I have a data set with death rate per 1000 birth as the response variable. The range of values for the response variable is [0.03 - 0.89]. The question to find association between death rate and other covariates.

Looking at the value what came to mind was beta regression.

For this particular data set, the values of the response variable lies between 0.03 and 0.89,but in general depending on the population I know that the death rate can be greater than 1.

Alternatively, I thought about count model, for count model I would need the values for the denominator to convert the death rates to death count, and model it using Poisson model. Unfortunately information about the denominator is not available and there is no reliable way to compute it.

Which brings me to my questions: is it still okay to go ahead and model the data using beta regression?

If not what other statistical method can I use that takes into consideration that the values can be bounded by 0 on the left-hand side but not bounded by 1 on the right hand side (though the right hand side value can not infinity)?


1 Answer 1


The Beta distribution is indeed defined in the $(0, 1)$. Hence, if you want to allow that the outcome can take a value greater than one, this cannot be done with the Beta model. However, if you would be willing to accept an upper bound, say that your outcome $Y$ is constrained in the interval $(0, A)$, then you could still transform to $(0, 1)$ by considering instead the outcome $Y^* = Y / A$. Nonetheless, note that the Beta distribution is defined in the $(0, 1)$, and thus, you still have a problem with the zeros.

As an alternative, you could consider treating the data as semi-continuous taking the value 0 or being positive. This is essentially a mixture model with a logistic regression submodel for the binary outcome $I(Y = 0)$ and a linear mixed model for the logarithm of the positive values.

If you want to do this in R, then you can use the GLMMadaptive package. An example illustrating an analysis of semi-continuous data can be found here.

  • $\begingroup$ thanks for your answer I will look into the suggested materials. $\endgroup$
    – tk001
    Oct 18, 2019 at 14:29

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