Correlation of Sum and Difference
Let $X$ and $Y$ be i.i.d. random variables, each with variance $\sigma^2$. Define $S = X + Y$ and $D = X - Y$.
What is the correlation $\rho(S, D)$?
As a follow-up: if $X$ and $Y$ are i.i.d. normal, can you say something stronger than just their correlation?
Hints
- Think about what the transformation $(X, Y) \to (X+Y, X-Y)$ does geometrically -- what kind of rotation is this?
- Use the bilinearity of covariance: $\text{Cov}(X+Y, X-Y)$ expands into four terms involving $\text{Var}$ and $\text{Cov}$.
- The i.i.d. assumption gives you $\text{Cov}(X,Y) = 0$ and $\text{Var}(X) = \text{Var}(Y)$. Substitute and watch the cancellation.
Worked Solution
How to Think About It: Before computing anything, think geometrically. Going from $(X, Y)$ to $(S, D) = (X+Y, X-Y)$ is a 45-degree rotation (scaled by $\sqrt{2}$). If $X$ and $Y$ are i.i.d., their joint distribution is symmetric around the $X = Y$ line. A rotation that aligns with that symmetry axis should decouple the two new coordinates -- so you'd expect zero correlation. That is the punchline, and the algebra just confirms the geometric intuition.
Quick Estimate: Since $X$ and $Y$ are i.i.d., the covariance matrix of $(X, Y)$ is $\sigma^2 I$. Any linear transformation of a vector with a scalar covariance matrix produces outputs whose cross-covariances depend only on the dot products of the transformation rows. The rows of the transformation $\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ are orthogonal, so the covariance between $S$ and $D$ is zero. Done -- correlation is $0$.
Approach: Use bilinearity of covariance to compute $\text{Cov}(S, D)$ directly.
Formal Solution:
Expand using bilinearity:
$$\text{Cov}(S, D) = \text{Cov}(X + Y,\; X - Y)$$
$$= \text{Cov}(X, X) - \text{Cov}(X, Y) + \text{Cov}(Y, X) - \text{Cov}(Y, Y)$$
Since $X$ and $Y$ are i.i.d., we have $\text{Cov}(X, Y) = 0$, $\text{Var}(X) = \text{Var}(Y) = \sigma^2$. So:
$$\text{Cov}(S, D) = \sigma^2 - 0 + 0 - \sigma^2 = 0$$
Since $\text{Cov}(S, D) = 0$:
$$\rho(S, D) = \frac{\text{Cov}(S,D)}{\sqrt{\text{Var}(S)\,\text{Var}(D)}} = 0$$
Note that $\text{Var}(S) = \text{Var}(D) = 2\sigma^2$, so the denominator is well-defined and positive.
Follow-up -- the normal case:
If $X, Y$ are i.i.d. normal, then $(S, D)$ is jointly normal (any linear transformation of a multivariate normal is again multivariate normal). For jointly normal random variables, uncorrelated implies independent. So $S$ and $D$ are not just uncorrelated -- they are fully independent.
This is strictly stronger. For non-normal i.i.d. variables, $S$ and $D$ are uncorrelated but may still be dependent.
Answer: $\rho(S, D) = 0$. The sum and difference of i.i.d. random variables are always uncorrelated. If additionally they are normal, $S$ and $D$ are independent.
Intuition
This problem illustrates a fundamental property of orthogonal linear transformations applied to uncorrelated random variables. When $X$ and $Y$ are i.i.d., their joint distribution has a "round" covariance structure ($\sigma^2 I$). Any two orthogonal directions you project onto will produce uncorrelated outputs. The sum and difference correspond to a 45-degree rotation -- an orthogonal transformation -- so the result is zero covariance. This is not a coincidence; it is the rotational symmetry of the covariance structure doing the work.
In practice, this shows up constantly. PCA on correlated assets amounts to finding orthogonal directions (principal components) that are uncorrelated by construction. The sum-and-difference decomposition is the simplest version of this: one component captures the "level" (both variables moving together), the other captures the "spread" (one moving relative to the other). Traders use this instinctively when they decompose a pairs trade into a directional bet and a spread bet -- those two bets are uncorrelated when the underlying legs have the same variance.