Correlation Under Shifts and Scales

The correlation between $X$ and $Y$ is $\rho$ . What is the correlation between $X + 5$ and $Y$ ? What is the correlation between $5X$ and $Y$ ?

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#Correlation ignores location

Correlation is covariance normalised by the two standard deviations,

\rho(X, Y) = \frac{\Cov(X, Y)}{\sd(X)\, \sd(Y)}. \tag{1}

Adding a constant moves $X$ bodily without changing how it varies around its mean. Covariance already subtracts that mean, so

\Cov(X + 5, Y) = \E\big[(X + 5 - \E[X + 5])(Y - \E Y)\big] = \E\big[(X - \E X)(Y - \E Y)\big] = \Cov(X, Y), \tag{2}

and $\sd(X + 5) = \sd(X)$ . Both the numerator and denominator are untouched, so

\rho(X + 5, Y) = \rho. \tag{3}

#Positive scaling cancels

Multiplying by $5$ scales the spread of $X$ , hence stretches covariance and standard deviation by the same factor,

\Cov(5X, Y) = 5\,\Cov(X, Y), \qquad \sd(5X) = 5\,\sd(X). \tag{4}

The two factors of $5$ divide out,

\rho(5X, Y) = \frac{5\,\Cov(X, Y)}{5\,\sd(X)\, \sd(Y)} = \rho. \tag{5}

Only the sign of the multiplier survives. A negative factor would flip $\rho$ to $-\rho$ , but $5 > 0$ leaves it alone.

Correlation measures the tilt of this cloud, not where it sits or how wide it is. Adding 5 slides the points sideways and multiplying by 5 stretches them out, yet the trend line keeps the same slope sign, so the correlation stays exactly rho. A negative multiplier would mirror the cloud and flip the sign.

#Result

Both correlations equal $\rho$ . Correlation is invariant under an increasing affine change of either variable, since it measures the shape of the linear relationship, not its position or units.