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The setup is as follows. We are given a martingale $X_0,X_1,...,X_k$. The difference $X_i-X_{i-1}$ is always between $[-1,1]$. Variance $D^2(X_i-X_{i-1}| X_{i-1})$ is something, but we can show that after taking expectation on $X_{i-1}$ then we get that $D^2(X_i-X_{i-1}) \leq X_0/k$. Therefore, I would like to get a tail bound $P( X_k - X_0 > t) \leq \exp(-t^2/ X_0 )$; constants in the exponent don't matter. There exists a similar bound $P( X_k - X_0 > t) \leq \exp(-\frac{t^2}{\sum_{i=1}^k c_i^2} )$, but here $c_i^2$ is the bound on supremum of $D^2(X_i-X_{i-1}| X_{i-1})$ for the worst-case $X_{i-1}$, so this one does not imply what I want.

How can I get a hold on a bound like that?

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  • $\begingroup$ Maybe it is standard, but what do you exactly denote by $D^2(\cdot) (I mean in terms of conditional expectation)? $\endgroup$ – Davide Giraudo Aug 24 '15 at 15:52
  • $\begingroup$ Sorry, Soviet notation for variance: $D^2(X | Y) = E( (X-E[X|Y])^2 | Y)$ $\endgroup$ – Marek Adamczyk Aug 24 '15 at 19:06
  • $\begingroup$ Thanks. How do you deduce that this is smaller than $X_0/k$? This implies that $X_0\geqslant 0$ which is not the case in general. But as I understand it is true in your situation. $\endgroup$ – Davide Giraudo Aug 24 '15 at 21:08
  • $\begingroup$ the martingale is actually composed of two random sequences, i.e. $X_t = Y_t + Z_t$. We start with $Y_0 = 0$, and $Z_0$ being some arbitrary value. At step $t+1$ we subtract from $Z_t$ quantity equal on expectation to $\frac{1}{k-t}Z_t$, and to $Y_t$ we add quantity on expectation equal to $\frac{1}{k-t}Z_t$. Since the change is always less than $1$, then we can argue that the variance is at most $\frac{1}{k-t}Z_t$ (maybe times two, forget the constants). Inductively one can show that $E Z_t = (1-t/k)Z_0$. $\endgroup$ – Marek Adamczyk Aug 24 '15 at 22:56
  • $\begingroup$ Also, as for the bound, little change. I need to upperbound $e^{\lambda \cdot Y_k}$ by $e^{(e^\lambda - 1) EY_k}$, or something similar that would yield the same asymptotic bound for the uppertail: $P(Y_k > (1+\delta) E Y_k) < (\frac{e^\delta}{(1+\delta)^{(1+\delta)}})^{E Y_k}$ $\endgroup$ – Marek Adamczyk Aug 24 '15 at 23:05
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Alas, the general bound you are hoping for cannot hold. Take $X_0:=1$ and continue the sequence as a simple random walk with probability $1/k$ and as the all ones sequence with probability $1-1/k$. Then for $t=\sqrt{k}$, the chance that $X_k-X_0>t$ is at least $C/k$ for some constant $C>0$.

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  • $\begingroup$ Hi Marek, Can you mark the question as answered? $\endgroup$ – Yuval Peres Dec 3 '15 at 11:19

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