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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1 Answer 1


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 $\mathbb E[T] = \infty$.

  • $\begingroup$ Makes me wondering. Is the probability of A a known value? $\endgroup$
    – SBF
    Commented Jun 11 at 18:34
  • $\begingroup$ 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! $\endgroup$
    – lorenzw
    Commented Jun 11 at 19:06
  • $\begingroup$ 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 $\delta > 0$ $\endgroup$
    – lorenzw
    Commented Jun 11 at 19:42
  • $\begingroup$ @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 $p>1$. $\endgroup$ Commented Jun 12 at 10:33
  • $\begingroup$ @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 $\endgroup$ Commented Jun 12 at 10:40

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