting nowhere.

Answer 1

$\newcommand{\var}{\operatorname{var}}\newcommand{\cov}{\operatorname{cov}}$ Notice that for $i\le j$ we have \begin{align} & \cov(B(t_i),B(t_j)) \\[10pt] = {} & \cov(B(t_i), (B(t_j)-B(t_i)) + B(t_i)) \\[10pt] = {} & \cov(B(t_i), B(t_j)- B(t_i)) + \cov(B(t_i),B(t_i)) \\[10pt] = {} & 0 + \var(B(t_i)) = t_i = t_{\min\{i,j\}} = \min\{t_i,t_j\}. \end{align}

Hence $$ \var\begin{bmatrix} B(t_1) \\ \vdots \\ B(t_n) \end{bmatrix} = \begin{bmatrix} t_1 & t_1 & t_1 & t_1 & \cdots & t_1 \\ t_1 & t_2 & t_2 & t_2 & \cdots & t_2 \\ t_1 & t_2 & t_3 & t_3 & \cdots & t_3 \\ t_1 & t_2 & t_3 & t_4 & \cdots & t_4 \\ \vdots & \vdots & \vdots & \vdots & & \vdots \\ t_1 & t_2 & t_3 & t_4 & \cdots & t_n \end{bmatrix}. $$ Call this matrix $V$. then the joint density of this random vector whose variance is $V$ is \begin{align} f(x_1,\ldots,x_n) = {} & \frac 1 {\sqrt{2\pi}^n}\cdot\frac 1 {(\det V)^{1/2}} \exp \left( \frac{-1} 2 x^\top V^{-1} x \right) \tag 1 \\[10pt] & \text{ where } x = \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix}. \end{align}

• Does $\det V$ have a nice closed form? I suspect it does. (Later note: It does; see below.)
• Does $V^{-1}$ have some nice pattern to it? Again, I suspect so. So maybe we're not done simplifying $(1)$, but $(1)$ is it.
• You may still have the task of proving that this is the density function ahead of you. Or maybe you can just cite some theorem in some book you've got.

PS: If I'm not mistaken, then when $t_i=i$ for $i=1,2,3,\ldots$ then

• $\det V = 1$. I think I could write a geometric justification for that. And:
• Computation suggests $V^{-1}$ has $2$ in every diagonal entry and $-1$ in every entry immediately adjacent to the diagonal, and $0$ elsewhere. I suspect I'll realize soon that there's a geometric reason for that.

PPS: Elementary row operations find the determinant: $t_1(t_2-t_1)(t_3-t_2) \cdots (t_n-t_{n-1})$. I think you can also give a geometric argument and a probabilistic argument for that result.