# model for continuous dependent variable bounded between 0 and 1

I'm attempting a multiple regression model where the predicted variable is runoff ratio - the ratio of watershed discharge to the precipitation input. This should generally be bounded [0,1], however, due to measurement error some values > 1 occur.

Originally, I modeled this with the predicted variable un-transformed, but logistic regression has been suggested to me, I also have heard Beta regression suggested. I'm not sure how to proceed, and if these transformations are appropriate to my data: My questions are: 1) Is a logistic regression appropriate for these data? and 2) If I were to proceed with logistic regression, would I need to convert the runoff ratios to proportions, or would I apply the logit to the values as they are?

Sorry if these are obtuse questions - I'm new to logit and most of the information I have found is for binary response variables.

Edited for suggested additions: As a simple version: I am modeling runoff ratio (rr) as an effect of precipitation (pcp) and antecedent water table position (ant):

rr ~ pcp + ant

rr is a continuous variable. I am not interested in the probability of specific values, rather I'm interested in the values themselves - both to assess the significance of the predictors and as a predictive model.

Conceptually, I was fine modeling it un-transformed. However, a simple linear regression allows predicted values outside of the physical range of [0,1]. As mentioned above, measurement error does lead to values >1, which I'll eventually have to deal with.

• I would suggest that you include in your question 1) The regression equation, 2) the names of the regressors and 3) What you want to measure. For example, Logistic regressions estimate the effect of the regressors on the probability that the dependent variable will acquire a specific value -is this what you want to do? etc – Alecos Papadopoulos Aug 7 '14 at 20:35
• Logistic regression is suitable of count proportions. Beta regression is sometimes used for continuous proportions. If you search on the term (here or more generally) you'll turn up a number of discussions, references and links. – Glen_b -Reinstate Monica Aug 7 '14 at 22:55

Let "$run$" be the runoff, as measured with error, so that the measured runoff ratio $rr$ is $run/pcp$. The stated model and its alternatives appear to be in the form

$$rr = \frac{run}{pcp} \sim F(\beta_{pcp} (pcp) + \beta_{ant} (ant) + \beta_0)$$

where $F$ is some family of distributions (such as Beta distributions) and the $\beta_{*}$ are coefficients to be estimated. The main problem with this is that unless the dispersion of the measurement error in $run$ is directly proportional to $pcp$, the structure of $F$ will be unnecessarily complicated. Why not algebraically rewrite the relationship as

$$run = \beta_{pcp} (pcp)^2 + \beta_{ant} (ant)(pcp) + \beta_0(pcp) + \varepsilon$$

where $\varepsilon$ represents the measurement error? The absence of several simple terms in this formula (such as one depending directly on $ant$ as well as a constant term) suggests that the proposed model may be artificially limited. Thus, ordinary regression (using $run$ or some re-expression thereof, such as a square or cube root, as the dependent variable) to fit a model like

$$run = \alpha_0 + \alpha_{pcp}(pcp) + \alpha_{ant}(ant) + \alpha_{pcp2}(pcp)^2 + \alpha_{ant,pcp}(ant)(pcp) + \varepsilon$$

would be a good way to begin an analysis. And if indeed the variance of $\varepsilon$ depends on $pcp$, that can be modeled in various straightforward ways. This approach seems more natural, realistic, and interpretable than hoping the ratio $rr$ would satisfy the more restrictive assumptions of Beta or Logistic regression.

• Very helpful, thanks. I will try the simpler, ordinary regression. I think that if modeling runoff, rather than runoff ratio, I will drop the (pcp)^2 term, as I have no physical reason to expect that to be a predictor. – Iceberg Slim Aug 7 '14 at 21:50
• If initially $run/pcp = pcp + ant$ etc then it is logically necessary to include also $(pcp)^2$ when the dependent variable is $run$ only -you essentially bring the denominator in the left-hand side to the right-hand side. – Alecos Papadopoulos Aug 7 '14 at 23:00
• @Alecos is right; and because of that, it might be wise to include the $(pcp)^2$ term in the initial model. If it is found to be insignificant (and especially if the estimated coefficient is small) then it can be dropped. The thrust of the analysis in this answer is that it suggests including both this quadratic term and an interaction term $(ant)(pcp)$ which might not otherwise normally be included in an initial model. Moreover, if such terms do turn out to be significant, then their coefficients have ready interpretations in terms of the original formulation of the $rr$ model. – whuber Aug 8 '14 at 13:58

Most (if not all) logistic regression routines require the numerator and denominator to be integers, which I do not think is your case.

The beta regression approach seems much more appropriate, though as you mention you will need to deal with the values that are greater than 1 since standard beta regression requires values in [0,1].

Is the precipitation amount used as both a predictor and as the denominator in your response variable (runoff ratio)? If so you may want to read:

Spurious Correlation and the Fallacy of the Ratio Standard Revisited Richard A. Kronmal, 1993, Journal of the Royal Statistical Society. Series A. Vol. 156, No. 3, 379-392.

Which talks about how relationships can be distorted when ratios are used/misused. You may do better with a model that predicts just the numerator of your ratio.

• It's true that the numerator and denominator are not integers. This has caused me some confusion, as I do not think I really have a proportion the way the term is usually used, rather I have the ratio between to physical values that, due to the physical system, should not exceed one. I'll take a look at that paper and consider modeling runoff volume rather than ratio. As far as beta regression, I'm not sure a beta function will fit the observed values, although maybe I misunderstand it. – Iceberg Slim Aug 7 '14 at 21:05
• +1. I think the key recommendation in your last sentence deserves elaboration, so I expanded on that in a separate answer. – whuber Aug 7 '14 at 21:27
• Beta Regression is (0,1), not [0,1] – user7340 May 5 '15 at 13:44