correlation of predictions and actuals

Question

I should say that this question has nothing to do with regression. I am familar with the Sums of Squares Decomposition when there is an intercept in a regression but I'm not referring to that in this question. The setting is the following:

Suppose I have a series of actuals, $Z_t$, and two sets of forecasts, $X_t$ and $Y_t$ where $t = 1, \ldots n$.

I can calculate the sample pearson correlation coefficient, $\rho$ as described here: https://en.wikipedia.org/wiki/Pearson_correlation_coefficient.

So, if I want to calculate the sample correlations, $cor(X_t,Z_t) = \rho_{xz}$ or $cor(Y_t, Z_t) = \rho_{yz}$, this is straightforward.

Also, besides the sample correlation, suppose I also want to calculate $SSE_{xz}$ and $SSE_{yz}$ where $SSE_{xz} = \sum_{t=1}^{n} (X_{t} - Z_{t})^2$ and $SSE_{yz} = \sum_{t=1}^{n} (Y_{t} - Z_{t})^2$.

My question is: Suppose one has $Z_{t}, X_{t}$ and $Y_{t}$, Can it ever be the case $SSE_{xz} < SSE_{yz}$ and $cor(X,Z) < cor(Y,Z)$.

We know that in the case of a regression with intercept, the above cannot happen. But I'm not talking about the regression framework when I'm posing my question. In fact, it may be the case that I should be using the term cross-correlation rather than correlation but, as far as formulae, that's just semantics. Thanks.

Answer 1

Toy example:

Z   X     Y
1   3   100
2   2   200
3   1   300

$cor(X,Z) = -1, cor(Y,Z) = 1$ while $SSE_{xz}<SSE_{yz}$

I'd say it's a matter of scale. Correlation accounts for it, while $SSE$ do not.

Another, not relying on scales example:

Take arbitrary $\varepsilon>0$ and

$X_1=Z_1, X_i=Z_i+\varepsilon$ for $i=2,3,\ldots$

$Y_i = Z_i+2\varepsilon$ for $i=1,2,3,\ldots$

$cor(X,Z)$ is slightly below 1 and $cor(Y,Z) = 1$ while $SSE_{xz}<SSE_{yz}$.