Can I predict values from a multiple regression model with fewer predictors than there are in the model? I have built a linear regression model with three predictor variables: the model predicts forest growth (y = stand volume) with stand age, stand basal area and site type (x1, x2 and x3 respectively). However, my goal is to predict the forest volume so that I only know the forest age and site type. Is there any way to make the prediction with the model when the stand basal area is unknown? I have also tried to build the model with only two predictor variables but it doesn't produce good values if the stand basal area is removed.
Do you have any tips? I couldn't find anything by googling except this: https://stackoverflow.com/questions/28528703/ols-predict-using-only-a-subset-of-explanatory-variables. It suggests setting the value of unused variable as 0.
EDIT:
So I have already trained my model and then tested it with a testing dataset with the predict() function. BOTH, the training AND the testing data sets indeed contained the value for area. But I'm now confused when I would have to find the volume for a certain-aged forest (in a future) as I don't know the area of the forest for example after 50 years. Thus, I can not add it to the model as a predictor variable. I added this in case someone thought that I haven't done the model predictions with a testing data set yet.
 A: That's very unusual, I can't see how one could be expected to estimate a volume without an area. I would suggest running predictions with a number of example values of area. For example, choose three values of area that you think are reasonable given your context, then run the model three times - each with a fixed area value (and variable x1 and x3 values). Then present prediction results as a range of volumes that could reasonably be expected from stand ages and site types.
A: Your regression models the mean conditional on having feature values, and if you don’t have values for those features, the conditional mean calculation breaks down. Thus, you need some value for that third covariate, and the predictions will not be the same for all such values.
However, you’re allowed to input whatever you want into the equation, so you can pick a value or even several values. For instance, you might want to make a prediction based on your two feature values and the mean value of that third feature (maybe the median value, maybe both). You might want to give a range of values, such as saying the range when you set that third feature to its $10$th and $90$th percentiles. You might consider extreme scenarios, such as what happens if that third feature takes its lowest or highest values (maybe go even a bit more extreme).
