0

Let $b\in\mathbb{R}^d$ and consider a $d$-dimensional Markov branching process. Define the stopping time $$\tau_b(n)=\min\{t\ge0:b\cdot X(t)\ge n\}$$ for $n\ge0$, where the $\cdot$ denotes the inner product between the two vectors. I would like to understand, at least intuitively, why does this inequality holds: $$|b\cdot X(\tau_b(n))-n|\le|b||\Delta X(\tau_b(n))|,$$ where $\Delta X(t)=X(t)-X({t^-})$ is the jump at time $t$, and $X({t^-})=\lim_{h\searrow0}X(t-h)$. On the RHS we use "a" vector norm: my source says that "we use | ยท | for the norm of both vectors and matrices. (The choice of matrix norm is irrelevant.)".

What I know is just that, by def. of $\tau_b$, the LHS will be non-negative, but I don't really get how to get this inequality with the jump of the process. Any idea?

Dada
  • 177
  • 6
  • 22

1 Answers1

1

$\tau_b$ is the first time $b\cdot X(t)$ is larger or equal to $n$. Since by definition $n \ge b \cdot X(\tau_b-)$ we have the inequality

$$ b \cdot X(\tau_b) -n \le b \cdot X(\tau_b) - b \cdot X(\tau_b-)=b\cdot \Delta (X(\tau_b)).$$

The LHS is nonnegative and therefore so is the RHS. Now apply Cauchy-Schwarz to conclude that the RHS is bounded above by $|b| |\Delta (X(\tau_b))|$.

Fnacool
  • 9,581