I am doing some regressions on real earnings as a function of some different variables and this came out:
Is this because earnings cannot be negative?
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You can answer the question for yourself with simple mathematics. If observed $y \ge 0$ and $\hat y$ denotes fitted $y$, then residual $e = y - \hat y$ must be $\ge -\hat y$. The line $e = - \hat y$ is thus a lower limit on your residuals. Despite your unconventional axis choice, it is clear that your data follow suit.
The underlying problem is presumably use of a standard linear model on data not suited to such. One way forward is a log-linear or Poisson(-like) model: (fortuitously but fortunately for the OP as a Stata user) there is a Stata-rich explanation in this blog posting. The posting should be of considerable interest to many users of statistics, however.
P.S. A standard residual plot has residuals on the vertical axis and fitted or predicted on the horizontal axis. The choice of axes is not here an arbitrary convention. A horizontal line indicating zero residuals is the natural reference line, as indicating behaviour matching a perfect model. As emphasised often by J.W. Tukey and others, the best references are linear, and the best linear references are horizontal, in the sense of being easiest to think about. In Stata there is a built-in post-estimation
rvfplot for use after
P.P.S. The graph flags a Stata user. Naturally use of Stata is quite secondary here to the main question.
There are two major aspects I see in the plot that I expect you might wonder about.
(I took the liberty of flipping your plot about to the way I'm more used to looking at them, with the random quantity on the y-axis.)
The first aspect is what looks like hard lower bound on the y-values (which is presumably 0), as you suggested.
The second is the fan-shape ("$<$") in the residuals. The two are related issues.
The spread seems to be linear in the mean - indeed, I'd guess proportional to it, but it's a little hard to tell from this plot, since your model looks like it's also biased at 0.
In that case, variance is proportional to the square of the mean, which suggests either taking logs (working with log-earnings would be a pretty common choice) or fitting a model with variance proportional to mean-squared (such as a Gamma GLM).
I disagree with Nick on this* - a Poisson-like model is unsuitable; a quasi-Poisson has variance proportional to mean, or standard deviation proportional to the square root of the mean, so its residual plot would look more parabolic. This one doesn't. As is common with financial data, the standard deviation is approximately proportional to the mean -- indeed it would be somewhat surprising if it were not, since it would imply that it would matter whether you worked in dollars or thousands of dollars
* Or perhaps I don't, since it seems we were having more a difference of terminology than substance.
If you have exact zeros in your data, neither of those suggestions would be suitable (at least not without some modification), but there are also zero-inflated models.
Working with a more appropriate model for the variance will likely improve other aspects of your model as well.