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Line 30: It is useful to consider the second definition to get the idea behind FVU. When trying to predict ''Y'', the most naïve regression function that we can think of is the constant function predicting the mean of ''Y'', i.e., <math>f(x_i)=\bar{y}</math>. It follows that the MSE of this function equals the variance of ''Y''; that is, ''SS<sub>E</sub>'' = ''SS<sub>T</sub>'', and ''SS<sub>R</sub>'' = 0. In this case, no variation in ''Y'' can be accounted for, and the FVU then has its maximum value of 1. More generally, ~~ASHEEM~~the FVU will be 1 if the explanatory variables ''X'' tell us nothing about ''Y'' in the sense that the predicted values of ''Y'' do not [[covariance\|covary]] with ''Y''. But as prediction gets better and the MSE can be reduced, the FVU goes down. In the case of perfect prediction where <math>\hat{y}_i = y_i</math>, the MSE is 0, ''SS<sub>E</sub>'' = 0, ''SS<sub>R</sub> = SS<sub>T</sub>'', and the FVU is 0. ==See also==

Fraction of variance unexplained: Difference between revisions