Let \(X\) and \(Y\) be identically distributed random variables with variance \(\sigma^2\in(0,\infty)\). Then the correlation coefficient between \(X\) and \(Y\) is
Show Hint
Expand E(X-Y)^2 using Var(X-Y) and the fact that identically distributed variables have equal means and variances, then solve for the covariance and divide by sigma squared.
Step 1: Start from the definition of correlation.
Since $X$ and $Y$ are identically distributed, both have variance $\sigma^2$, so \[ \rho=\frac{\mathrm{Cov}(X,Y)}{\sigma^2} \] The whole problem is just about writing $\mathrm{Cov}(X,Y)$ in terms of $E(X-Y)^2$.
Step 2: Expand $E(X-Y)^2$ directly using expectations.
\[ E(X-Y)^2=E(X^2)-2E(XY)+E(Y^2) \]
Step 3: Rewrite each term using variance and covariance.
Since $X,Y$ have the same mean $\mu$, $E(X^2)=\sigma^2+\mu^2$ and $E(Y^2)=\sigma^2+\mu^2$, and $E(XY)=\mathrm{Cov}(X,Y)+\mu^2$. Substituting, \[ E(X-Y)^2=(\sigma^2+\mu^2)-2(\mathrm{Cov}(X,Y)+\mu^2)+(\sigma^2+\mu^2)=2\sigma^2-2\,\mathrm{Cov}(X,Y) \] (the $\mu^2$ terms cancel completely, which makes sense since correlation should not depend on the common mean at all).
Step 4: Isolate the covariance.
\[ \mathrm{Cov}(X,Y)=\sigma^2-\frac{1}{2}E(X-Y)^2 \]
Step 5: Divide by $\sigma^2$ to get $\rho$.
\[ \rho=\frac{\mathrm{Cov}(X,Y)}{\sigma^2}=1-\frac{E(X-Y)^2}{2\sigma^2} \]
Step 6: Sanity check with a simple case.
If $X=Y$ exactly, then $\rho$ should be $1$ (perfect correlation), and indeed $E(X-Y)^2=E(0)=0$, giving $\rho=1-0=1$. This matches, which supports the formula.
Step 7: Rule out the other listed formulas.
Formula (B) has an extra factor of $4$ compared to ours (it uses $2E(X-Y)^2/\sigma^2$ instead of $E(X-Y)^2/(2\sigma^2)$), which would break the sanity check above whenever $E(X-Y)^2\neq0$. Formula (C) is missing the factor of $2$ in the denominator. Formula (D) uses $E(X+Y)^2$, which mixes in the mean $\mu$ and does not cancel the way our derivation needs. Only our derived formula survives.
