Upcrossings

Last updated Nov 1, 2022

Suppose $(x_{n})_{n=0}^{\infty} \subset \mathbb{R}$ is a Sequence and $a < b \in \mathbb{R}$. Inductively define for $i \in \mathbb{N}$:

$$\begin{align*} & \tau_{0} = 0\\ &\sigma_{i} = \inf {j \in \mathbb{N} : j > \tau_{i-1}, x_{j} \leq a}\\ &\tau_{i} = \inf {j \in \mathbb{N} : j > \sigma_{i}, x_{j} \geq b} \end{align*}$$ Then the number of Complete Upcrossings, $U_{N}^{}[a, b]$, of $x_{1}, \dots x_{N}$ for $N \in \mathbb{N}$ is $$U_{N}^{}[a, b] = \sup {i \in \mathbb{N} : \tau_{i} \leq N}$$ Furthermore we call $$U_{\infty}^{}[a, b] = \lim\limits_{N \to \infty} U_{N}^{}[a, b] \in \bar{\mathbb{R}}$$

1. We are measuring the number of times that our sequence has fallen down to $a$, then come back up to $b$.
2. Our defintion for $U_{\infty}^{}$ makes sense since $U_{N}^{}[a, b]$ is a Monotone Sequence and Monotone Sequences converge to their Extremum.

Let $(\Omega, \mathcal{B}, \mathbb{P})$ be a Probability Space and let $(\mathcal{F}{n}){n=1}^{\infty}$ be a Discrete-Time Filtration on $\mathcal{B}$. Let $(X_{n}){n=1}^{\infty}$ be a $(\mathcal{F}{n}){n=1}^{\infty}$-Adapted Process. Let $a < b \in \mathbb{R}$. Then the following Random Variables are Stopping Times for $k \geq 1$: $$\begin{align*} N{0} &:= 0\\ N_{2k-1} &:= \inf\limits {m > N_{2k-2} : X_{m} \leq a}\\ N_{2k} &:= \inf\limits {m > N_{2k-1} : X_{m} \geq b}\\ \end{align*}$$ We define the $\mathcal{F}{n}$-Random Variable $$U{n}[a,b] = \sup\limits {k \geq 0 : N_{2k} \leq n}$$ to be the number of Complete Upcrossings by time $n \in \mathbb{N}$.

We check that $N_{2k-1}$ and $N_{2k}$ are Stopping Times by Induction. First observe that, letting $n \in \mathbb{N}$, $$\begin{align*} [N_{1} = n] &= [X_{1} > a, \dots, X_{n-1} > a] \cap [X_{n} \leq a] \in \mathcal{F}{n}\\ \end{align*}$$ by virtue of being a Filtration, so $N{1}$ is a Stopping Time. Now suppose $N_{1}, \dots, N_{2k-1}$ are Stopping Times for $k \geq 1$ . Then, letting $n \in \mathbb{N}$, $$\begin{align*} [N_{2k} = n] &= [X_{N_{2k-1}+1} < b, \dots, X_{n-1} < b] \cap [X_{n} \geq b] \in \mathcal{F}{n}\\ &\bigcup\limits{m = 1}^{n-1} [X_{m+1} < b, \dots, X_{n-1} < b] \cap [X_{n} \geq b] \cap [N_{2k-1} = m] \in \mathcal{F}{n} \end{align*}$$ by virtue of $N{2k-1}$ being a Stopping Time and $(\mathcal{F}{n}){n=1}^{\infty}$ being a Filtration. Thus, $N_{2k}$ is a Stopping Time. A similar approach will show that $N_{2k+1}$ must also be a Stopping Time.

Now we check that $U_{n}[a,b] \in \mathcal{F}{n}$. Let $k \in \mathbb{N}$ $$[U{n}[a,b] = k] = [N_{2k} \leq n] \cap [N_{2k+2} > n] \in \mathcal{F}{n}$$ which holds because $N{2k}$ and $N_{2k+2}$ are both Stopping Times. $\blacksquare$

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