Chebyshev Bound for Bivariate Normal Difference
Let $(X, Y) \sim \text{BVN}(0, 0, \sigma^2, \sigma^2, \rho)$ -- a bivariate normal with zero means, equal marginal variances $\sigma^2$, and correlation $\rho$.
Using Chebyshev's Inequality, find an upper bound on $P(|X - Y| \geq \sigma)$ when $\sigma^2 = 4$ and $\rho = 3/4$.
Hints
- Chebyshev requires the random variable to be centered -- verify that $E[X - Y] = 0$ before applying it.
- Use the variance formula $\text{Var}(X - Y) = \text{Var}(X) + \text{Var}(Y) - 2\,\text{Cov}(X,Y)$, and note that $\text{Cov}(X,Y) = \rho\sigma^2$ for a bivariate normal.
- The Chebyshev bound becomes $P(|X-Y| \geq \sigma) \leq \text{Var}(X-Y)/\sigma^2 = 2(1-\rho)$; plug in $\rho = 3/4$.
Worked Solution
How to Think About It: Chebyshev's Inequality says that for any random variable $Z$ with finite mean and variance, $P(|Z - \mu_Z| \geq k) \leq \text{Var}(Z)/k^2$. To apply it to $|X - Y| \geq \sigma$, we need two things: (1) confirm that $X - Y$ is centered (mean zero), and (2) compute $\text{Var}(X - Y)$. The correlation $\rho$ enters through the covariance term in the variance of a difference, which is the main calculation here.
Quick Estimate: At $\rho = 3/4$, the two variables are highly correlated -- $X - Y$ has much less variance than if $X$ and $Y$ were independent. With independent normals, $\text{Var}(X - Y) = 2\sigma^2$, giving a Chebyshev bound of 2. But with $\rho = 3/4$, the variance shrinks to $2\sigma^2(1 - 3/4) = \sigma^2/2$, so the bound should be $1/2$. That is below 1, so it is an informative (non-trivial) bound.
Formal Solution:
First, check centering: $E[X - Y] = E[X] - E[Y] = 0 - 0 = 0$, so $X - Y$ has mean zero. Chebyshev's Inequality applies directly:
$$P(|X - Y| \geq \sigma) \leq \frac{\text{Var}(X - Y)}{\sigma^2}$$
Compute $\text{Var}(X - Y)$ using the variance of a linear combination:
$$\text{Var}(X - Y) = \text{Var}(X) + \text{Var}(Y) - 2\,\text{Cov}(X, Y)$$
Since $\text{Cov}(X, Y) = \rho \sigma^2$:
$$\text{Var}(X - Y) = \sigma^2 + \sigma^2 - 2\rho\sigma^2 = 2\sigma^2(1 - \rho)$$
Substituting into Chebyshev:
$$P(|X - Y| \geq \sigma) \leq \frac{2\sigma^2(1 - \rho)}{\sigma^2} = 2(1 - \rho)$$
With $\rho = 3/4$:
$$P(|X - Y| \geq \sigma) \leq 2\left(1 - \frac{3}{4}\right) = 2 \times \frac{1}{4} = \frac{1}{2}$$
Answer: $P(|X - Y| \geq \sigma) \leq \dfrac{1}{2}$. The general formula is $2(1 - \rho)$, which decreases toward 0 as $\rho \to 1$ (perfectly correlated variables have $X = Y$ almost surely, so their difference is zero).
Intuition
This problem illustrates how correlation tightens the distribution of a difference. When $\rho = 0$ (independent), the bound is $2(1 - 0) = 2$, which is vacuous -- probabilities cannot exceed 1. When $\rho = 1/2$, the bound is 1, still vacuous. Only at $\rho > 1/2$ does Chebyshev give a non-trivial bound for this specific threshold. This is a reminder that Chebyshev is a weak inequality -- it holds for any distribution, which is its strength, but it pays for generality with looseness.
The $2(1 - \rho)$ formula has a direct interpretation: it is just $\text{Var}(X - Y)/\sigma^2$, the variance of the difference scaled by the threshold variance. High correlation means the difference is tightly concentrated around zero, so the probability of a large deviation is small. In risk management, this is the principle behind pairs trading -- two highly correlated assets have a near-zero long-short P&L in normal conditions, and large deviations signal a potential mean-reversion opportunity.