A Worked Profile Interval Analysis for Count Rates

Why Profile a Count-Rate Ratio?

For count data, the number of events observed is not itself a rate: the groups may have different amounts of time at risk, person-time, or other exposure. A Poisson regression with a log-exposure offset handles that difference directly. When the scientific target is a ratio of two rates, a profile likelihood interval provides a way to assess plausible values of that ratio while re-estimating the baseline rate at each value.

The example below has only five observed events in total. That sparsity makes it useful to compare a likelihood-ratio interval with a Wald interval, but it also calls for caution: the usual likelihood-ratio cutoff is an asymptotic approximation, not a guarantee of exact 95% coverage.

Model and Profile Likelihood

Let \(Y_0\) and \(Y_1\) be independent Poisson counts from two groups, with known exposures \(t_0>0\) and \(t_1>0\). Write their rates per unit exposure as \(\lambda_0\) and \(\lambda_1\). Use group 0 as the reference and parameterize the rates by a baseline log rate \(\alpha\) and a log rate ratio \(\beta\):

Y_0\sim\operatorname{Poisson}(t_0e^\alpha),\qquad Y_1\sim\operatorname{Poisson}(t_1e^{\alpha+\beta}),\qquad R=\frac{\lambda_1}{\lambda_0}=e^\beta.

The exposures enter as offsets: the expected counts are exposure multiplied by rate. For observed counts \(y_0,y_1\), omit the factorial terms, which do not depend on the parameters. The log-likelihood is

\ell(\alpha,\beta) =y_0\log(t_0)+y_0\alpha-t_0e^\alpha +y_1\log(t_1)+y_1(\alpha+\beta)-t_1e^{\alpha+\beta} +\text{constant}.

For a fixed candidate \(\beta\), profiling means maximizing this log-likelihood over the nuisance parameter \(\alpha\). Differentiating with respect to \(\alpha\) gives

\frac{\partial\ell}{\partial\alpha} =y_0+y_1-e^\alpha(t_0+t_1e^\beta).

If \(y_0+y_1>0\), setting this derivative to zero yields the unique maximum in \(\alpha\):

\widehat{\alpha}(\beta) =\log\left(\frac{y_0+y_1}{t_0+t_1e^\beta}\right).

Profile likelihood: For each fixed candidate \(\beta\), the profile log-likelihood is \(\ell_p(\beta)=\ell(\widehat{\alpha}(\beta),\beta)\). Its maximizer is the unrestricted maximum-likelihood estimate \(\hat\beta\). A profile likelihood-ratio confidence set retains candidate values whose likelihood-ratio statistic is no larger than the chosen cutoff.

The maximizing baseline rate changes as the candidate rate ratio changes. Substituting a single baseline estimate from the unrestricted fit at every candidate value would therefore fail to profile the nuisance parameter.

Worked Example: Profiling a Sparse Count-Rate Model

Suppose group 0 contributes \(t_0=8\) person-time units and has \(y_0=4\) events. Group 1 contributes \(t_1=1\) unit and has \(y_1=1\) event. The observed rates are \(4/8=0.5\) and \(1/1=1\), so the estimated rate ratio is \(1/0.5=2\).

For any fixed \(R=e^\beta>0\), the profiled baseline rate is

\widehat{\lambda}_0(R)=e^{\widehat{\alpha}(\beta)} =\frac{y_0+y_1}{t_0+t_1R} =\frac{5}{8+R}.

The corresponding fitted means are \(\widehat\mu_0(R)=8\widehat{\lambda}_0(R)=40/(8+R)\) and \(\widehat\mu_1(R)=R\widehat{\lambda}_0(R)=5R/(8+R)\). At the unrestricted estimate \(R=2\), these means are \(4\) and \(1\), exactly the observed counts. At a different candidate ratio, such as \(R=1\), the fitted means are \(5(8/9)=40/9\) and \(5(1/9)=5/9\). In both cases the total fitted count remains 5, while its allocation between groups changes with the candidate ratio.

Likelihood-Ratio Curve and Confidence Limits

Let \(n=y_0+y_1\), and suppose first that both observed counts are positive. At a candidate \(R\), the fitted fraction of the total expected count assigned to group 1 is

p(R)=\frac{t_1R}{t_0+t_1R}.

The unrestricted fitted fraction is \(\hat p=y_1/n\). Subtracting the profiled log-likelihood from its unrestricted maximum gives the likelihood-ratio statistic

Q(R)=2\{\ell_p(\hat\beta)-\ell_p(\log R)\} =2\left[ y_0\log\left(\frac{y_0}{n\{1-p(R)\}}\right) +y_1\log\left(\frac{y_1}{np(R)}\right) \right].

This expression is also obtained by conditioning on the total count: independent Poisson counts, conditional on \(Y_0+Y_1=n\), have a binomial allocation with group 1 probability \(p(R)\). For the profile calculation, the key point is that \(Q(R)\) measures the loss in maximized likelihood at a fixed ratio, after the baseline rate has been re-estimated.

Likelihood-ratio interval: Under the usual regularity conditions for likelihood-ratio asymptotics, an approximate \(100(1-\gamma)\%\) profile interval for \(R\) is the set of positive values satisfying \(Q(R)\leq \chi^2_{1,1-\gamma}\), where \(\chi^2_{1,1-\gamma}\) is the \(1-\gamma\) quantile of a chi-square distribution with one degree of freedom. Its endpoints solve \(Q(R)=\chi^2_{1,1-\gamma}\) when the retained set is a single interval with two finite boundary points.

