Legitimacy of Regressing Actual Values on Predicted Values for Better Residual Sum of Squares? A coworker of mine recently performed an analysis where after training a simple linear model, he regressed the actual Y values against his model's predicted Y values, and applied this regression's slope and intercept back to the original predictions to get better point-to-point agreement (residual sum of squares) while maintaining the same $R^2$.
I'm not as versed in statistics as he, but I've never heard of such a technique before, and can't find any information online supporting the legitimacy of it. It seems a bit like cheating since he's using the true values post-modeling to forcibly get better point-to-point agreement, but he justifies it by saying the model's ability to predict variance ($R^2$) remains the same, he's just adjusting for the model underestimating the point values.
Is this approach valid?
The following is a toy example of such a calculation:
Y_Actual    Y_Predicted 
0.2         0.1         
0.4         0.2         
0.7         0.3 

RSquare:    0.987   
RSS:        0.210   

Regress Y_Actual on Y_Predicted:    
Slope:  0.395   
Intercept:  0.029

Apply regression for new Y'_Predicted via:
Y'_Predicted = (Y_Predicted - intercept)/slope
Y_Actual    Y_Predicted Y'_Predicted
0.2         0.1         0.18
0.4         0.2         0.43
0.7         0.3         0.69

RSquare:    0.987       0.987
RSS:        0.210       0.002

As we can see, Y'_Predicted has the same $R^2$ as Y_Predicted, but much better Residual Sum of Squares:

 A: I believe your colleague made a mistake in claiming that the $R^2$ stays the same.
This is because $R^2$ is defined as: 

$R^2 = 1- {(SS_{res})\over(SS_{tot})}$

The reason they have to be different is because both 1 and $SS_{tot}$ are fixed by the data set. The only thing that changes between the two models is $SS_{res}$. So this seems to suggest there is an issue with your colleagues model or model fit. 
If we just plug the two into each other what we get is:

$Y = \alpha*x + \beta$

Then based on this Y he performed the following regression:

$Y_2 = \alpha_2*Y + \beta_2$

So if we plug this all into one equation we get:

$Y_2 = \alpha_2*(\alpha*x + \beta) + \beta_2$

We can simplify this a little bit to show the issue:

$Y_2 = \alpha_2*\alpha*x + \alpha*\beta + \beta_2$

This is of course just the equation for a line with:

$slope = \alpha_2*\alpha$

and

$intercept = \alpha*\beta + \beta_2$

So if your colleague's first model had been fit correctly no improvement should be possible by regressing the predictions on the actual values because a linear model fitting is already doing that. 
So I suspect he made a mistake in fitting his data/setting up his model. 
Now as to why the $R^2$ can appear to stay unchanged is because there are multiple definitions of $R^2$ and they are only sometimes the same. The definition that I quoted above is useful for the regression context. But it should be identical to simply squaring the correlation coefficient. However, this only holds for a successful least squares fit. So if there is a mistake/issue the two will not be equal. However, it's not clear without knowing which software was use which approach to calculate the $R^2$ was used so the program may just be showing the square of the correlation coefficient.
