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The Coefficient of Determination $R^2=SSR/SSTO$ and Coefficient of Correlation $r=\pm \sqrt{R^2}$ where the slop of the fitted regression line determine the positive or negative.

However, I also remember another correlation coefficient https://en.wikipedia.org/wiki/Pearson_correlation_coefficient.

My question was that: Were they equivalent? If not what was the difference?

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For the simpler linear model, $r^2$, which is the square of sample Pearson correlation coefficient, is equivalent to $R^2$. The prove is straightforward as you can find in @Benjamin's answer. For multiple regression, $R^2$ equals the square of Pearson correlation coefficient between the actual $y$ values and the fitted values, $\hat{y}$.

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When an intercept is included in linear regression(sum of residuals is zero), they are equivalent. $$ \begin{eqnarray*} ρ(y_i,\hat y_i)&=&\frac{cov(y_i,\hat y_i)}{\sqrt{var(y_i)var(\hat y_i)}}\\&=&\frac{\sum_{i=1}^n{(y_i - \bar{y})(\hat y_i - \bar{y})}}{\sqrt{\sum_{i=1}^n{(y_i - \bar{y})^2}\sum_{i=1}^n{(\hat y_i - \bar{y})^2}}} \\&=&\frac{\sum_{i=1}^n{(y_i -\hat y_i+\hat y_i- \bar{y})(\hat y_i - \bar{y})}}{\sqrt{\sum_{i=1}^n{(y_i - \bar{y})^2}\sum_{i=1}^n{(\hat y_i - \bar{y})^2}}}\\&=&\frac{\sum_{i=1}^n{(y_i -\hat y_i)(\hat y_i - \bar{y})}+\sum_{i=1}^n{(\hat y_i- \bar{y})^2}}{\sqrt{\sum_{i=1}^n{(y_i - \bar{y})^2}\sum_{i=1}^n{(\hat y_i - \bar{y})^2}}} \end{eqnarray*} $$ $$ \begin{eqnarray*} \sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)&=&\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)(\beta_0+\beta_1x_i-\bar y)\\&=&(\beta_0-\bar y)\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)+\beta_1\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)x_i \end{eqnarray*} $$

In Least squares regression, the sum of the squares of the errors is minimized. $$ SSE=\displaystyle\sum\limits_{i=1}^n \left(e_i \right)^2= \sum_{i=1}^n\left(y_i - \hat{y_i} \right)^2= \sum_{i=1}^n\left(y_i -\beta_0- \beta_1x_i\right)^2 $$ Take the partial derivative of SSE with respect to $\beta_0$ and setting it to zero. $$ \frac{\partial{SSE}}{\partial{\beta_0}} = \sum_{i=1}^n 2\left(y_i - \beta_0 - \beta_1x_i\right)^1 (-1) = 0 $$ So $$ \sum_{i=1}^n \left(y_i - \beta_0 - \beta_1x_i\right)^1 (-1) = 0 $$ Take the partial derivative of SSE with respect to $\beta_1$ and setting it to zero. $$ \frac{\partial{SSE}}{\partial{\beta_1}} = \sum_{i=1}^n 2\left(y_i - \beta_0 - \beta_1x_i\right)^1 (-x_i) = 0 $$ So $$ \sum_{i=1}^n \left(y_i - \beta_0 - \beta_1x_i\right)^1 x_i = 0 $$ Hence, when an intercept is included in linear regression(sum of residuals is zero), $$ \begin{eqnarray*} ρ(y_i,\hat y_i)&=&\frac{\sum_{i=1}^n{(y_i -\hat y_i)(\hat y_i - \bar{y})}+\sum_{i=1}^n{(\hat y_i- \bar{y})^2}}{\sqrt{\sum_{i=1}^n{(y_i - \bar{y})^2}\sum_{i=1}^n{(\hat y_i - \bar{y})^2}}}\\&=&\frac{0+\sum_{i=1}^n{(\hat y_i- \bar{y})^2}}{\sqrt{\sum_{i=1}^n{(y_i - \bar{y})^2}\sum_{i=1}^n{(\hat y_i - \bar{y})^2}}}\\&=&\sqrt{\frac{\sum_{i=1}^n{(\hat y_i- \bar{y})^2}}{\sum_{i=1}^n{(y_i- \bar{y})^2}}}\\&=&\sqrt{\frac{SSR}{SST}}\\&=&\sqrt{R^2} \end{eqnarray*} $$

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Were they equivalent? If not what was the difference?

Go to the section "In least squares regression analysis" on the Wiki page you link to. It shows that they are equivalent in simple linear regression and why.

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