I was hoping to get some input on people's thought process when looking at residual plots of a linear regression to assess the fit / whether assumptions are met. I included a plot of a model I'm currently working on as an example.
I was always taught when looking at such a plot, it's best not to be so picky i.e. we're looking for serious deviations. Generally the first thing I usually look for is any apparent trend (i.e. what I want is a shapeless cloud of points centered around 0), and then look for evidence of uneven vertical spread, and then check qq plot for normality. Upon initial inspection of my plots, I thought the qq plot looked "normal enough" and there was no apparent trend in my residuals vs. fitted plot. However, it did look like there was evidence of uneven vertical spread (i.e. heteroskedasticity). Now it seems to me like what I thought was uneven vertical spread was really just a downward sloping trend. What do you guys think?
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The residual plot shows that you have a lower-bound on your response variable, which is contrary to the assumptions of the standard Gaussian linear regression (I am assuming that this is the model you fit). Your description of the data does not disclose the variables and their allowable values, but presumably you have a response variable that gives only positive values.
In such cases you often get a better model fit using a log-linear model, where you take the logarithm of the response variable before fitting the regression. If you are using a response variable with a known lower-bound (e.g., zero), I would suggest you fit this alternative model. That is likely to conform far better to the regression assumptions than your present model.
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$\begingroup$ That is exactly the case! my response variable is a type of count. hours until a certain "event". I thought a generalized linear model would be more appropriate as opposed to a survival analysis because the entire sample had the "event" in question - which i figured would result in bias in the analysis. Thank you for the suggestion on log-linear model $\endgroup$– cliftjc1Commented Mar 19, 2020 at 0:23
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$\begingroup$ I'm trying to fully understand the logic behind what led you to that correct diagnosis based off of what you saw in the graph, could you be a little more explicit in your thought process? $\endgroup$– cliftjc1Commented Mar 19, 2020 at 0:25
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2$\begingroup$ In the residual plot to the left, you can clearly see the lower-bound of the residuals at the bottom (shown as a diagonal boundary on the residual values). This kind of pattern frequently occurs when you fit a bounded response variable using a standard Gaussian linear regression. Since you have a response variable that is a count variable, I would recommend you try a negative binomial GLM as a starting point for analysis. This should give you a much better fit than a standard linear regression. $\endgroup$– BenCommented Mar 19, 2020 at 3:05
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$\begingroup$ thanks for that, I will research negative binomial GLM. follow up question(s): (1) I now understand to equate the downward sloping trend seen in the bottom left of the residual graph with a left-bounded response variable.. but I'm still struggling with the intuition behind it. So the bounded nature of the response restricts the residuals in such a way that the vertical spread of negative residuals increases as predictions increases? cause we get farther away from our bound point? (2) if my response was bound at the top, would I see a similar downward slope in the top right of the graph? $\endgroup$– cliftjc1Commented Mar 19, 2020 at 14:34
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1$\begingroup$ If you have the response bound $y \geqslant L$ then you have the residual bound $r \geqslant L - \hat{y}$, and since $\hat{y}$ is the horizontal axis of the residual plot, you see the diagonal lower bound on the residuals in that plot. If instead you had an upper bound $y \leqslant U$ then you would get the residual bound $r \leqslant U-\hat{y}$, so yes, you would see a similar diaogonal upper bound at the top of the residual plot. $\endgroup$– BenCommented Mar 19, 2020 at 22:13