Use squared correlation in regression without intercept [duplicate]

If I want to compare the goodness-of-fit of two regression models, with and without intercept, is it valid to compare the squared correlation coefficient between the fitted values and the data? Since the squared correlation would get back the $R^2$ for the model with intercept, it seems to make sense to compute the squared correlation for the model without intercept and use that to make comparison. I am not entirely sure whether this is legitimate. Am I missing anything? If this approach is not valid, what methods can I use to make such comparisons?

marked as duplicate by kjetil b halvorsen, Peter Flom♦ regression StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 26 '18 at 10:50

• I think this is a good idea. You just lose the interpretation of the $R^2$ as the "proportion of variability explained by the model". – Stéphane Laurent Jan 11 '13 at 21:09
• Sorry I misunderstood the question. I didn't see your goal is to compare the two models. As said by ttnphns below, you would get the same correlation for both models. – Stéphane Laurent Jan 12 '13 at 10:47

$R^2 = 1-SS_{residuals}/SS_{total}$, where "SS" is "sum of squares".
In the model with intercept included, $SS_{total}$ is the sum of squares about the dependent variable's mean. In the model with intercept suppressed, $SS_{total}$ is the sum of squares about 0, i.e. the sum of squares in the non-centered dependent variable. Therefore, one cannot directly compare the two R-squares.
• I am aware that $R^2$ for a regression model without intercept does not have the valid interpretation. That's why I am asking whether it's valid to compare squared correlation instead of $R^2$. – user13587 Jan 11 '13 at 20:27