# Do perfectly correlated data have the same z-score?

Consider two time series, $\{X(t),Y(t)\in \Re |t\geq0\}$, whose correlation is 1. That is, $$\rho_{X,Y}=\frac{\mathbb{E}\left[\left(X - \mathbb{E}[X]\right)\left(Y-\mathbb{E}[Y]\right)\right]}{\sqrt{\mathbb{V}[X]\mathbb{V}[Y]}} = 1.$$ Now consider each series' respective z-score, defined (for $X(t)$) as $$z_{X}(t) = \frac{X(t) - \mathbb{E}[X]}{\sqrt{\mathbb{V}[X]}}.$$ Is there any way to show that, because $\rho_{X,Y} = 1$, $z_{X}(t) = z_{Y}(t) \;\forall t$?

• Yes. See first paragraphs in stats.stackexchange.com/a/135231/3277. If two vectors have Pearson r = 1 it means their standardized values are identical, so euclidean distance between these standardized vectors is 0. – ttnphns Jun 20 '18 at 8:19