Solved – R implementation of coefficient of partial determination

anovarregression

Does anyone have suggestions or packages that will calculate the coefficient of partial determination?

The coefficient of partial determination can be defined as the percent of variation that cannot be explained in a reduced model, but can be explained by the predictors specified in a full(er) model. This coefficient is used to provide insight into whether or not one or more additional predictors may be useful in a more fully specified regression model.

The calculation for the partial r^2 is relatively straight forward after estimating your two models and generating the ANOVA tables for them. The calculation for the partial r^2 is:

(SSEreduced – SSEfull) / SSEreduced

I've written this relatively simple function that will calculate this for a multiple linear regression model. I'm unfamiliar with other model structures in R where this function may not perform as well:

partialR2 <- function(model.full, model.reduced){
    anova.full <- anova(model.full)
    anova.reduced <- anova(model.reduced)

    sse.full <- tail(anova.full$"Sum Sq", 1)
    sse.reduced <- tail(anova.reduced$"Sum Sq", 1)

    pR2 <- (sse.reduced - sse.full) / sse.reduced
    return(pR2)

    }

Any suggestions or tips on more robust functions to accomplish this task and/or more efficient implementations of the above code would be much appreciated.

Best Answer

Well, r^2 is really just covariance squared over the product of the variances, so you could probably do something like cov(Yfull, Ytrue)/var(Ytrue)var(Yfull) - cov(YReduced, Ytrue)/var(Ytrue)var(YRed) regardless of model type; check to verify that gives you the same answer in the lm case though.

http://www.stator-afm.com/image-files/r-squared.gif

Related Solutions

Solved – Coefficient of Determination with Multiple Dependent Variables

This is a first attempt at an answer.

Source
I used your data for X, Y1, and Y2.

There is a 1:1 relationship here. A particular value of X, gives particular values of Y1 and Y2. The Y values can be thought of as a single point located in a 2d space. $Y=\left[ y_1,y_2 \right]$

Procedure:

enter the data into excel (excuse any typos)
compute the mean, slope, and intercept using normal methods
compute error between mean and actual for each row
compute error between linear fit and actual for each row
compute sum of squares for the mean-error column
compute sum of squares for the line-error column
compute the ratio of the sums in steps 5 and 6
subtract that value from 1, and compare to the provided R^2

Results from approach is shown here:

enter image description here

Compute of ratio for RSS shown here:

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Graph of data shown here (yes, y1 label is poorly placed): enter image description here

If you have a column of error, and a mean value of the target, then you can compute a Pearson R^2 statistic.

Some relevant references:

Multivariate multiple regression in R

ANOVA – Comparison with R-Squared (Explained Variance)

If you are willing to assume normally distributed errors, you could simply apply a likelihood ratio test as the models you want to compare are nested.

Edit: Assuming normal errors $\epsilon$ (with variance = 1), the likelihood function of the parameters in the model $$y_i = \theta_1x_{1i} + \theta_2x_{2i} + \theta_3x_{3i} + \epsilon_i$$ is given by $$\mathcal L(\theta) = (2\pi)^{-\frac 32}\exp\left(-\frac 12\Vert y - X\theta\Vert_2^2\right),$$ where $y$ is the n×1 vector of $y_i$'s, $X$ is the n×3 matrix of $(x_{1i}, x_{2i}, x_{3i})'$ and $\theta = (\theta_1,\theta_2,\theta_3)'$. The likelihood ratio test for testing, say, $\theta_3 = 0$ is given by $$\text{LRT} = \frac{\sup_{\theta\in\mathbb R^2\times\{0\}}\mathcal L(\theta)}{\sup_{\theta\in\mathbb R^3}\mathcal L(\theta)},$$ which is approximately $\chi^2$-distributed with 1 degree of freedom.

In fact, this test does not test whether the improvement in the goodness of fit in terms of the $R^2$ is significant, however, it tests whether the improvement in terms of the likelihood, which could be seen as some measure of goodness of fit (see, for instance, the Nagelkerke $R^2$, McFadden $R^2$, Cox and Snell's $R^2$), is statistically significant.

The $R^2$, in the above example, has a Beta distribution with shape parameters $1$ and $(n - 3)/2$. However, the linear combination of two dependent (!) Beta distributed random variables is, I guess, no standard distribution. I don't think that following this track will help you achieving what you want. This is why I suggested to test the difference in likelihoods using the likelihood ratio test.

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