# Expected stopping time of summation of ranking supermartingale is finite

Given a stochastic process (that takes on real numbers) $$X_n$$, which is a ranking supermartingale, which we defined as $$\mathbb{E}(X_{n+1}-X_n|\mathcal{F_n})\leq \epsilon\ < 0$$. Let now $$Y_n$$ be the summation of this stochastic process: $$Y_n = \sum_{0\leq i \leq n} X_i$$.

I want to show, that the process $$Y_n$$ is expected to be smaller or equal than $$0$$ after a finite number of steps. I think I can express this through the stopping time: $$T:=\inf\{n≥0:Y_n\leq 0\}$$. I then would need to show, that $$\mathbb{E}(T)< \infty$$.

I have shown by induction, that the the inequality $$\mathbb{E}(Y_{n+m}|\mathcal{F_n}) \leq Y_n+X_nm+ \frac{m(m+1)}{2}\epsilon$$ holds. What my idea was, was to solve the equivalence $$Y_n+X_nm+ \frac{m(m+1)}{2}\epsilon = 0$$ for $$m_0$$. Since $$m_0$$ would have a positive, non-infinite solution when $$Y_n$$ is positive (since $$\epsilon$$ is negative), either $$Y_n$$ is already stopped, or expected to be stopped in $$n+m$$ steps. The problem however is, that even when $$m$$ is finite, $$n+m$$ might not be. Does my argument still hold, since I can say, that at any point $$n$$ in time the process is expected to be stopped in $$m$$ time steps in the future?

I am stuck here, and feel like maybe my approach is inappropriate as a whole. Any guidance in the right direction would be highly apprechiated.

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This is wrong without further assumptions on $$(X_n)_{n\geq 1}$$. Here is a counterexample.

Let $$(U_n)_{n\geq 1}$$ be independent random variables such that for every $$n\geq 1$$, $$U_n = 0$$ with probability $$1-2^{-n}$$ and $$U_n = \varepsilon 2^n$$ with probability $$2^{-n}$$. Then $$\mathbb E[U_n] = \varepsilon$$.

By Borel-Cantelli, the event $$A = \{ 0 = U_1 = U_2 = \dots \}$$ has positive probability.

Now define $$X_n = 1 + U_1 + \dots + U_n$$, and choose the filtration $$\mathcal F_n = \sigma(U_1, \dots, U_n)$$. Clearly $$(X_n)_{n\geq 0}$$ satisfies your conditions. Yet on the event $$A$$, we have $$X_n = 1$$ for every $$n\geq 0$$, and thus $$Y_n = n$$, hence $$T=\infty$$. And event $$A$$ has positive probability. This implies in particular that $$\mathbb E[T] = \infty$$.

• Makes me wondering. Is the probability of A a known value?
– SBF
Commented Jun 11 at 18:34
• the probabilities $1-(1/(n+1)^2)$ and $(1/(n+1)^2)$ with the values 0 and $\epsilon (n+1)^2$ would make sense for me, as the infinite product of the first probability converges to 0.5. Thanks for the answer! Commented Jun 11 at 19:06
• Would a sufficient condition for my assumption holding be, that the probability of $X_n$ decreasing would be larger than constant? I.e. $\mathbb{P}(X_{n+1} - X_{n}<0|\mathcal{F_n}) \geq \delta$ for $\delta > 0$ Commented Jun 11 at 19:42
• @lorenzw the probas $(n+1)^{-2}$ would work too. As for $\mathbb P(X_{n+1}-X_n<0|\mathcal F_n) \geq \delta$, no, it will not work, because you can choose for example $U_n = -2^{-n-1}$ with probability $2^{-n-1}$ (taking the probability from the event $\{U_n=0\}$). Instead, you could try looking at conditions like $\mathbb E[(X_{n+1}-X_n)^p | \mathcal F_n] \leq c < \infty$ for every $n$, for some $p>1$. Commented Jun 12 at 10:33
• @SBF yes, although not an interesting one, $\prod_{n\geq 1}(1-2^{-n}) \approx 0.288788$. Some softwares may have something to say about the value: wolframalpha.com/input?i=product+%281-2%5E%28-n-1%29%29 Commented Jun 12 at 10:40