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Suppose $X_1$ and $X_2$ are iid from an arbitrary distribution with variance $\sigma^2$. How can we derive an upper bound for: $$P(|X_1-X_2|\ge\delta)$$

One simple idea is Chebyshev's Inequality, however, when $\delta<\sqrt{2}\sigma$, we have $$P(|X_1-X_2|\ge\delta)\le\frac{var(X_1-X_2)}{\delta^2}=\frac{2\sigma^2}{\delta^2},$$ which literally tells nothing. Clearly, this inequality can be improved in this case.

How can we get a better bound? Thank you.

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  • $\begingroup$ This isn't how you use Chebyshev; for that, you need a random variable minus its expectation. You instead found a bound for $|(X_1 - X_2) - E[(X_1 - X_2)]|$. At a glance, the only guarantee you have is the tail bound from Markov's inequality, which is pretty weak: $E(|X_1 - X_2|)/\delta$. Do you have any assumptions about the mean (other than finiteness)? $\endgroup$
    – chang_trenton
    Commented Oct 22, 2020 at 7:33
  • $\begingroup$ In this case $E[X_1-X_2]=0$. $\endgroup$
    – Claucisco
    Commented Oct 22, 2020 at 18:31

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