For the 95% interval, the conventional cutoff is \(\chi^2_{1,0.95}\approx3.841\). In the sparse example, \(n=5\), \(\hat p=1/5\), and \(p(R)=R/(8+R)\). Thus the endpoint equation is

2\left[ \log\left(\frac{0.2}{p}\right) +4\log\left(\frac{0.8}{1-p}\right) \right]=3.841, \qquad R=\frac{8p}{1-p}.

Worked Example: Solving the Profile Endpoints

For the observed counts \(y_0=4,y_1=1\), solve the endpoint equation first on the probability scale. The two roots are approximately \(p=0.01263\) and \(p=0.62823\). Substitution into \(R=8p/(1-p)\) gives

R_{\mathrm{lower}} =\frac{8(0.01263)}{1-0.01263} \approx0.1023,\qquad R_{\mathrm{upper}} =\frac{8(0.62823)}{1-0.62823} \approx13.52.

As a check on the endpoint calculation, substituting either rounded probability into the left side of the endpoint equation gives a value close to 3.841. The resulting approximate 95% profile interval is \((0.102,13.52)\). It includes 1, so the rate-ratio value of no difference is not excluded by this interval. The limits are markedly asymmetric around the point estimate 2 on the rate-ratio scale.

Worked Example: Evaluating the Profile Curve at a Candidate Ratio

For the same data, evaluate the likelihood-ratio statistic at \(R=1\). The corresponding allocation probability is \(p(1)=1/(8+1)=1/9\). Hence

\begin{aligned} Q(1) &=2\left[ \log\left(\frac{0.2}{1/9}\right) +4\log\left(\frac{0.8}{8/9}\right) \right]\\ &=2\left[\log(1.8)+4\log(0.9)\right]\\ &=2\left[0.58779-0.42144\right] \approx0.333. \end{aligned}

Since \(0.333<3.841\), the candidate \(R=1\) lies inside the 95% profile likelihood set. This calculation illustrates how a point on the profile curve is assessed: re-fit the nuisance baseline at that candidate, then compare the resulting likelihood with the unrestricted maximum.

Comparison With a Wald Interval

A common large-sample approximation for the standard error of the estimated log rate ratio in this two-group model is

\widehat{\operatorname{SE}}(\hat\beta) =\sqrt{\frac{1}{y_0}+\frac{1}{y_1}}.

It follows by estimating the variance of each log rate from its Poisson count and adding the two independent contributions. A Wald interval is formed symmetrically around \(\hat\beta=\log(\hat R)\), then exponentiated to return to the rate-ratio scale.

Worked Example: Calculating the Wald Interval

Here \(\hat R=2\), \(\hat\beta=\log 2\), and

\widehat{\operatorname{SE}}(\hat\beta) =\sqrt{\frac14+\frac11} =\sqrt{1.25} \approx1.118.

Using 1.96 as the normal critical value, the Wald limits on the log scale are

\log(2)\pm1.96(1.118) \approx -1.498,\ 2.884.

Exponentiating these limits gives

\left( \exp\{\log(2)-1.96(1.118)\}, \exp\{\log(2)+1.96(1.118)\} \right) \approx(0.2235,\ 17.89).

The Wald interval is also very wide, but its lower endpoint is about \(0.224\), compared with the profile endpoint about \(0.102\); its upper endpoint is about \(17.89\), compared with the profile endpoint about \(13.52\). The two methods give different shapes and limits because the profile curve reflects the likelihood at each candidate ratio, whereas the Wald method uses a local quadratic approximation around the estimate.

Why the Difference Matters

With sparse counts, the log-rate-ratio estimate can be far from normally distributed, and the estimated standard error can be large relative to the estimate. The Wald method nevertheless imposes symmetry on the log scale and relies on a local approximation. Profiling uses the likelihood over a range of candidate values and can therefore produce a differently shaped interval. That does not make a standard-cutoff profile interval exact: the chi-square calibration is still approximate, and its accuracy can also be limited in very small samples.

The exposure adjustment is essential to both methods. In the worked data, the raw count ratio \(y_1/y_0=1/4\), but the rate ratio is \((y_1/t_1)/(y_0/t_0)=1/0.5=2\). Ignoring exposure reverses the substantive impression.

Common mistake: Treat \(y_1/y_0\) as the rate ratio, even when the exposures differ.
Correction: Compare rates, or equivalently use the offset model; here the estimate is \((1/1)/(4/8)=2\), not \(1/4\).
Common mistake: Profile the ratio while holding the nuisance baseline rate fixed at its unrestricted estimate.
Correction: At every candidate ratio, maximize over the baseline rate; in this model, \(\widehat\lambda_0(R)=(y_0+y_1)/(t_0+t_1R)\).
Common mistake: Interpret a profile interval as automatically exact because it uses a likelihood ratio.
Correction: The usual chi-square cutoff is an asymptotic calibration. Sparse counts warrant careful interpretation and, where appropriate, a separately justified alternative calibration.

Key takeaway: A count-rate profile interval is built by fixing a candidate rate ratio, re-estimating the baseline rate, and retaining candidates whose likelihood loss is within the chosen cutoff. Exposure belongs in the model, and sparse-data profile and Wald intervals can differ substantially.

Check Your Understanding

Answer each question using the model and calculations developed above.

For independent Poisson counts with exposures \(t_0,t_1\), write the fitted baseline rate at a fixed candidate rate ratio \(R\).
Why is the raw count ratio not generally an estimate of the rate ratio when exposures differ?
In the worked data, what allocation probability \(p(R)\) corresponds to the candidate \(R=1\)?
What equation defines the endpoints of the approximate 95% profile interval?
Why can the profile interval and the Wald interval have different endpoints even when they use the same observed counts?