Likelihood Ratio Test

Study guides: Exam ASTAM, Exam MAS-I, Exam MAS-II

A Likelihood Ratio Test (LRT) compares two nested models by how much better the larger one fits: it takes the ratio of maximized likelihoods and rejects the restricted model when the ratio is too small. It is the general-purpose test behind the analysis of Deviance used to compare Generalized Linear Models.

$\Lambda = \frac{\sup_{\theta \in \Theta_0} L(\theta)}{\sup_{\theta \in \Theta} L(\theta)}$

$-2\ln\Lambda = 2\left[\ell(\hat\theta) - \ell(\hat\theta_0)\right] \ \dot\sim\ \chi^2_{r}$

$r$ is the number of restrictions — the difference in free parameters between the full and reduced models — and the $\chi^2$ approximation is asymptotic (Wilks' theorem)
Reject $H_0$ (the reduced model) when $-2\ln\Lambda$ exceeds the upper- $\alpha$ point of $\chi^2_r$ (at $\alpha = 0.05$ , the $0.950$ column of the CAS MAS-I table, which gives lower-tail areas): a large value means the extra parameters bought a real improvement in fit
For a GLM this statistic is the difference in deviances divided by the dispersion, $-2\ln\Lambda = (D_{\text{reduced}} - D_{\text{full}})/\phi$ , with $D$ the unscaled deviance and $\phi$ the Dispersion Parameter (equivalently, the difference in scaled deviances). That is why the analysis of deviance table is an LRT in disguise. For Poisson and binomial models $\phi = 1$ and the statistic is the deviance difference $\Delta D$ itself
Models must be nested and fitted to the same data; comparing non-nested models needs AIC or BIC instead
By the Neyman–Pearson lemma the LRT is the most powerful test of a simple null against a simple alternative — it maximizes the power at any given level $\alpha$
The Wald test (coefficient over its standard error, as printed in Parameter Estimate Tables) and the score test are asymptotically equivalent to the LRT. Their small-sample properties are not known, though some simulation studies suggest the LRT may do better than the other two in small samples
When $\phi$ is unknown, estimate it from the full model and use the $F$ form $F = \dfrac{(D_{\text{reduced}} - D_{\text{full}})/r}{\hat\phi_{\text{full}}}$ , built from the unscaled deviances, on $(r,\ n - p_{\text{full}})$ degrees of freedom. For a normal linear model the deviances are residual sums of squares and this $F$ is exact

Testing a Group of Predictors A Poisson GLM with $4$ parameters has log-likelihood $\ell = -412.6$ . Adding a 3-level territory variable (2 extra parameters) gives $\ell = -407.1$ . Test at $\alpha = 0.05$ whether territory belongs in the model, given $\chi^2_{0.05, 2} = 5.99$ .

Answer $-2\ln\Lambda = 2\left[-407.1 - (-412.6)\right] = 2(5.5) = 11.0$ Since $11.0 > 5.99$ , reject the reduced model — territory significantly improves the fit. Equivalently, the deviance drops by $11.0$ on $2$ degrees of freedom.

Comparing an Exponential to a Gamma Fit A Gamma model fitted to $n = 300$ claim sizes has $\ell = -2{,}401.2$ ; the exponential special case ( $\alpha = 1$ ) has $\ell = -2{,}403.0$ . Does the extra shape parameter earn its place at $\alpha = 0.05$ ?

Answer One restriction, so compare with $\chi^2_{0.05,1} = 3.84$ : $-2\ln\Lambda = 2\left[-2{,}401.2 + 2{,}403.0\right] = 3.6$ $3.6 < 3.84$ , so fail to reject — the exponential is an adequate simplification of the Gamma here. The call is close: the statistic depends only on the log-likelihood gap, not on $n$ , and a gap above $3.84/2 = 1.92$ would have rejected.