# A possible generalization of Chebyshev's inequality

Let $$X$$ and $$Y$$ be identically and independently distributed random variables.

By Chebyshev's inequality we know that for any fixed $$k > 0$$ $$\mathbb{P}(X-Y > k) \leq \frac{\mathbb{E}[(X-Y)^2]}{k^2} = \frac{2\mathrm{Var}(X)}{k^2}.$$

My question is what happens if we allow $$k$$ above to be random? Specifically, is there a meaningful bound for the following quantity: $$\mathbb{P}(X-Y > Y).$$ Ideally, there would exist some constant $$C > 0$$ such that $$\mathbb{P}(X-Y > Y) \leq C\mathrm{Var}(X).$$ Is there any reason to expect such an inequality would even hold?

In Chebyshev the constant depends only on $$k$$ and not on the distribution of the random variables. What should $$C$$ depend on in the new inequality? It cannot be independent of the distribution. But then you can simply take $$C=\frac {P\{X-Y >Y\}} {Var (X)}$$. To show that you cannot have a universal constant $$C$$ take $$X=Y=-1$$. An example with $$var (X) \neq 0$$: let $$X,Y$$ be i.i.d. normal with mean $$0$$ and variance $$\frac 1 n$$. Then LHS is indpendent of $$n$$ and is strictly positive wheras RHS $$\to 0$$; for large enough $$n$$ the inequality does not hold.
• Great examples! can you think you can come up with such an example when $X$ and $Y$ are positive almost surely. – Cain Oct 9 '18 at 5:25