Is this a martingale sequence?

I have a sequence of random variables $X_1, X_2, \ldots X_N$ such that $|X_i| \leq R \ \forall \ i$, satisfying $$|E[X_n|X_1,X_2,\ldots X_{n-1}]| \leq |X_{n-1}|,$$

Can I construct a sub/super-martingale sequence from this? Intuitively, it makes sense that the random variables are constrained to be not from far the previous one in absolute/distance sense. But can I view this as some sort of martingale or perhaps construct a sequence (or sequences) which satisfy the definition of (sub/super)martingale(s)?

• How do such sequences arise? – Iosif Pinelis Apr 24 '18 at 14:29
• Suppose $Y_i$ is an estimator for $Y$. Define $X_i = Y_i - Y$. I have an iterative randomized algorithm which estimates $Y_i$ and $i^{th}$ iteration and ensures $$|E[X_n|X_1,X_2,\ldots X_{n-1}]| \leq \alpha |X_{n-1}|, \alpha < 1$$. I want to show that this is some standard stochastic process (like martinagale) and apply concentration tools to show that $X_n$ goes to $0$ with high probability. – Enayat Apr 28 '18 at 19:43
• I think this is a good question, but I don't have a good answer to it at this point. – Iosif Pinelis Apr 29 '18 at 3:28
• Just a minor comment; it is easy to show that this inequality does not imply a concentration (or convergence for that matter); if you have $X_n$s to be iid mean 0, this inequality is always trivially true... – E-A Oct 31 '18 at 23:59

If you aim at using Jensen's inequality, you need an even and concave function $f$ increasing on $\mathbb R_+$ to define a supermartingale. Then by Jensen's inequality: $$E[f(X_n)|X_1,\ldots,X_{n-1}]\leq f(E[X_n|X_1,\ldots,X_{n-1}])\leq f(X_{n-1}).$$
To be more precise, since your inequality does not depend on the sign of your random variable, you cannot use anything else than even functions (you don't have an inequality for $X_{n-1}$). Then using a concave and increasing function on $\mathbb R_+$ allows you to use Jensen's inequality in the right order. A convex and decreasing function would lead to a submartingale.
Unfortunately I don't think such functions exist. The issue is in zero. An even function increasing on $\mathbb R_+$ cannot be concave on $\mathbb R$. Indeed if $f$ is even and concave on $\mathbb R_+$ for any $x>0$, $$f(0)=f(-\frac12 x+\frac12 x)\geq \frac12 f(-x)+ \frac12 f(x)=f(x).$$ Hence, if $f$ is monotonous on $\mathbb R_+$, it is non increasing.