How to determine whether $\Bbb P ( B_{t_1} \in [x - c , x + c ], \ldots , B_{t_n} \in [x -c , x + c ])$ is decreasing in $x$?

Question

Let $\phi_t (z) := \frac{1}{\sqrt{2\pi t}} e^{-\frac {z^2} {2t}}$. By intuition $\omega : [0,\infty ) \to [0,1]$ \begin{align} \omega (x) &:= \Bbb P ( B_{t_1} \in [x - c , x + c ], \ldots , B_{t_n} \in [x -c , x + c ])\\ &=\int_{x-c}^{x+c} \phi_{t_1} (y_1) \int_{x-c}^{x+c} \phi_{t_2-t_1} (y_2 - y_1) \ldots \int_{x-c}^{x+c} \phi_{t_n -t_{n-1}} (y_n - y_{n-1}) \text d y_n \ldots\text d y_1 \end{align} should be decreasing in $x$. For $n=1$ this is easily done by derivating in $x$. Does anyone see a better way for this case here?

Answer 1

For $n=1$, the expression

$$\omega (x) = \int_{x-c}^{x+c} \phi_{t_1} (y)= \int_{-c}^c \phi_{t_1} (y+x)\text d y$$

is decreasing in $x\in[0,\infty)$, since $\phi_t\colon\Bbb R\to(0,\infty)$ is symmetric around the origin and strictly decreasing (increasing) on the positive (negative) half-axis. This is essentially the same reason why the argument with taking the derivative works, too.

For $n>1$, we might use induction. We have

\begin{align} \omega (x) &= \int_{-c}^c \phi_{t_1} (y_1+x) \int_{-c}^{c} \phi_{t_2-t_1} (y_2 - y_1) \ldots \int_{-c}^{c} \phi_{t_n -t_{n-1}} (y_n - y_{n-1}) \text d y_n \ldots\text d y_1 \\ &= \int_{-c}^c \phi_{t_1} (y_1+x) g(y_1) \text d y_1, \end{align}

where both $g\colon\Bbb R\to(0,\infty)$ and $\phi_t\colon\Bbb R\to(0,\infty)$ are symmetric around the origin and strictly decreasing (increasing) on the positive (negative) half-axis.

Note that $\omega$ may be defined in the same way for negative $x$ as well, and that the resulting function will be decreasing in $|x|$.

Answer 2

$\newcommand{\PRx}[2]{\Bbb{P}_{#1} (#2)}$

Let $c>0$ and $G\subset (0,\infty)$ a finite set of time points. The function $\omega : \Bbb R \to [0,1]$ given by $\omega (x) := \Bbb P_x ({\vert{B_s}\vert \leq c\ \forall s \in G})$ is radial nonincreasing.

The strategy is induction over $\vert G \vert$. For $\vert G \vert =1$ the derivative of $\omega (x)$ is $\phi_t (x+c) - \phi_t (x-c) \leq 0$. Let $G$ be arbitrary, $s = \min G$ and write $\tilde{G} := \{t-s : t \in G\setminus \{s\} \}$. Then \begin{align*} \omega (x) = \int_{-c}^c \phi_s (y-x) \Bbb P_y ({\vert{B_r}\vert\leq c \ \forall r\in \tilde{G}}) \text d y = \int_{-c}^c \phi_s (y-x) \tilde{\omega} (y) \text d y \end{align*} where $\tilde{\omega}(y) := \PRx{y}{\vert{B_r}\vert\leq c \ \forall r\in \tilde{G}}$ is radial nonincreasing by assumption. Let $x\geq 0 $. Then by differentiation of parameter integrals \begin{align*} \omega ' (x) &= \int_{-c}^c \frac{\partial}{\partial x} \phi_s (y-x) \tilde{\omega} (y) \text d y = \int_{-c}^c \frac{y-x}{s} \phi_s (y-x) \tilde{\omega} (y) \text d y\\ &= \int_{-c-x}^{c-x} \frac{y}{s} \phi_s (y) \tilde{\omega} (y+x) \text d y \end{align*} which is clearly non-positive if $x\geq c$. In the other case rewrite above as \begin{align*} &\int_0^{c-x} \frac{y}{s} \phi_s (y) \tilde{\omega} (y+x) \text d y + \int_{-c-x}^0 \frac{y}{s} \phi_s (y) \tilde{\omega} (y+x) \text d y\\ &= \int_0^{c-x} \frac{y}{s} \phi_s (y) \tilde{\omega} (x+y) \text d y - \int_0^{c+x} \frac{y}{s} \phi_s (y) \tilde{\omega} (x -y) \text d y\\ &= \int_0^{c-x} \frac{y}{s} \phi_s (y) \tilde{\omega} (x +y) \text d y - \int_0^{c-x} \frac{y}{s} \phi_s (y) \tilde{\omega} (x -y) \text d y -\int_{c-x}^{c+x} \frac{y}{s} \phi_s (y) \tilde{\omega} (x -y) \text d y\\ &= \int_0^{c-x} \frac{y}{s} \phi_s (y) ( \tilde{\omega} (x +y) - \tilde{\omega} (x -y) )\text d y -\int_{c-x}^{c+x} \frac{y}{s} \phi_s (y) \tilde{\omega} (x -y) \text d y\\ &\leq 0 \end{align*} since $\tilde{\omega} (x +y) - \tilde{\omega} (x -y) \leq 0$ due to the fact that $\vert{x-y}\vert \leq \vert{x+y}\vert$. Therefore the assertion follows by induction.