Correlation Coefficient vs Coefficient of Determination I am doing regression analysis on two stocks. The correlation coefficient is -0.7190 and the coefficient of determination is 0.5170. I am confused on how to interpret this. Is this correct...when stock A goes up 1%, 50% of the time stock B will go down 0.72%?
 A: The usual way of interpreting the coefficient of determination $R^{2}$ is as the percentage of the variation of the dependent variable $y$ $(Var(y))$ that one is able to explain with the explanatory variables. You can find the exact interpretation and derivation of the coefficient of determination $R^{2}$ on this Economic Theory Blog website.
However, a less known interpretation of the coefficient of determination $R^{2}$ is to interpret it as the Squared Pearson Correlation Coefficient between the observed values $y_{i}$ and the fitted values $\hat{y}_{i}$. You can find the proof that the coefficient of determination is the equivalent of the Squared Pearson Correlation Coefficient between the observed values $y_{i}$ and the fitted values $\hat{y}_{i}$ on this Economic Theory Blog website.
A: If you do a regression $y=\beta_1 x + \beta_2$ (so with one independent variable), then the squared of the correlation coefficient is equal to the coefficient of determination.  So (-0.7190)^2 is around 0.5170 (rounding errors). 
The coeffcient of determination tells you that 51.7% of the variance in the dependent variable $y$ is explained by the regression.  
if $x$ goes up by one unit, then $y$ goes up by $\beta_1$ units. $\beta_1$ can be estimated using ordinary least squares and can be found in the output of R function 
summary(lm(y~x))

So e.g. 
reg <- lm(mpg ~ hp, data = mtcars)
summary (reg)

yields
Call:
lm(formula = mpg ~ hp, data = mtcars)

Residuals:
    Min      1Q  Median      3Q     Max 
-5.7121 -2.1122 -0.8854  1.5819  8.2360 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 30.09886    1.63392  18.421  < 2e-16 ***
hp          -0.06823    0.01012  -6.742 1.79e-07 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 3.863 on 30 degrees of freedom
Multiple R-squared:  0.6024,    Adjusted R-squared:  0.5892 
F-statistic: 45.46 on 1 and 30 DF,  p-value: 1.788e-07

So if $hp$ increases by one unit, then $mpg$ decreases (decrease because negative sign) by 0.06823.  
The coefficient of determination is $R^2=0.6024$ and (there is only one independent variable) the correlation is -0.776 (square root of 0.6024 and the sign is negative because the coefficient of hp is negative (-0.06823) ). 
