First, define the following stopping times \[eqt\_j\] t\_0&=&0,\
t\_j&=&{t&gt;t\_[j-1]{}: m(t\_j)&gt;m(t\_[j-1]{})},  j=1,2,…. Let $$S(j)=\sum_{k=1}^{M} v_k^*(t_j)$$ where v\_k\^\*(t\_j)={

[rl]{} v\_k(t\_j), & v\_k(t\_j)2,\
0, &

. where the stopping times are defined by (\[eqt\_j\]).

\[prop\_a\_men\] $S(j+1)\le S(j)$. Moreover, there is an integer $K=K(M)$ and an $\eps>0$ such that $$\P(S(j+K)-S(j)\le -1\| v(t_j))\ge \eps$$ on the event $S(j)>0$.

For simplicity, let us write $v_k$ for $v_k(t_j)$. Take some non-zero element $a\ge 1$ in the sequence of $v_k$ at time $t_j$. Whenever it is followed by a consecutive chunk of $0$s, namely $$\dots  a\ \underbrace{0\ 0\ \dots \ 0}_m \dots$$ at time $t_{j+1}$ this becomes either $$\dots  a\ \underbrace{z_1\ z_2\ \dots \ z_m}_m\dots$$ or $$\dots  a-1\ \underbrace{z_1\ z_2\ \dots \ z_m}_m\dots,$$ where $z_i\in\{0,1\},$ and the latter occurs if the second $0$ from the left is chosen before the first one. On the other hand, if $a$ is succeeded by a non-zero element, say ‘$\dots\  a\ b\ \dots$’ at time $t_{j+1}$ this becomes either ‘$\dots\ a-1 \ b \ \dots $’ or ‘$\dots\ a-1
\ b-1 \ \dots $’. In all cases, this leads to $S(j+1)\le S(j)$.

Secondly, from the previous arguments we see that if there is at least one $a\ge 2$ in the sequence of 