I am performing a meta-analysis on the response of rodent abundance to clear-cut logging. I have data from multiple sites, across multiple years, and for different species of rodents, and am using these data to compare the abundance of rodents in clear-cuts to the abundance of rodents in un-logged nearby 'control' sites.

The predictor variable is the size of the clearcut (e.g, 1ha, 2ha, etc.). The response variable is calculated as the proportional difference in abundance, between the treatment (i.e., clear-cut) site, $N_T$, and the control site, $N_C$, as follows: $$R = \frac{(N_T-N_C)}{(N_T+N_C)}$$ This response variable is a common choice in meta-analyses, and results in a variable that is bounded between -1 and 1.

I am wondering if there is a regression model that can model this relationship, something similar to a logistic regression model. A simple linear regression could work, but I'm wondering if there are other choices. If possible, I would prefer to use a model that does not require me to transform the data.

  • 3
    $\begingroup$ Since transformations are straightforward and powerful methods, could you explain why you prefer not to apply a transformation? This seems like an artificial restriction that can only hamper your ability to perform the analysis. $\endgroup$
    – whuber
    Dec 29, 2016 at 21:04
  • 2
    $\begingroup$ Mainly for consistency with previously-published meta-analyses in the same field that don't use transformations. That being said, if the only option is to use a transformation, I will go with that. I'm just curious if there's a non-transformation option. $\endgroup$ Dec 29, 2016 at 21:07

3 Answers 3


You can always use beta regression (Ferrari and Cribari-Neto, 2004). It's a model for response variable bounded in $(0, 1)$, but you can easily transform your variable by taking $\frac{Y+1}{2}$ (I know you said you do not want to transform, but it's a really basic transformation).

Moreover, such model still makes perfect sense since what you are estimating is mean of non-standard beta distribution parametrized by mean $\mu_i$ and precision $\phi$. The standard beta regression model can be used for variable with any $(a,b)$ bounds by using above transformation

$$ g(\mu_i) = x_i^{T}\beta \\ \tfrac{y_i-a}{b-a} \sim \mathcal{B}(\mu_i, \phi) $$

where $g$ is a link function (e.g. logistic function). Such model is equivalent to

$$ h(\mu_i) = x_i^{T}\beta \\ y_i \sim \mathcal{B}_{a,b}(\mu_i, \phi) $$

where $\mathcal{B}_{a,b}$ is is beta distribution bounded in $(a,b)$ and $h$ is a link function used for mapping from given range (e.g. logistic, or hyperbolic tangent functions together re-scaling and shifting if needed).

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

  • 1
    $\begingroup$ @berg2294 logistic regression can be used for binary data or proportions of successes in samples of known size, beta regression is used for continous bounded data -- those are different problems. $\endgroup$
    – Tim
    Dec 29, 2016 at 21:05
  • $\begingroup$ See this Cross Validated page for ways to proceed if the response variables include values of 0 or 1. See this page if a mixed model including random effects is needed. $\endgroup$
    – EdM
    Dec 29, 2016 at 21:09
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    $\begingroup$ An important note about using beta regression is that I believe it is required that $y_i \in (a, b)$ (where $a$ and $b$ are the bounds, in this case -1, 1), rather than $y_i \in [a,b]$. From the OP's question, it's not quite clear if that's the case? I would think $N_T$ is discrete and could easily be 0. $\endgroup$
    – Cliff AB
    Dec 29, 2016 at 23:41
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    $\begingroup$ The main problem I see with applying Beta regression is that if $N_T+N_C$ varies considerably over the data, then the model will need to allow the precision $\phi$ to vary, too. After all, low $N_T+N_C$ means poor precision in the responses and high $N_T+N_C$ means good precision. The ability to capture such heteroscedasticity directly by means of transforming the responses back to the original, raw, $(N_C,N_T)$ pairs is one advantage of returning to the original data. Another advantage is it will allow modeling of multiple $N_T$ values associated with any given $N_C$ (control observation). $\endgroup$
    – whuber
    Aug 29, 2017 at 15:25
  • $\begingroup$ @whuber what would you suggest as an alternative? $\endgroup$
    – Tim
    Aug 29, 2017 at 17:17

There now exist a proposed advancement of classical beta regression, referred to a 'Boosted Beta Regression', that may be of value in this meta-analysis with locational data.

