If we have one independent variable, then the given regression coefficient $\beta$ = Pearson's $r$. If we have multiple independent variables, how can calculate Pearson's $r$ for each variable if only $\beta$ values are given for each independent variable? (Assume I have T1, T2, T3 independent variables and I have $\beta_1$, $\beta_2$, $\beta_3$, i.e path coefficients or regression coefficients given. So, how can I calculate $r$ value for each independent variable?
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$\begingroup$ You can't deduce correlation from slope. Also, could you edit the first sentence to make it clearer? $\endgroup$– Richard HardyCommented Mar 13, 2016 at 10:47
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$\begingroup$ Yes, it's done. Like its not slope but its the regression coefficient ß. $\endgroup$– Karthik SharmaCommented Mar 13, 2016 at 11:07
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$\begingroup$ I was going through my old answers and noticed this one was not accepted. Do you perhaps need further clarification? $\endgroup$– Richard HardyCommented Feb 12, 2017 at 13:02
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1 Answer
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If we have one independent variable, then the given regression coefficient $\beta$ = Pearson's $r$.
This is not correct. A simple argument is that regression coefficients are not bounded between [-1,1] (e.g. $\beta=15$ is nothing extraordinary) while the correlation coefficient $r$ is.
If we have multiple independent variables, how can calculate Pearson's $r$ for each variable if only $\beta$ values are given for each independent variable.
You cannot. $\beta$s do not imply $r$s, i.e. for a fixed set of $\beta$s you may have different $r$s.
answered Mar 13, 2016 at 11:29
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$\begingroup$ With a single IV, the standardised regression coefficient is equal to the correlation coefficient, so the OP was not entirely wrong on this point. $\endgroup$ Commented Apr 4, 2022 at 9:00
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$\begingroup$ @TimBainbridge, you are welcome to provide such an answer. $\endgroup$ Commented Apr 4, 2022 at 9:17
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$\begingroup$ I realised a previous comment of mine was a little silly, so I've deleted it. I think this is a sufficient and correct answer. Sorry! $\endgroup$ Commented Apr 5, 2022 at 7:44
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$\begingroup$ @TimBainbridge, no worries! I did not think it was all that silly, and you are indeed welcome to post a better answer. $\endgroup$ Commented Apr 5, 2022 at 8:04