I heard a claim today that a correlation of (say) 0.5 between variables X and Y means "if we look at a value of X that is one standard deviation above its mean, we would expect the corresponding value of Y to be 0.5 standard deviations above its mean". It wasn't immediate to me whether this is true or false.

On further reflection: By definition, $$ \text{Cov}(X,Y) = \rho_{XY} \sigma_X \sigma_Y $$ and, by conditioning, $$ \text{Cov}(X,Y) = \int E[(X - \mu_X)(Y - \mu_Y) \mid X - \mu_X = k\sigma_X] P(X - \mu_X = k\sigma_X)\; dk. $$

If the claim is true, the above is $$ \int k\sigma_X k \rho_{XY}\sigma_Y P\left(\frac{X-\mu_X}{\sigma_X} = k\right)\; dk $$ Assuming a normal approximation holds, this is (using $Z \sim \mathcal{N}(0,1)$) $$ \rho_{XY}\sigma_X \sigma_Y \int k^2 P(Z = k)\; dk = \rho_{XY} \sigma_X \sigma_Y \text{Var}(Z) = \rho_{XY} \sigma_X \sigma_Y $$ which is consistent with the definition.

So is this a reasonable way to interpret the correlation? (The above seems to show that it is true on average, but whether it's true conditioning on $X$ isn't clear to me.)


1 Answer 1


Another identity that may drive it home is this:

$$\beta_{OLS} = \frac{\text{Cov}(X,Y)}{\text{var}(X)} = \frac{\rho \sigma_y}{\sigma_x}$$

Or $\rho = \beta_{OLS} \sigma_x / \sigma_y$. So, if you use this "strong causal" interpretation of the $\beta_{OLS}$, the correlation interpretation is just a scaled version of the same thing.


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