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Suppose I have a supermartingale

$$ \mathbb{E}[X_{n+1} \mid X_n, \dots , X_2, X_1] \leq X_n $$

There are 2 other conditions:

  • bounded difference: $|X_{n+1} - X_n| \leq A$ with probability $1$ (almost surely) for all $n$, where $A$ is a constant

  • all $X_n$ are lower bounded: $X_n \geq B$ with probability $1$ (almost surely) for all $n$, where $B$ is a constant

My question is, is there a way to bound the variance $\mathrm{Var}[X_n]$ for all $n$?

Ideally the bound on the variance $\mathrm{Var}[X_n]$ should $\to 0$ as $n \to \infty$ and $A \to 0$. So are there other (possibly known) conditions that are needed?

  • 0
    Should you not prescribe the variance of $X_1$? – 2018-02-28
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    At the moment, there's a trivial counterexample obtained by setting $X_{i+1} = X_i$ for all $i>1$. Then $A=0$ but as $n \to \infty$, $\text{Var}[X_n]$ stays at $\text{Var}[X_1]$, whatever that was. – 2018-03-02

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