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The title question here is a bit awkward because I'm really asking if this illustration I've drawn is true:

My interpretation of Beta Regression

Suppose we have a Beta Regression of one predictor, X, which is used to model both the response, y, and the variation of y. So, for example, at X=x1, x2 and x3, we would generate the point predictions mu1, mu2 and mu3 as I have illustrated. Since we are also modelling the precision over the predictor, the variance of the response at each of these locations also changes. I have tried to illustrate that.

As I understand this answer from several years ago, the variances that would be obtained from predict.betareg() in R are about the response. I feel like it therefore follows that we're saying:

  1. y1|X=x1 ~ B(mu1, var1)
  2. y2|X=x2 ~ B(mu2, var2)
  3. y3|X=x3 ~ B(mu3, var3)

as I have annotated my graphic. Is this actually the case or have I misunderstood something?

I think I am right because I can obtain the same quantiles in R that predict.betareg() generates by converting the mu and phi values that predict.betareg() also generates into shape1, shape2 values and using those within qbeta(). However, I am concerned that this may be a coincidence... I would like to stress that my question is not about the code because I can see that the code seemingly confirms this interpretation (it is, in fact, how I came to have this interpretation), but the theory.

As an implication, if this is true does that mean we could validate the appropriateness of a Beta Regression by generating multiple residuals for each x value (e.g. using y3 - realisations from B(mu3,var3)) to infer the residual distribution for the model and then seeing if the actual residuals from the point predictions belong to that distribution? I've looked into Beta Regression before (there wasn't even a Wikipedia article at the time, but I see that there is now), and I really struggled to find an explanation of model diagnostics then.

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    $\begingroup$ The residual distribution differs for every value of x, so I don't see any use for the pooled residuals here. $\endgroup$ Commented Jul 5, 2023 at 11:06

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Beta regression is a generalization of the generalized linear model, and it assumes that the model takes the form of a linear predictor $\eta_i$, that is transformed using the link function $g$ (e.g. logistic function), to obtain the mean $\mu_i$ of the beta regression

$$\begin{align} \eta_i &= \mathbf{X}\beta \\ E[y_i|\mathbf{X}] &= \mu_i = g(\eta_i) \\ y_i &\sim \mathcal{B}(\mu_i, \phi) \end{align}$$

where $\mathcal{B}$ is the re-parametrized beta distribution, with parameters $\mu$ for the mean and $\phi$ for precission. In this sense, beta regression is defined in a similar way as other regression models that predict the expected value. The model is described in greater detail by Ferrari and Cribari-Neto (2004).

Commenting on your question, notice also that the classic beta regression model assumes that there is a single $\phi$ parameter (no $i$ subscript), so it does not change for different observations. It is possible though to have varying $\phi_i$ parameters, as noticed in the comment below by Achim Zeileis.

As for validating the model, it is not the case that we need to test every assumption of the model otherwise it is not applicable. If you search our site for "test regression assumptions" you will find many threads, on many different models. What you will learn from the threads is that in some cases it is useful to conduct formal tests, in some, it is not. In many cases, we look at the diagnostic plots for this purpose to see if there are no serious problems, but minor discrepancies usually are not a concern.


Ferrari, S., & Cribari-Neto, F. (2004). Beta regression for modelling rates and proportions. Journal of Applied Statistics, 31(7), 799-815.

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  • $\begingroup$ The classic beta regression model has a fixed precision parameter $\phi$. But very often the extended model is used where both $\mu$ and $\phi$ depend on regressors. This is also available in betareg. $\endgroup$ Commented Jul 6, 2023 at 0:01
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    $\begingroup$ @AchimZeileis thanks, edited for the correction. $\endgroup$
    – Tim
    Commented Jul 6, 2023 at 7:39

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