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In the species-only model, the log odds for adult species A is the intercept, so ${\hat{\theta }}_A=0.91191$ with standard error 0.07824. This also follows directly by pooling its vials: there are 570 other-species eggs among 799 eggs eaten, giving ${\hat{p}}_A=570/799$ andFor a regular maximum-likelihood estimator, asymptotic normality of a maximum likelihood estimator gives ${\hat{\theta }}_A\stackrel{\mathrm{approx}}{\sim }N({\theta }_A,I_A^{-1})$, with information $I_A=N_A{p}_A(1-p_A)$. Substitution of ${\hat{p}}_A$ gives $\mathrm{se}({\hat{\theta }}_A)=\sqrt{1/570+1/229}\simeq 0.07824$. An approximate 95% Wald confidence interval isThis interval treats the selected species-only model as fixed; it does not account separately for uncertainty from choosing that model.

For independent identically distributed observations $Y_1,\dots ,Y_n$, the likelihood function and log-likelihood areA maximum-likelihood estimator is a measurable choice ${\hat{\theta }}_n\in {\mathrm{argmax}}_{\theta \in \Theta }{L}_n(\theta )$, when a maximizer exists. It need not be unique. The one-observation score function is $s_{\theta }(y)={\nabla }_{\theta }\log f(y,\theta )$, and the Fisher information matrix isUnder the regularity assumptions, differentiation beneath the integral gives ${\mathbb{E}}_{\theta }{s}_{\theta }={\nabla }_{\theta }\int f=0$. Differentiating again gives the second information identity. Independent observations have total Fisher information $nI(\theta )$.

Assume fixed parameter dimension, an interior true parameter ${\theta }_0$, a positive-definite matrix $I({\theta }_0)$, the standard differentiability and integrability conditions, and statistical consistency of ${\hat{\theta }}_n$. Then the asymptotic normality of a maximum likelihood estimator isHere $I$ is the information per observation. Thus the leading covariance matrix of the estimator itself is $I({\theta }_0{)}^{-1}/n$.

To prove this, statistical consistency places the estimator in an interior ball about ${\theta }_0$ with probability tending to one, where its score function vanishes. Write $d_n={\hat{\theta }}_n-{\theta }_0$ and use the integral first-order Taylor formula for a vector map:This integral matrix is needed in a vector problem; one does not have to assert a common scalar mean-value point for every score component. PutThe regular local uniform law of large numbers, continuity of the expected Hessian matrix, and statistical consistency give $J_n\stackrel{P}{\to }I({\theta }_0)$. The multivariate central limit theorem givesThe limiting information matrix is nonsingular, so $J_n^{-1}\stackrel{P}{\to }I({\theta }_0{)}^{-1}$. Solving the Taylor identity and applying the Slutsky theorem proves the displayed normal limit, since $I^{-1}I{I}^{-1}=I^{-1}$. Boundary parameters or singular information are outside this regular theorem.

Let $x$ be the parameter-contrast vector for the specified game: it has entries $+1$ for ${\beta }_i,{\beta }_j,{\beta }_{\{i,j\}}$, entries $-1$ for ${\beta }_k,{\beta }_{\ell },{\beta }_{\{k,\ell \}}$, and zero elsewhere. PutUnder the stated regularity assumptions, asymptotic normality of a maximum likelihood estimator giveswhere $i_N(\beta )/N\to I(\beta )$. Since $g^{\prime }(\eta )=p(1-p)$, the delta method givesEquivalently, for large $N$,with $p=e^{x^T\beta }/(1+e^{x^T\beta })$.

Under the usual regularity conditions, consistency and asymptotic normality of a maximum likelihood estimator givewhere $I({\theta }_0)$ is the one-observation Fisher information matrix. In a scalar model this reads $N(0,I({\theta }_0{)}^{-1})$.