Why Test Statistics Separate in Sparse Data
Wald, score, and likelihood-ratio tests are often introduced as three routes to a similar large-sample conclusion. Their agreement is approximate: it relies on a likelihood that is well behaved near the null and has enough information for a local quadratic approximation to be useful. With only a few successes or events, the likelihood may be asymmetric, the fitted value may be close to a boundary, or the unrestricted estimate may lie far from the null. In these settings, the three statistics can differ substantially.
The point is not that one formula is universally correct and the others are defective. They measure departure from the null in different ways. Wald uses the estimated parameter and curvature at the unrestricted fit; the score test uses the slope and curvature at the null; the likelihood-ratio test compares the maximized likelihoods. Sparse data can make those reference points and approximations matter.
A Binomial Model: Three Different Reference Points
Let \(Y\sim\operatorname{Binomial}(n,p)\), with observed count \(y\), and test \(H_0:p=p_0\), where \(0<p_0<1\). Write \(\hat p=y/n\). For an interior estimate \(0<y<n\), the canonical parameter is the logit \(\eta=\log\{p/(1-p)\}\), with null value \(\eta_0=\log\{p_0/(1-p_0)\}\). The Wald statistic below is specifically the Wald statistic for \(\eta\).
To derive these expressions, ignore the binomial coefficient, which is constant in \(p\), and write \(\ell(p)=y\log p+(n-y)\log(1-p)\). The score in the canonical parameter is \(U(\eta)=y-np\), and the information is \(I(\eta)=np(1-p)\). Thus the score statistic is \(U(\eta_0)^2/I(\eta_0)=n^2(\hat p-p_0)^2/\{np_0(1-p_0)\}\). At the unrestricted maximum, \(p=\hat p\), so the estimated variance of \(\hat\eta\) is \(1/\{n\hat p(1-\hat p)\}\); substituting it into the Wald formula gives \(W_\eta\). Finally, twice the log-likelihood difference between \(p=\hat p\) and \(p=p_0\) gives \(G^2\) above. This derives the three statistics and also shows why the probability-scale and logit-scale Wald statistics need not agree.
Worked Example: One Success in Five Binomial Trials
Suppose \(n=5\), \(y=1\), and the null is \(p_0=0.5\). Then \(\hat p=0.2\). The canonical estimate is \(\operatorname{logit}(0.2)=\log(0.2/0.8)=-\log 4\), while \(\operatorname{logit}(0.5)=0\).
Canonical-logit Wald. The estimated information is \(5(0.2)(0.8)=0.8\). Therefore
Score. The null information is \(5(0.5)(0.5)=1.25\), and the score at the null is \(1-5(0.5)=-1.5\). Hence
Likelihood ratio. Substitution into the binomial deviance gives
Using a one-degree-of-freedom chi-square reference distribution, the approximate p-values are about \(0.215\), \(0.180\), and \(0.165\), respectively. The probability-scale Wald statistic would be \(5(0.2-0.5)^2/\{0.2(0.8)\}=2.8125\), not \(1.53745\). These are materially different calculations even in this small example; the choice of Wald parameterization must be stated.
What Happens at a Boundary?
When \(y=0\) or \(y=n\), the binomial maximum-likelihood estimate is on the boundary of the probability parameter space. A finite logit estimate no longer exists. The score and likelihood-ratio statistics can still be calculated for an interior null value, but the ordinary canonical-logit Wald statistic cannot be formed from a finite unrestricted estimate and its standard error.
Worked Example: No Successes in Five Trials
Suppose \(n=5\), \(y=0\), and test \(p_0=0.2\). The fitted probability is \(\hat p=0\), which is a boundary estimate.
The score statistic is
The likelihood-ratio statistic is
Their one-degree-of-freedom chi-square p-value approximations are about \(0.264\) and \(0.135\). The usual canonical-logit Wald calculation is unavailable: \(\operatorname{logit}(\hat p)=-\infty\), not a finite estimate with a finite estimated standard error. The probability-scale Wald expression also has a zero denominator at \(\hat p=0\). The finite score and likelihood-ratio values do not make their chi-square approximations exact; with five trials, their calibration should be treated cautiously.
