Coefficient of Determination/regression formula question

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MizzouNoel

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Hi, so I am wondering something.

If I run a bunch of linear regressions for a single y variable (Example: Running "points differential" in linear regressions with "rebound%", "field goal%", "turnover%", etc.), what would it do if I multiply the formula for each by the coefficient of determination (r squared)? I've thought about using this as a way of "weighting" statistics in a formula and trying to create a formula that predicts production based on several factors by doing this, but want to know if it would give me meaningless numbers?

Further example:

If the regression equation for the first x-value is "y = 1x + 2" and r squared is "0.5"
and the regression equation for the second x-value is also "y = 1x + 2" but the r squared value is "1",

If the value for both x's is 1, but I multiply the resulting y-value by the coefficient of determination, the first x would give me a weighted value of 1.5 while the second would be worth 3. Is that valid?

Is there a better way to do what I'm trying to do? Gracias
 
MizzouNoel said:
Hi, so I am wondering something.

If I run a bunch of linear regressions for a single y variable (Example: Running "points differential" in linear regressions with "rebound%", "field goal%", "turnover%", etc.), what would it do if I multiply the formula for each by the coefficient of determination (r squared)? I've thought about using this as a way of "weighting" statistics in a formula and trying to create a formula that predicts production based on several factors by doing this, but want to know if it would give me meaningless numbers?

Further example:

If the regression equation for the first x-value is "y = 1x + 2" and r squared is "0.5"
and the regression equation for the second x-value is also "y = 1x + 2" but the r squared value is "1",

If the value for both x's is 1, but I multiply the resulting y-value by the coefficient of determination, the first x would give me a weighted value of 1.5 while the second would be worth 3. Is that valid?

Is there a better way to do what I'm trying to do? Gracias
Click to expand...

Is there a better way? Possibly, but I think that is what the professionals are looking for also. Have you seen the movie 'Money Ball' - it's about baseball but the idea is the same.

Normally, as I have understood it, one would expect R2 to be between 0 and 1 with a value of one indicating a 'perfect fit' and a value of zero being a 'get out of here' fit. So weighting by R2 looks to be a decent idea. You certainly can't weight a bad fit worse than zero and accepting a 'perfect fit' 100% is hard to beat. You could weight the high end (near 1) more and the low end (near zero) less by using some sort of function, i.e. W = \(\displaystyle \sqrt {1 - (R2-1)^2}\), where W is the weighting factor but I've no idea how that would work out.
 
Thank you, that's what I was thinking too. Intuitively, it seems like a logical way to express how meaningful a regression is and reflect that into a formula, but I don't necessarily know what kinks (if any) would have to be modified in that design. Perhaps standard deviation would come into play to reflect the range of normal expected outcomes?

Like, you could display the results as a range of y values (given "x", you would expect y to be either -1 standard deviation from the mean, or +1 standard deviation, and then you multiply both of those y-values by their r squared values in order to find the range of reasonable predicted outcomes for the final weighted outcome value? I am making up some of these terms as I go and am not 100% sure that anything I've said makes sense haha, I just want opinions on why or why not this stuff could be useful)
 
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