Bayesian logistic regression for presence-only data

Abstract

Presence-only data are referred to situations in which a censoring mechanism acts on a binary response which can be partially observed only with respect to one outcome, usually denoting the presence of an attribute of interest. A typical example is the recording of species presence in ecological surveys. In this work a Bayesian approach to the analysis of presence-only data based on a two levels scheme is presented. A probability law and a case-control design are combined to handle the double source of uncertainty: one due to censoring and the other one due to sampling. In the paper, through the use of a stratified sampling design with non-overlapping strata, a new formulation of the logistic model for presence-only data is proposed. In particular, the logistic regression with linear predictor is considered. Estimation is carried out with a new Markov Chain Monte Carlo algorithm with data augmentation, which does not require the a priori knowledge of the population prevalence. The performance of the new algorithm is validated by means of extensive simulation experiments using three scenarios and comparison with optimal benchmarks. An application to data existing in literature is reported in order to discuss the model behaviour in real world situations together with the results of an original study on termites occurrences data.

Appendix

Property 1

Under a stratified random sampling design adjusted for presence-only data, with non-overlapping strata ${\mathcal {U}}$ and ${\mathcal {U}}_p$, the inclusion probabilities in the sample are given by

$$\begin{aligned} \rho _{0}=\frac{n_{0u}}{(1-\pi )N} \end{aligned}$$

for the stratum of cases and by

$$\begin{aligned} \rho _{1}=\frac{n_{1u}+n_{p}}{2\pi N}\end{aligned}$$

for the stratum of controls.

Proof

By definition one has

$$\begin{aligned} P(C=1 \mid y)=\, & {} P(C=1 \mid y,Z=0)P(Z=0 \mid y)\\ & {} \quad+ P(C=1 \mid y,Z=1)P(Z=1 \mid y). \end{aligned}$$

The term $P(Z=z \mid y)=\frac{P(Z, Y)}{P(Y)}$ can be computed easily from Table 1. In particular, one has

$$\begin{aligned}P(Z=0 \mid Y=0)=\frac{N_0}{N_0}=1,\end{aligned}$$

$$\begin{aligned}P(Z=1 \mid Y=0)=\frac{0}{N_0}=0,\end{aligned}$$

$$\begin{aligned}P(Z=0 \mid Y=1)=\frac{N_1}{2N_1}=\frac{1}{2},\end{aligned}$$

$$\begin{aligned}P(Z=1 \mid Y=1)=\frac{N_1}{2N_1}=\frac{1}{2}.\end{aligned}$$

Now it is easy to derive

$$\begin{aligned} \rho _{0}=\, & {} P(C=1 \mid Y=0)\\= & {} \dfrac{n_{0u}}{N_0} \times 1+ 0\\= & {} \dfrac{n_{0u}}{(1-\pi )N}\\ \end{aligned}$$

and

$$\begin{aligned} \rho _{1}=\, & {} P(C=1 \mid Y=1)\\= & {} \dfrac{n_{1u}}{N_1} \times \frac{1}{2}+\dfrac{n_{p}}{N_1} \times \frac{1}{2}\\= & {} \dfrac{n_{1u}+n_p}{2 \pi N}. \end{aligned}$$

$\square$

Proposition 1

Let us consider the population ${\mathcal {U}}$ augmented with its subset ${\mathcal {U}}_p$. Then, under the assumption that the stratum variable $Z$ is conditionally independent of $X$ given $Y$, the conditional probability of presence in the design population ${\mathcal {U}}_D$ is given by

$$\begin{aligned} P(Y=1|x)=\frac{2\pi ^*(x)}{1+\pi ^*(x)}. \end{aligned}$$

Proof

The hypothesis of conditional independence results in

$$\begin{aligned} P(Z|y,x)=P(Z|y), \end{aligned}$$

which can be express also as

$$\begin{aligned} \dfrac{P(Y|z,x)P(Z|x)}{P(Y|x)}=\dfrac{P(Y|z)P(Z)}{P(Y)}. \end{aligned}$$

Let us consider the case with $Y=1$ and $Z=0$, one has

$$\begin{aligned} \dfrac{P(Y=1|Z=0,x)P(Z=0|x)}{P(Y=1|x)}=\dfrac{P(Y=1|Z=0)P(Z=0)}{P(Y=1)}. \end{aligned}$$

The probabilities enclosed in the second term can be derived from Table 1 and one has

$$\begin{aligned} \dfrac{\pi ^*(x)P(Z=0|x)}{P(Y=1|x)}=\dfrac{\frac{N_1}{N}\frac{N}{N+N_1}}{\frac{2N_1}{N+N_1}}=\frac{1}{2}. \end{aligned}$$

(16)

In the case $Y=1$ and $Z=1$ one, similarly, obtains

$$\begin{aligned} \dfrac{P(Z=1|x)}{P(Y=1|x)}=\dfrac{\frac{N_1}{N_1}\frac{N_1}{N+N_1}}{\frac{2N_1}{N+N_1}}=\frac{1}{2}. \end{aligned}$$

(17)

From (17) it is obtained $P(Y=1|x)=2\,P(Z=1|x)$ and by substituting into (16), one can derive that $P(Z=0|x)=\dfrac{1}{1+\pi ^*(x)}$ and hence $P(Z=1|x)=\dfrac{\pi ^*(x)}{1+\pi ^*(x)}$. Now, it is easy to obtain that

$$\begin{aligned} P(Y=1|x)=\dfrac{2\pi ^*(x)}{1+\pi ^*(x)}. \end{aligned}$$

$\square$

Corollary 1

Under the assumption that, given $Y$, the inclusion into the sample ($C=1$) is conditionally independent of the covariates $X$, one has

$$\begin{aligned} P(Y=0|C=1,x)\,P(C=1|x)=\frac{1-\pi ^*(x)}{1+\pi ^*(x)}\,\rho _0 \end{aligned}$$

and

$$\begin{aligned} P(Y=1|C=1,x)\,P(C=1|x)=\frac{2\pi ^*(x)}{1+\pi ^*(x)}\,\rho _1. \end{aligned}$$

