Bounds on the difference of correlated random variables Given two highly correlated random variables $X$ and $Y$, I'd like to bound the probability that the difference $ |X - Y| $ exceeds some amount:
$$ P( |X - Y| > K) < \delta $$
Assume for simplicity that:


*

*The correlation coefficient is known to be "high", say : 
$ \rho_{X,Y}= {covar(X,Y)} / {\sigma_X \sigma_Y}  \geq 1 - \epsilon $

*$X,Y $ are  zero mean: $ \mu_x = \mu_y = 0 $

*$-1 \leq x_i, y_i \leq 1$ (or  $ 0 \leq x_i, y_i \leq 1$ if
that's any easier)

*(If it makes things easier, let's say $X,Y $ have identical variance: $\sigma_X^2 = \sigma_Y^2  $)
Not sure how feasible it is to derive a bound on the difference given only the above information (I certainly couldn't get anywhere). A specific solution (if any), mandatory additional restrictions to impose on the distributions, or just advice on an approach would be great.
 A: Even without those simplifying assumptions, a bound can be obtained by combining a couple of usual tools:


*

*The variance of the difference of two correlated variables. It allows us to turn a two variables problem into an univariate problem.

*Chebyshev's inequality. It puts a bound on the probability of exceeding a given value.


In some detail:
$$\sigma^2_{X-Y}=\sigma^2_X+\sigma^2_Y-2·cov(X,Y)$$
$$cov(X,Y)=\sigma_X·\sigma_Y·\rho_{XY}$$
$$\sigma^2_{X-Y}=\sigma^2_X+\sigma^2_Y-2·\sigma_X·\sigma_Y·\rho_{X,Y}$$
According to Chebyshev's inequality, for any random variable $Z$:
$$ \Pr(|Z-\mu|\geq k\sigma) \leq \frac{1}{k^2}$$
Then (and using that $\mu_{X-Y}=\mu_X-\mu_Y)$:
$$ \Pr(|X-Y-\mu_X+\mu_Y|\geq k·\sqrt{\sigma^2_X+\sigma^2_Y-2·\sigma_X·\sigma_Y·\rho_{X,Y}}) \leq \frac{1}{k^2}$$
We can use the proposed simplifying assumptions to get a simpler expression. When:
$$\rho_{X,Y}= {covar(X,Y)} / {\sigma_X \sigma_Y}  = 1 - \epsilon $$
$$\mu_x = \mu_y = 0$$
$$\sigma_X^2 = \sigma_Y^2 = \sigma^2$$
Then:
$$\sigma^2_X+\sigma^2_Y-2·\sigma_X·\sigma_Y·\rho_{X,Y} = 2·\sigma^2·(1-(1-\epsilon)) = 2\sigma^2\epsilon$$
And therefore:
$$\Pr(|X-Y|\geq k·\sigma\sqrt{2\epsilon}) \leq \frac{1}{k^2}$$
Interestingly, this result holds even if $\epsilon$ is not small, and if the condition for correlation changes from $=1-\epsilon$ to $\geq 1-\epsilon$, the result doesn't change (because it's already an inequality).
