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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%?

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  • $\begingroup$ not directly related to the question so a mere remark; beware, from a practitioner point of view, using Pearson correlation is usually not a good idea to measure association between two stocks $\endgroup$
    – mic
    Mar 21, 2016 at 8:13

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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.

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    $\begingroup$ +1: It's worth mentioning that $R$ is referred as the multiple correlation coefficient. $\endgroup$
    – Francis
    Mar 21, 2016 at 7:06
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    $\begingroup$ This is helpful but dependent on an external site. If that blog site were to disappear, the answer would lose much of its value. I don't know about "less known". I'd regard it as standard to explain up front that they are one and the same in plain regression. Many people are taught regression after correlation, so emphasis on the link is then natural. (Whether that's a good thing can be set on one side.) $\endgroup$
    – Nick Cox
    Mar 21, 2016 at 9:40
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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) ).

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    $\begingroup$ +1. I suspect you meant to write "one independent variable" every place that "one dependent variable" appears. $\endgroup$
    – whuber
    Aug 7, 2015 at 13:12
  • $\begingroup$ @whuber: you are right, it must be 'independent variable' of course, I changed it in the text and give (+1) for correcting the error $\endgroup$
    – user83346
    Aug 7, 2015 at 13:42
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    $\begingroup$ The psychology of this terminology is interesting--I frequently make the same mistake. I have taken to using the term "response variable" instead of "dependent variable" in order to avoid confusing myself and others by the similarity of "dependent" and "independent" (although I worry about implying a causal relationship). I also like "explanatory" instead of "independent," but a client recently objected to that terminology (even in the form "potential explanatory variable") because the coefficients were insignificant--and so the variables were considered not "explanatory"! $\endgroup$
    – whuber
    Aug 7, 2015 at 13:47
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    $\begingroup$ Further objections to "dependent" and "Independent" are that the terms are overloaded and in any case are so close that beginning students (and indeed others) can easily get confused between them. Since 2015 when @whuber wrote I think "response" has held up well as good terminology in several fields, but is also being challenged by "outcome". Either term seems excellent to me. $\endgroup$
    – Nick Cox
    Jan 14 at 16:05
  • $\begingroup$ @Nick Many times after initially writing "outcome" I withdraw that term in favor of "response" out of concern it could (validly) be considered to be the entire vector of observed values in any observational setting. In effect, "outcome" does not adequately distinguish the explanatory from the response variables. $\endgroup$
    – whuber
    Jan 14 at 16:08

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