Per this 2013 research article available online to quote, in part:

In the statistical literature, beta regression has been established as a powerful technique to model percentages and proportions [10]. Also, the method has been used in a variety of research fields [3], [8], [12]. There are applications, however, where classical beta regression methodology still has a number of limitations:

  1. Scientific databases often involve large numbers of potential predictor variables that could be included in a regression model. Consequently, if maximum likelihood estimation is used to fit a beta regression model, the model may become too complex and may thus overfit the data. This usually leads to a large variance and to a high uncertainty about the predictor-response relationships. As a consequence, techniques for variable selection in beta regression models are needed.

  2. Statistical models often suffer from multicollinearity problems, meaning that predictor variables are highly correlated. Also, observations of the response variable may be affected by spatial correlation, which is, for example, a common problem in ecology [13], [14]. To date, these issues have not been incorporated into beta regression methodology.

  3. In many applications, predictor-response relationships are nonlinear in nature [15], [16]. This means that the linear predictor of the classical beta regression model needs to be replaced by a more flexible function that allows for an appropriate quantification of nonlinear predictor effects. Although Simas et al. [17] have recently suggested an approach to incorporate nonlinear effects into beta regression models, this approach requires the functional form of the predictor-response relationships (e.g., quadratic or exponential) to be specified in advance. In cases where the functional forms of predictor effects are unknown, a more flexible approach based on smooth nonlinear effects is desirable.

  4. Percentage outcomes that are based on the binomial model are often overdispersed, meaning that they show a larger variability than expected by the binomial distribution. Classical beta regression models conveniently account for overdispersion by including a precision parameter to adjust the conditional variance of the percentage outcome (see the next section for details). On the other hand, it is often observed that overdispersion depends on the values of one or more predictor variables [17]. In the context of a beta regression model, this implies that is not constant but needs to be regressed to the predictor variables. This issue makes variable selection even more complicated because analysts need to identify the predictor variables that affect.

I welcome opinions of the article claims.

DISCLOSURE: I do not own or have any relationship with said authors or vendors selling this or other competing software.


We can consider the transformed variable

$$0.5(R+1) = \frac{N_T}{N_T+N_C}$$

which follows a scaled binomial distribution with parameter $p$ conditional on the total $N = N_T+N_C$.

So the conditional mean and 2nd moment are

$$\text{E}(0.5(R+1)|N) = p$$ $$\text{E}((0.5(R+1))^2|N) = p(1-p)/N + p^2$$

The unconditional mean and variance will relate to the expectation value of $\text{E}[N]$

$$\text{E}(0.5(R+1)) = p$$ $$\text{E}((0.5(R+1))^2) = p(1-p)/E[N] + p^2$$ $$\text{Var}(0.5(R+1)) = p(1-p)/E[N] $$

Let's assume $N_C$ and $N_T$ to be Poisson distributed with rates $\lambda_C$ and $\lambda_T$, then $E[N] = \lambda_C+\lambda_T$ and $p= \lambda_C/(\lambda_C+\lambda_T)$ or $1/E[N] p/\lambda_C$ and we can rewrite the above

$$\text{E}(0.5(R+1)) = p$$ $$\text{Var}(0.5(R+1)) = \frac{1}{\lambda_C} p^2(1-p)$$

Now we can describe $1+0.5R$ with a generalized linear model, using a quasi-likelihood as family with relationship $\sigma^2 = c \mu^2(1-\mu)$.

In a similar way we can describe $R$ directly as a generalized linear model as it just involves a shift and translation.

$$\text{E}(R) = 2p-1$$ $$\text{Var}(R) = \frac{4}{\lambda_C} p^2(1-p)$$

and now $\sigma^2 = c (\mu+1)^2(1-\mu)$.


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