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In the thread from correlation coefficient to conditional probability, a simple formula is given for calculating a conditional probability from a correlation coefficient: $$\frac {1} {2} + \frac {\arcsin \rho} {\pi} ,$$

in which $\rho$ is the correlation coefficient. Now, this goes for a simple population correlation coefficient between two variables. What I have, is a sample and multiple regression; for example, in the sample, dependent variable Y is predicted by independent variables A, B and C. If SPSS gives me the multiple R for Y, can I use that in the same way as in the formula above to calculate a conditional probability for Y? If not, how can I calculate that probability? Or should I use the (standardized or unstandardized) regression coefficients (betas) of A, B and C to calculate conditional probability, instead of using the multiple R? I don't have individual cases, just means, standard deviations, N of cases and a (Pearson) correlation matrix for all variables. With this as input, SPSS is able to do multiple regression, from which i get the multiple R and the regression coefficients. I'm using this in my own analysis of data, reported in a number of articles about the prediction of reading problems.

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  • $\begingroup$ What do you mean by "multiple R"? "Multiple R squared"? $\endgroup$ – Ferdi Jun 13 '17 at 19:51
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    $\begingroup$ Your question appears to be predicated on things that just aren't true. That formula applies only to a bivariate normal distribution and only to a specific probability (that both variables exceed their means). There can be no general formula to translate correlation into conditional probability. Since you have data and the software to work with them, why do you seek such formulas? What is the statistical problem you are trying to address? $\endgroup$ – whuber Jun 13 '17 at 19:56
  • $\begingroup$ By "multiple R" I mean "multiple R", not "multiple R squared". The question is about converting a correlation coefficient into a conditional probability, and R is a (multiple) correlation coiefficient. $\endgroup$ – JaapV Jun 16 '17 at 11:04
  • $\begingroup$ From the answer of whuber, I see that the case is more complicated than I thought. I am seeking such a formula simply because I want to look at the data in different ways. With reading problems in children, let's say at the age of 8 years, it's useful to be able to predict them from the children's earlier scores on one or more variables, at 4 or 5 years, but it's also useful if you can say, based on the scores on predicting variables, "These children have a chanced of x % of developing reading problems". That is why I'm interested in conditional probability. $\endgroup$ – JaapV Jun 16 '17 at 11:10
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    $\begingroup$ An expression like $P (y_1>y_2|a_1>a_2)$ is over-complicating the analysis for the goal you wish to achieve. You would have to integrate over all possible pairs $a_1,a_2$ for which $a_1>a_2$ and sum the probabilities $P(a_1,a_2)$ multiplied with $P(y_1>y_2|a_1,a_2) $. You could instead use your regression coefficients to express the increase in the expected outcome as function of the independent variables. Also your 'conditional probability for Y' sounds much like Y is a binary variable, and you can do a logistic regression, directly computing probabilities for Y as function of A, B and C. $\endgroup$ – Sextus Empiricus Dec 9 '18 at 11:10
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Partially answered in comments:

Your question appears to be predicated on things that just aren't true. That formula applies only to a bivariate normal distribution and only to a specific probability (that both variables exceed their means). There can be no general formula to translate correlation into conditional probability. Since you have data and the software to work with them, why do you seek such formulas? What is the statistical problem you are trying to address? – whuber

AS for your additional question in comments: With reading problems in children, let's say at the age of 8 years, it's useful to be able to predict them from the children's earlier scores on one or more variables, at 4 or 5 years, but it's also useful if you can say, based on the scores on predicting variables, "These children have a chanced of x % of developing reading problems". This could be addressed by, for instance, logistic regression, or maybe better, if you have a reading score at 8 years and not only a binary response, usual (linear) regression or some variant.

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