Sparse Poisson Counts and Extreme Fitted Means
The same issue arises in a Poisson model. Let \(Y\sim\operatorname{Poisson}(\mu)\), where \(\mu=E\lambda\) is the expected count for exposure \(E>0\) and rate \(\lambda>0\). To test a specified mean \(\mu_0>0\), use the canonical log-mean parameter \(\theta=\log\mu\). For an observed count \(y>0\), the unrestricted estimate is \(\hat\mu=y\), which may be far from \(\mu_0\) when counts are sparse.
Worked Example: One Event When the Null Mean Is Ten
Consider a Poisson count with exposure \(E=1\), observed count \(y=1\), and null mean \(\mu_0=10\). The unrestricted mean estimate is \(1\), a tenfold difference from the null.
Wald. The fitted log-mean is \(\log 1=0\), the null log-mean is \(\log 10\), and the estimated information is \(1\). Thus
Score. At the null, the score is \(1-10=-9\) and the information is \(10\), giving
Likelihood ratio. The deviance is
The corresponding one-degree-of-freedom chi-square p-value approximations are approximately \(0.021\), \(0.0044\), and \(0.00025\). A single count has led to very different apparent evidence against the null. The log-mean estimate differs from the null by \(\log(10)\), while the estimated information at the fit is only \(1\). A local quadratic approximation centered at the fit is therefore being used over a large parameter displacement. The small-count setting also makes reliance on a large-sample chi-square calibration questionable.
Diagnosing Disagreement
A useful diagnosis asks where each statistic gets its information and how far the fit lies from the null. The Wald statistic relies on the unrestricted estimate and its local estimated variance. When the estimate is near a boundary, its standard error may be unstable or undefined; when curvature at the fit is weak, a large parameter displacement can produce a Wald statistic that poorly represents the shape of the likelihood elsewhere. The score statistic evaluates slope and curvature at the null, so its behavior can differ when the null is far from the maximum. The likelihood-ratio statistic uses the full log-likelihood difference, but its usual chi-square reference is also an asymptotic approximation.
The local equivalence of the three tests under regular large-sample conditions does not imply finite-sample identity. In particular, a small p-value from one statistic and a larger p-value from another should not be resolved by silently choosing the preferred answer. Report the test and parameterization, inspect the fitted and null values, and consider whether the asymptotic approximation is credible for the available data.
Common Mistakes
- Calling every Wald formula “the” Wald statistic. In the binomial example, the canonical-logit statistic is \(1.53745\), whereas the probability-scale statistic is \(2.8125\). State the parameter used for the Wald calculation.
- Treating boundary estimates as ordinary interior fits. With zero binomial successes, the fitted logit is not finite. Do not substitute it into an interior Wald formula and report a finite statistic.
- Assuming a finite statistic guarantees a reliable p-value. Score and likelihood-ratio statistics can be computed at a boundary, but the chi-square calibration remains approximate and may be poor in sparse samples.
- Expecting asymptotic equivalence to mean equal answers in small samples. Equivalence is a large-sample, local result. Sparse counts and large shifts from the null are precisely situations where its approximation can be weak.
Check Your Understanding
Use the formulas and diagnostics in this tutorial to answer the following questions.
- For a binomial model with \(n=5\), \(y=1\), and \(p_0=0.5\), why is the probability-scale Wald statistic not the canonical-logit Wald statistic?
- For zero successes in \(n\) binomial trials, explain why the score statistic is finite for an interior null probability while the canonical-logit Wald statistic is unavailable.
- In the Poisson example with \(y=1\) and \(\mu_0=10\), which statistic uses information evaluated at the null, and what is that information?
- Why can a large difference between the fitted and null log-means make local quadratic reasoning questionable?
- Does a finite likelihood-ratio statistic imply that its chi-square p-value is an exact small-sample p-value? Explain.