6
$\begingroup$

I'm reading a section on conditional logistic models in which a heterogeneous Poisson process is used to make inferences in disease mapping. Basically, the likelihood of a Poisson process is used to describe the events $\{s\}$ within a spatial region $T$

$$L(\{s\}|\Psi) = \frac{1}{m!}\prod_{i=1}^{m}\lambda(s_{i}|\Psi)\exp\{\Lambda_{T}\} $$

where $\Lambda_{T} = \int_{T}\lambda(u|\Psi)du$. $\lambda(s)$ is called the intensity. This quantity determines the rate of case events (detected cases of disease, for example) and $\Lambda_{T}$ is the intensity over the region $T$.

Usually for case events, $\lambda(s|\Psi)$ is given by $\lambda_{0}(s|\Psi_{0})\lambda_{1}(s|\Psi_{1})$ in which $\lambda_{0}(s|\Psi_{0})$ is a "spatially-varying function of the population at risk of the disease in question" and $\lambda_{1}(s|\Psi_{1})$ includes appropriate predictors.

However, in the case of modeling cases and controls, I don't understand what calculation is performed. I quote the relevant paragraph:

When a bivariate realization of cases and controls are available it is possible to make conditional inference on this joint realization. Define the case events as $s_{i} : i = 1, ..., m$ and the control events as $s_i : i = m + 1, ...., N$ where $N = m + n$ the total number of events. Associated with each location is a binary variable ($y_i$) which labels the event either as a case ($y_i = 1$) or a control ($y_i = 0$). Assume also that the point process models governing each event type (case or control) is a heterogeneous Poisson process with intensity $\lambda(s|ψ)$ for cases and $\lambda_0 (s|ψ_{0})$ for controls. The superposition of the two processes is also a heterogeneous Poisson process with intensity $λ_0(s|ψ_0) + λ(s|ψ) = λ_0 (s|ψ_0 )[1+λ_1 (s|ψ_1 )]$. Conditioning on the joint realization of these processes, then it is straightforward to derive the conditional probability of a case at any location as

$$Pr(y_{i}=1) = \frac{λ_0(s_i |ψ_0 )λ_1(s_i |ψ_1 )}{λ_0 (s_i |ψ_0 )[1 + λ_1 (s_i |ψ_1 )]} = p_{i}$$

$$Pr(y_{i}=0) = \frac{1}{1 + λ_1 (s_i |ψ_1)} = 1-p_{i}$$

What is the author calculating in these equations? It seems to be the probability of a case and a control respectively. However, if a Poisson process is modeling each event type, I don't see traces of a Poisson process there.

Furthermore, using the previous equations, the likelihood is:

$$L(\Psi_{1}|s) = \prod_{i\in \text{cases}}p_{i}\prod_{i\in \text{controls}}(1-p_{i})$$

which seems very reasonable except for the conditioning on $s$ instead of conditioning on the parameters $\Psi$.

I would appreciate any help.

UPDATE: You can find this section available in Google Books using this link.

$\endgroup$
1
+50
$\begingroup$

If you assume that control events arrive with rate $\lambda_0$ and case events arrive with rate $\lambda$, then the total number of events that occur happen with rate $\lambda_0 + \lambda$. Now, conditional on an event occurring, then you could calculate the probability that it was a case to be,

$P(Y=1 \vert N=1) = P(Y=1, N=1)/P(N=1)$ $ = P(1 \text{ case event and } 0 \text{ control events})/P(N=1)$ $ = P(1 \text{ case event})* P(0 \text{ control events})/P(N=1) \text{ (independence of case and control events) }$ $= [\lambda^1e^{-\lambda}/1! * \lambda_0^0e^{-\lambda_0}/0!]/ [ (\lambda_0+\lambda)^1e^{-(\lambda+\lambda_0})/1!] \text{ (poisson process probabilities)}$
$ = \lambda e^{-\lambda} e^{-\lambda_0}/[ (\lambda_0+\lambda)e^{-(\lambda+\lambda_0)} = \lambda/(\lambda_0+\lambda) = \lambda_0\lambda_1/(\lambda_0(1+\lambda_1)) =\lambda_1/(1+\lambda_1) $

Recall that for a poisson process with rate $\lambda$, P(k events) = $\lambda^ke^{-\lambda}/k!$

Similarly, the conditional probability that it was a control can be found to be $1/(1+\lambda_1)$

Notice here that the probability of case or control conditional on an event occurring depends only on $\lambda_1$ and $1+\lambda_1$ could be thought of as the relative increase in rates of events for the cases. If $\lambda_1$ is your parameter of interest, then there is nothing wrong with using the conditional logistic model to estimate the parameters that comprise the linear predictor $\lambda_1$.

The assumptions in the calculation of the conditional probabilities are contingent upon the stated superposition of the two independent Poisson processes. The likelihood, $L(\Psi_1 \vert s, N) = P( s \vert \Psi_1, N)$, is actually the conditional probability of the observed data $s$, the case and control events, given the parameters $\Psi_1$ and the total number of observed events N. Note that I do realize that the notation of the likelihood function is confusing how it is written, but a likelihood is not the conditional probability of the parameters given the data, but is actually the probability of the data given the parameters.

Furthermore, the beauty of this approach is that you don't have to estimate the parameters for the poisson process for the controls, $\Psi_0$, if all that you really care about is the parameters in $\Psi_1$.

This link may be helpful as well, theorems 1, 2, and 3 are relevant.

$\endgroup$
  • 1
    $\begingroup$ @Robert My apologies for any confusion that may have arisen from the first post. Similar calculations to these arise in other contexts. The link to the book was very helpful in clarifying the context of your question. I hope this answer aids in your understanding. $\endgroup$ – jsk May 10 '14 at 0:31
  • $\begingroup$ Excellent answer. One question: When you write the joint distribution $P(Y=1, N=1) = P(\text{1 case event and 0 con trol event})$, why do you get $\lambda \exp(-\lambda)\exp(-\lambda_{0})$? Actually, I think I have the same question about $P(N)$. $\endgroup$ – Robert Smith May 10 '14 at 1:48
  • $\begingroup$ @RobertSmith I'll go ahead and edit my response to address your question shortly. $\endgroup$ – jsk May 10 '14 at 2:23
  • $\begingroup$ Great. Thank you very much! $\endgroup$ – Robert Smith May 10 '14 at 2:41
  • $\begingroup$ Oh... right. I missed the fact that "0 control events" still means $P(k=0)$. Thank you! $\endgroup$ – Robert Smith May 10 '14 at 2:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.