The question is how to calculate $r$ from $r^2$. Now, I know that its just as simple as taking a square root but is it as simple as that?
Solved – Calculate the correlation coefficient from the coefficient of determination
correlationr-squared
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This is true that $SS_{tot}$ will change ... but you forgot the fact that the regression sum of of squares will change as well. So let's consider the simple regression model and denote the Correlation Coefficient as $r_{xy}^2=\dfrac{S_{xy}^2}{S_{xx}S_{yy}}$, where I used the sub-index $xy$ to emphasize the fact that $x$ is the independent variable and $y$ is the dependent variable. Obviously, $r_{xy}^2$ is unchanged if you swap $x$ with $y$. We can easily show that $SSR_{xy}=S_{yy}(R_{xy}^2)$, where $SSR_{xy}$ is the regression sum of of squares and $S_{yy}$ is the total sum of squares where $x$ is independent and $y$ is dependent variable. Therefore: $$R_{xy}^2=\dfrac{SSR_{xy}}{S_{yy}}=\dfrac{S_{yy}-SSE_{xy}}{S_{yy}},$$ where $SSE_{xy}$ is the corresponding residual sum of of squares where $x$ is independent and $y$ is dependent variable. Note that in this case, we have $SSE_{xy}=b^2_{xy}S_{xx}$ with $b=\dfrac{S_{xy}}{S_{xx}}$ (See e.g. Eq. (34)-(41) here.) Therefore: $$R_{xy}^2=\dfrac{S_{yy}-\dfrac{S^2_{xy}}{S^2_{xx}}.S_{xx}}{S_{yy}}=\dfrac{S_{yy}S_{xx}-S^2_{xy}}{S_{xx}.S_{yy}}.$$ Clearly above equation is symmetric with respect to $x$ and $y$. In other words: $$R_{xy}^2=R_{yx}^2.$$ To summarize when you change $x$ with $y$ in the simple regression model, both numerator and denominator of $R_{xy}^2=\dfrac{SSR_{xy}}{S_{yy}}$ will change in a way that $R_{xy}^2=R_{yx}^2.$
Suppose you run a regression of $Y$ on regressor matrix $X$ with error term $\varepsilon$, i.e. \begin{align} Y = X\beta + \varepsilon \end{align} where $Y$, $\beta$, and $\varepsilon$ are $n\times1$ vectors and $X$ is a $n\times p$ matrix. Using Ordinary Least Squares (OLS), you estimate $\beta$ as $\hat{\beta}$ and obtain $\hat{y} = X\hat{\beta}$. Denote $\bar{y} = n^{-1}\sum_{i=1}^ny_i$, i.e. $\bar{y}$ is the average value of the entries in $y$.
Define the Total Sum of Squares (TSS) as $TSS:=\sum_{i=1}^n (y_i - \bar{y})^2$. This is the total square variation of $y$ without explaining any of this variation using $X$. One can further define the Residual Sum of Square (RSS) as $RSS:=\sum_{i=1}^n (y_i - \hat{y_i})^2$ and the Explained Sum of Square (ESS) as $ESS:= \sum_{i=1}^n (\hat{y_i} - \bar{y})^2$. RSS is called this way because it gives the variation of $y$ after using the fitted value $\hat{y_i}$ (instead of the mean) as predictor. ESS is the remaining (unexplained) variation after fitting the model.
$R^2$ is defined as $R^2:=1-RSS/TSS$. In the special case of a linear regression (and only then!) does this definition coincide with taking the square of the (estimated) correlation coefficient $r$. Finally, to answer your question, it is easy to just consult the above formula for $R^2$: Because it holds that $TSS = RSS + ESS$, one can rewrite $R^2$ as $R^2 = ESS/TSS = \frac{ESS}{n}/\frac{TSS}{n}$. Crucially, note that $\frac{ESS}{n}$ would be the * unexplained variance* and $\frac{TSS}{n}$ the total variance. This way, $R^2$ can be thought of as indicating the amount of 'explained variance/variation' that $X$ has with respect to $Y$.
Best Answer
A Pearson correlation is really only defined for two variable problems. If you happen to run a regression to obtain $R^2$, yes the square root will just convert back to the correlation. In a multivariate context, however, the square root of $R^2$ really doesn’t tell you much.