Is it possible that the fraction of variance unexplained becomes greater than 1? I read this definition of fraction of variance unexplained from https://en.wikipedia.org/wiki/Fraction_of_variance_unexplained:
$$FVU = \frac{VAR_\text{err}}{VAR_\text{tot}}$$
Support the real data is $x_i$ and the regression estimate is $y_i$, then if $x_i$ and $y_i$ are completely uncorrelated, then 
$VAR_\text{error} = VAR(x - y) = VAR(x) + VAR(y)$,
but 
$VAR_\text{tot} = VAR(x) \le VAR_\text{error}$
Therefore if 
$VAR(y)\ne 0$
then
$FVU > 1$
I am wondering if I make anything wrong here?
Thanks!
 A: Lets assume a simple model, $Y=\beta_0 + \beta_1X+\epsilon$. The absolute value of Pearson's correlation coefficient is bounded by $1$,
$$ 
|\rho_{X,Y}| = \left|\frac{cov(X,Y)}{\sigma_X \sigma_X}\right| \le 1
$$
you can easily show it by using Cauchy-Schwartz inequality. To estimate $\rho$ you use the sample equivalent $r^2=S_{XY}/(S_{XX}S_{YY})$. By using the LS estimators, you can show that for the simple model $r^2 = R^2$, that is, square of the Person's statistic equals the "proportion of explained variance" that have to lie in $[0,1]$.
The key term  here is the use of the Least Squares estimators. If you are using some other unrelated method to predict $Y$, then you definitely  can get such "anomalies" like $SSE>SST$, but in this case the very use of the $R^2$ is questionable. Sure, you can construct some generalized or alternative measure of $R^2$ that would satisfy its initial mathematical properties, but it won't be the basic $SSR/SST$ proportion from the LS simple linear model.   
