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Since standardized betas are correlation coefficients in bivariate regression, is it the case that standardized betas in multiple regression are partial correlations?

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Short answer: No. –  Glen_b Feb 17 '13 at 2:51
    
Because the value of a beta can be anything (including of absolute value greater than $1$), it cannot generally be interpreted as a partial correlation. –  whuber Feb 17 '13 at 3:00

2 Answers 2

Longer answer.

If I have this right --

Partial correlation:

$$ r_{y1.2} = \frac{r_{y1}-r_{y2}r_{12}}{\sqrt{(1-r^2_{y2})(1-r^2_{12})}} $$

equivalent standardized beta:

$$ \beta_1 = \frac{r_{y1}-r_{y2}r_{12}}{\sqrt{(1-r^2_{12})}} $$

As you see, the denominator is different. Indeed, since the additional term in the denominator of the first thing is between 0 and 1 (inclusive), it looks like $\beta_1$ will almost always be smaller than $r_{y1.2}$ though they could be equal if the two "independent variables" (and here's why I dislike that term) are ... well, independent - or at least uncorrelated.

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+1. But not least fact is that the numerator is the same. This implies that both coefficients are just different ways to standardize the raw regression coefficient b. –  ttnphns Feb 17 '13 at 8:05

I've in another question the following covariance matrix C for the three variables X,Y,Z given:
$$ \text{ C =} \small \left[ \begin{array} {rrr} 1&-0.286122&-0.448535\\ -0.286122&1&0.928251\\ -0.448535&0.928251&1 \end{array} \right] $$ From its cholesky-decomposition L $$ \text{ L =} \small \left[ \begin{array} {rrr} X\\Y\\Z \end{array} \right] = \left[ \begin{array} {rrr} 1&.&.\\ -0.286122&0.958193&.\\ -0.448535&0.834816&0.319215 \end{array} \right] $$ we can directly retrieve the partial correlation between Y,Z wrt. X as $ \small corr(Y,Z)_{\cdot X} = 0.958193 \cdot 0.834816 $ Now if we have the variables ordered such that the dependent variable is Z then the betas are computed by inverting the square-submatrix of the range in L which is populated by the independent variables X,Y: $$ L_{X,Y} = \small \left[ \begin{array} {rrr} 1&.\\ -0.286122&0.958193 \end{array} \right] $$ and its inverse, which is inserted into a 3x3 identity-matrix to form the matrix $M$: $$ M = \small \left[ \begin{array} {rrr} 1&.&.\\ 0.298605&1.043631&.\\ .&.&1 \end{array} \right] $$ Then the betas occur by the matrix-multiplication $ \beta = L \cdot M $

$$ \beta =\small \left[ \begin{array} {rrr} X\\Y\\Z \end{array} \right] = \small \left[ \begin{array} {rrr} 1&.&.\\ .&1&.\\ -0.199254&0.871240&0.319215 \end{array} \right] $$

which indicates, that the $\beta_X$ contribution for $Z$ is $\small \beta_X=-0.199254$ and the $\beta_Y$ contribution for $Z$ is $\small \beta_Y=0.871240$ . The unexplained variance in Z is the bottom-right entry squared: $\small resid^2= (0.319215)^2$
We see in $M$ that -being an inverse of a partial cholesky-matrix- it can contain values bigger than $1$ - and as well the Beta-matrix can then have entries bigger than 1.

So - to come back to your question- the partial correlation between $Y$ and $Z$ were the product of the entries in the second column of the L-matrix. The $ \small \beta_Y$ however is the product of the entry in the second column of the Z-row with the inverse of that in the Y-row and the relation between the concepts of partial correlation and $\small \beta$ can be described by this observation.

Additional comment: I find it a nice feature, that we get by this also the compositions of $X$ and $Y$ in terms of $X$ and $Y$ - which of course are trivially 1. It is also obvious, how we would proceed, if we had a second dependent variable, say $W$, and even that scheme can smoothly be extended to compute/show the coefficients of the generalization to the canonical correlation - but that's another story....

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