0
$\begingroup$

Let $X=(X_t:t\ge 0)$ be a stochastic process (martingale in general) starting at $X_0=0$. For $T>0$ and $a<b$, let $U^T_{a,b}(X)$ be the number of upcrossings of $X$ across the interval $[a,b]$ over $[0,T]$, i.e. $U^T_{a,b}(X)$ is the supremum of the nonnegative integers $n\in\mathbb N$ such that there exist times $s_k,t_k\in [0,T]$ satisfying $$s_1<t_1<s_2<t_2<\cdots<s_n<t_n$$ and for which $X_{s_k}\le a<b\le X_{t_k}$. My question is the following: For any continuous and non-decreasing function $h:\mathbb R_+\to [0,\infty]$ with $h(0)=0$ (may not be strictly increasing), set $Y=(Y_t:=X_{h(t)}: t\ge 0)$. Does $U^T_{a,b}(X)=U^{h(T)}_{a,b}(Y)$ hold?

$\endgroup$

1 Answer 1

0
$\begingroup$

We have $U_{a,b}^T(Y) = U_{a,b}^{h(T)}(X)$.

By construction, the function $h$ maps $[0,T]$ onto $[0,h(T)]$. More precisely, each $t' \in [0,h(T)]$ can be written $h(h^\leftarrow(t'))$, where $h^\leftarrow(t') := \inf\{t \in [0,T] : h(t) \ge t'\} \in [0,T]$. The function $h^\leftarrow$ thus defined is strictly increasing (but not necessarily continuous).

Hence, if $s'_1<t'_1<\ldots<s'_n<t'_n$ provide $n$ upcroassings of $X$ in $[0,h(T)]$ then $h^\leftarrow(s'_1)<h^\leftarrow(t'_1)<\ldots<h^\leftarrow(s'_n)<h^\leftarrow(t'_n)$ provide $n$ upcroassings of $Y$ in $[0,T]$.

Conversely, if $s_1<t_1<\ldots<s_n<t_n$ provide $n$ upcroassings of $Y$ in $[0,T]$, then $h(s_1)<h(t_1)<\ldots<h(s_n)<h(t_n)$ provide $n$ upcroassings of $X$ in $[0,h(T)]$ (the inequalities cannot be equalities because of the conditions $X(h(s_i)) \le a < b \le X(h(t_i))$.

$\endgroup$
1
  • $\begingroup$ Thanks a lot for the answer. The definition of the inverse of $h$ actually helps $\endgroup$
    – Fawen90
    Commented Oct 31, 2022 at 15:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .