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Consider two sequences of (not necessarily independent) Bernoulli random variables $X_1, X_2, \ldots, X_n$ and $Y_1, Y_2, \ldots, Y_n$. Suppose that for any $i$, we have $\Pr[X_i = 1] = \Pr[Y_i = 1] = p_i$, but the actual value of $p_i$ is determined only after observing $\{X_1, \ldots, X_{i-1}, Y_1, \ldots, Y_{i-1} \}$. That is, each $p_i$ is also a random variable.

Intuitively it seems that the two sums $\sum_i X_i$ and $\sum_i Y_i$ should typically be very close to each other. Are any such bounds known? Is there e.g. a simple argument to show that with high probability $\sum_i X_i = \sum_i Y_i \pm O(\sqrt{n \log n})$?

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    $\begingroup$ Are you assuming that given $p_i$, $X_i$ and $Y_i$ are independent of each other? $\endgroup$
    – Marcus M
    Mar 6, 2020 at 0:04
  • $\begingroup$ @MarcusM Intuitively, I don't think that should matter much for the sums whether $X_i$ and $Y_i$ are independent since they have the same success probabilities. Would it help if we assume they are independent? $\endgroup$
    – Mathman
    Mar 6, 2020 at 0:10
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    $\begingroup$ You're right it shouldn't matter, actually. I'll add an answer. $\endgroup$
    – Marcus M
    Mar 6, 2020 at 0:12

1 Answer 1

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Define the martingale $M_n = \sum_{i = 1}^n (X_i - Y_i)$ with the filtration $\mathcal{F}_n = \sigma( \{X_j,Y_j\}_{j=1}^n )$. Then $|M_{n + 1} - M_n| = |X_{n+1}-Y_{n+1}| \leq 1$ and it is indeed a martingale since $$\mathbb{E}[M_{n+1}\,|\,\mathcal{F}_n]= \mathbb{E}[X_{n+1} - Y_{n+1}\,|\,\mathcal{F}_n] + M_{n} = M_n\,. $$

By Azuma's inequality, we then have $$\mathbb{P}(|M_n| \geq \lambda) \leq 2 \exp\left( -\frac{\lambda^2}{2n}\right)\,.$$

Taking $\lambda \gg \sqrt{n}$ shows your desired behavior.

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  • $\begingroup$ Very cute. Thanks for the super fast response. $\endgroup$
    – Mathman
    Mar 6, 2020 at 0:21

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