# Assessing performance of linear model with fixed slope

I have a very simple linear model that I need to fit with a fixed slope (I'm interested in estimating the intercept). How would I go about describing the performance of the fit, since the usual r^2 doesn't apply. I'm reasonably handy with the math, so even a point in the right direction would be helpful. Doing it in R (lm with offset), in case there's a handy solution there

• What exactly do you want to use this measure to do? Would mean square error work? May 19, 2014 at 12:25
• Mean s1uare error could work, especially compared with MSE of a standard lm. I have theoretical reason to expect a fixed slope, which would then make the intercept physically meaningful. I'm interested in seeing if my data match this assumption! May 20, 2014 at 19:39

Perhaps the best way would be to examine the distribution of the residuals. You get them from lm (it's one of lm's list components). The root mean square error (as Glen_b mentioned) gives you a sense of how far your line is off on average. You could also plot the residuals in a histogram to get a more complete sense of how badly the line fits. Perhaps the most interesting look would be to plot the residuals by the predictor(s) which has(ve) fixed slope to see if there is some trend there.

However, your assumption that $R^2$ is useless is incorrect.. See the definition here. $R^2$ is really just a way to understand how much better/worse your line is compared to the worst possible 'best guess' line: the mean of the observations. It's possible for your line to have an $R^2$ that is not in $[0,1]$, since your fixed slope line is not guaranteed to be better than the mean of the observations like a typical regression line would be (e.g., the observations are all on a horizontal line but your fixed slope is nonzero).

• Thanks, I'm not sure where the prescription against r^2 comes from, I don't see anything in the math that prevents it's use. I suppose I could also compare the errors to those from a floating slope lm. May 20, 2014 at 19:38
• @IcebergSlim in your question you seem to claim that you can't use $R^2$ and that's what I was alluding to.
– user44764
May 20, 2014 at 19:40
• Agreed - just wondering why everywhere I read seemed to say not only not to use it, but that it was somehow fundamentally meaningless. May 20, 2014 at 19:54
• @IcebergSlim Well besides being tricky for standard analyses, it isnt' really that telling/useful in your case since you aren't automatically doing better than the mean. With that said, it still has meaning as a way to compare yourself to the mean of the dependent variable.
– user44764
May 20, 2014 at 20:05

Let's say you can group your explanatary varibles into two categries, $X_1$ and $X_2$.

$X_1$ corresponds to the fixed slope and $X_2$ corresponds to the intercept. Since $X_2$ corresponds to the intercept, $X_2$ is a factorial variable.

Your linear regression model, theoretically, would be like, using Least Square Estimation.

$Y=b_0+X_1b_1+X_2b_2+X_1X_2b_{12}$

Since the slope is fixed, we have $b_{12}=0$.

We have $=b_0+X_1b_1+X_2b_2$, and the slope is $b_1$, and the intercept is $b_0+X_2b_2$.

Your observation can be grouped according to the fatorial explanatory variable $X_2$.

The linear estimations of all the groups have the same slope with different intercepts.