Let's say I have a series of some values.

When I transform the series now by simple shrinkage, for instance, multiplying the original value by 0.5 and adding 0.5 constant .

Then, are both series correlated perfectly with 1 as coefficient?

  • 6
    $\begingroup$ In statistics, "shrinkage" does not mean what you think it does. Standard terms for adding and multiplying by constants are affine transformation, recentering and rescaling, and changing the units of measurement. $\endgroup$
    – whuber
    Dec 28, 2016 at 20:17

1 Answer 1


As I understand you just look for $cor(x,\frac{x+1}{2})=1$ that is right. So in general case $cor(x,\alpha x+\beta)=1$ if $\alpha>0$ and $cor(x,\alpha x+\beta)=-1$ if $\alpha<0$.

Also it is very easy to proof. $cor(x,\alpha x+\beta)=\frac{cov(x,\alpha x+\beta)}{\sqrt{Var(x)}\sqrt{Var(\alpha x+\beta)}}=\frac{ \alpha cov(x,x)}{\alpha\sqrt{Var(x)}\sqrt{Var(x)}}=\frac{ \alpha cov(x,x)}{\alpha\sqrt{Var(x)}\sqrt{Var(x)}}=\frac{cov(x,x)}{\sqrt{Var(x)}\sqrt{Var(x)}}=1$

So in general case $cor(\alpha_{1} x+\beta_{1},\alpha_{2} y+\beta_{2})=sgn(\alpha_{1})*sgn(\alpha_{2})cor(x,y)$. It means that linear dependence measure between two values does not depends on units of measure (or shrinkage) that is why it is more comfortable then covariation.


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