Proof

In general we have

$$\begin{aligned} P(Y|C=1,x)=\frac{P(C=1|y,x)\,P(Y|x)}{P(C=1|x)}. \end{aligned}$$

(18)

From the conditional independence between $C=1$ and $X$ given $Y$, the (18) becomes

$$\begin{aligned} P(Y|C=1,x)=\frac{P(C=1|y)\,P(Y|x)}{P(C=1|x)}, \end{aligned}$$

hence

$$\begin{aligned} P(Y|C=1,x)P(C=1|x)=P(C=1|y)\,P(Y|x). \end{aligned}$$

Recalling that $P(Y=1|x)=\frac{2\pi ^*(x)}{1+\pi ^*(x)}$ and the definitions of $\rho _0=P(C=1|Y=0)$ and $\rho _1=P(C=1|Y=1)$ the proofs for $Y=0$ and $Y=1$ can be derived. $\square$

Property 2

Under the case-control design adjusted for presence-only data the logistic regression function $\phi _{pod}(x)$ is given by

$$\begin{aligned} \phi _{pod}(x)={\text {log}}\frac{\pi ^*(x)}{1-\pi ^*(x)}+{\text {log}}\frac{n_{1u}+n_p}{n_{0u}}-{\text {log}}\frac{\pi }{1-\pi }. \end{aligned}$$

Proof

By the definition of logistic regression function for presence-only data and by simple algebra one has

$$\begin{aligned} \phi _{pod}(x)=\, & {} {\text {logit}} P(Y=1|C=1,x)\\=\, & {} {\text {log}} \frac{P(Y=1|C=1,x)}{P(Y=0|C=1,x)}\\=\, & {} {\text {log}} \frac{P(Y=1|C=1,x)P(C=1 \mid x)}{P(Y=0|C=1,x)P(C=1 \mid x)}\\=\, & {} {\text {log}} \frac{\frac{2\pi ^*(x)}{1+\pi ^*(x)}\,\rho _1}{\frac{1-\pi ^*(x)}{1+\pi ^*(x)}\,\rho _0}\\=\, & {} {\text {log}} \frac{2\pi ^*(x) \frac{n_{1u}+n_p}{2 \pi N}}{[1-\pi ^*(x)]\frac{n_{0u}}{(1-\pi )N}}\\=\, & {} {\text {log}} \frac{\pi ^*(x) \frac{n_{1u}+n_p}{\pi }}{[1-\pi ^*(x)]\frac{n_{0u}}{1-\pi }}\\=\, & {} {\text {log}}\frac{\pi ^*(x)}{1-\pi ^*(x)}+{\text {log}}\frac{n_{1u}+n_p}{n_{0u}}-{\text {log}}\frac{\pi }{1-\pi }. \end{aligned}$$

$\square$

Proposition 2

Using the approximation (10) of the ratio (8), the posterior predictive probability of occurrence for an unobserved response $Y=y$ in the sub-sample $S_u$ is approximated by the probability law ${\mathbb {P}}$ that generates the data at the population level, that is

$$\begin{aligned} P(Y=1|Z=0,C=1,x) \approx \pi ^*(x). \end{aligned}$$

Proof

From the conditional independence between $Z$ and $X$ given $Y$, the predictive probability of occurrence in $S_u$ is given by

$$\begin{aligned} P(Y=1|Z=0,C=1,x)=\dfrac{P(Z=0|Y=1,C=1)P(Y=1|C=1,x)}{P(Z=0|C=1,x)}. \end{aligned}$$

From Table 2 one has that $P(Z=0|Y=1,C=1)= \frac{n_{1u}}{n_p+n_{1u}}$ and hence

$$\begin{aligned} P(Y=1|Z=0,C=1,x)= \dfrac{n_{1u}}{n_p+n_{1u}}\,\dfrac{P(Y=1|C=1,x)}{P(Z=0|C=1,x)}. \end{aligned}$$

Now, recalling that in the general case one has

$$\begin{aligned} P(Y=1|C=1,x) \approx \frac{\left( 1+\frac{n_p}{n_{1u}}\right) \, {\text {exp}}\{\phi (x)\}}{1+\left( 1+\frac{n_p}{n_{1u}}\right) {\text {exp}}\{\phi (x)\}} \end{aligned}$$

(19)

and

$$\begin{aligned} P(Z=0|C=1,x) \approx \frac{1+{\text {exp}}\{\phi (x)\}}{1+\left( 1+\frac{n_p}{n_{1u}}\right) {\text {exp}}\{\phi (x)\}}, \end{aligned}$$

(20)

by substituting (19) and (20) in (19), one obtains

$$\begin{aligned} P(Y=1|Z=0,C=1,x)\approx \,& {} \dfrac{n_{1u}}{n_p+n_{1u}} \,\dfrac{\left( 1+\frac{n_p}{n_{1u}}\right) \, {\text {exp}} \{ \phi (x) \}}{1+ {\text {exp}} \{ \phi (x) \}} \\=\, & {} \dfrac{{\text {exp}}\{\phi (x)\}}{1+{\text {exp}}\{\phi (x)\}}\\=\, & {} \pi ^*(x). \end{aligned}$$

$\square$