Let $N_1(t)$ be a delayed renewal process and $N_2(t)$ be an ordinary renewal process such that $N_1(t)\geq_{st}N_2(t)$. Consider a renewal process $Z(t)$ with the same inter-arrivall distribution as $N_2(t)$. We know that $N_1(t)$ and $N_2(t)$ and $Z(t)$ are independent. Show that: $$\mathbb{P}\left(\max_{t \in [0,T]}\{Z(t)-N_1(t)\}\geq a\right) \leq \mathbb{P}\left(\max_{t \in [0,T]}\{Z(t)-N_2(t)\}\geq a\right),\,\,\,\,\,\,\forall a \in \mathbb{R}\,\,(1)$$
My effort: Intuitively, it makes sense. Eq (1) is equivalent to: $$\mathbb{P}\left(\min_{t \in [0,T]}\{-Z(t)+N_1(t)\}\geq b\right) \geq \mathbb{P}\left(\min_{t \in [0,T]}\{-Z(t)+N_2(t)\}\geq b\right),\,\,\,\,\,\forall b\in \mathbb{R}\,\,\,(2)$$
First, note that we can say $-Z(t) \geq_{st} Z(t)$. From the problem statement, $N_1(t)\geq_{st}N_2(t)$. Since $Z(t)$, $N_1(t)$, and $N_2(t)$ are independent, can we conclude?:
$$N_1(t)-Z(t)\geq_{st}N_2(t)-Z(t)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(3)$$
If (3) is correct, then it yields:
$$\mathbb{E}\left[f(N_1(t_1)-Z(t_1),\ldots,N_1(t_n)-Z(t_n))\right] \geq \mathbb{E}\left[f(N_2(t_1)-Z(t_1),\ldots,N_2(t_n)-Z(t_n))\right], \,\,\,\,(4)$$
for all non-decreasing functions $f(x_1,\ldots,x_n)$, points $t_1,\ldots,t_n$, and $n$. Consider the non-decreasing function $f(x_1,\ldots,x_n) = \mathbb{I}_{\min\limits_{1 \leq i \leq n}\{x_i\}\geq b}$, where $\mathbb{I}$ is the indicator function. Thus, (4) yields (2), so (1) is proved. So, my proof is compete as long as (3) is proved. Is that right? In other words, if random vectors $\underline{A}=(A_1,\ldots,A_n),\underline{B}=(B_1,\ldots,B_n),$ satisfy $\underline{A} \leq_{st} \underline{B}$,and $\underline{A}$ and $\underline{B}$ and $\underline{C}$ random vectors are mutually independent, can we conclude?:
$$\underline{A}-\underline{C} \leq_{st} \underline{B}-\underline{C}$$