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Sufficiency and Information · Tutorial 78 of 981

Likelihood Curvature and Parameter Uncertainty

Learn to read local likelihood curvature as a measure of uncertainty and use it to construct approximate standard errors and intervals.

Intermediate 10 min read

What You'll Learn

  • Define observed information as the negative second derivative of the log likelihood.
  • Derive the local quadratic approximation around an interior maximum.
  • Explain why the reciprocal curvature gives an approximate variance.
  • Distinguish observed information from expected Fisher information.
  • Calculate curvature-based standard errors in Bernoulli, Poisson, and normal models.
  • Identify conditions and limitations of curvature-based uncertainty estimates.

Why Curvature Measures Uncertainty

An estimate identifies a location in parameter space, but it does not by itself indicate how precisely the data determine that location. The shape of the log likelihood near its maximum supplies that information: a sharp peak means that small changes in the parameter reduce the likelihood quickly, while a broad peak means that a wider range of values remains plausible. This tutorial develops that relationship for a single scalar parameter.

Let \(Y\) denote the observed data, let \(\theta\) be a scalar parameter, and let \(\ell(\theta)\) be the log likelihood. Suppose \(\hat{\theta}\) is an interior maximizer and that \(\ell\) has a continuous second derivative near \(\hat{\theta}\). At an interior maximum, \(\ell'(\hat{\theta})=0\). The second derivative therefore describes the leading change in log likelihood near the estimate.

Definition: The observed information at \(\theta\) is \(j(\theta)=-\ell''(\theta)\). It depends on the observed data. The expected Fisher information is \(I(\theta)=E_\theta[j(\theta)]\), when this expectation exists. At a local maximum, \(j(\hat{\theta})>0\) indicates downward curvature of the log likelihood.

The distinction is important. Observed information describes the curvature for the data actually collected; expected information averages that curvature over repeated data generated at \(\theta\). The earlier material on Fisher information established the score-variance and expected-information identities under their regularity conditions. Here the focus is how either kind of information translates into an uncertainty scale.

The Local Quadratic Approximation

For a twice continuously differentiable log likelihood, Taylor expansion around the interior maximum gives, for \(\theta\) sufficiently close to \(\hat{\theta}\),

$$ \ell(\theta)=\ell(\hat{\theta})+\ell'(\hat{\theta})(\theta-\hat{\theta}) +\frac{1}{2}\ell''(\hat{\theta})(\theta-\hat{\theta})^2 +o\!\left((\theta-\hat{\theta})^2\right). $$

Because the score \(\ell'(\hat{\theta})\) is zero and \(\ell''(\hat{\theta})=-j(\hat{\theta})\), this becomes

$$ \ell(\theta)\approx \ell(\hat{\theta}) -\frac{1}{2}j(\hat{\theta})(\theta-\hat{\theta})^2. $$

Exponentiating the approximation gives a local likelihood proportional to \(\exp\{-j(\hat{\theta})(\theta-\hat{\theta})^2/2\}\). This has the shape of a normal density, centered at \(\hat{\theta}\), with variance \(1/j(\hat{\theta})\). It motivates the curvature-based standard error \(\{j(\hat{\theta})\}^{-1/2}\). This is a local approximation, not an assertion that the likelihood is itself a probability distribution for \(\theta\).

Result: If \(\hat{\theta}\) is an interior maximum with \(j(\hat{\theta})>0\), the local quadratic likelihood approximation has variance scale \(1/j(\hat{\theta})\), and standard-error scale \(1/\sqrt{j(\hat{\theta})}\). These are generally approximate uncertainty measures, not exact finite-sample variance and standard error.

The same calculation can be expressed as a local likelihood-ratio comparison:

$$ 2\{\ell(\hat{\theta})-\ell(\theta)\} \approx j(\hat{\theta})(\theta-\hat{\theta})^2. $$

Thus, values of \(\theta\) a few curvature-based standard errors from the estimate produce a noticeable decrease in log likelihood. This local picture gives a practical way to interpret the information number: greater information means greater curvature and a smaller uncertainty scale.

Worked Example: Curvature in a Bernoulli Sample

Suppose \(n=10\) independent Bernoulli observations contain \(x=7\) successes. Estimate the success probability \(p\) and calculate the observed-information standard error.

Ignoring terms not involving \(p\), the log likelihood is \(\ell(p)=7\log p+3\log(1-p)\), for \(0<p<1\). Its first derivative is

$$ \ell'(p)=\frac{7}{p}-\frac{3}{1-p}. $$

Setting this equal to zero gives \(7(1-p)=3p\), so the maximum-likelihood estimate is \(\hat p=7/10=0.7\). The second derivative and observed information are

$$ \ell''(p)=-\frac{7}{p^2}-\frac{3}{(1-p)^2}, \qquad j(p)=\frac{7}{p^2}+\frac{3}{(1-p)^2}. $$

At the estimate,

$$ j(0.7)=\frac{7}{0.49}+\frac{3}{0.09} =14.2857\ldots+33.3333\ldots =47.6190\ldots. $$

The curvature-based variance estimate is \(1/47.6190\ldots=0.021\), and the standard error is \(\sqrt{0.021}=0.1449\) (approximately). The expected information for \(n\) Bernoulli observations is \(10/\{p(1-p)\}\), which at \(p=0.7\) is also \(47.6190\ldots\). Here observed and expected information agree at the estimate because the observed success proportion is \(0.7\).

From Curvature to Sampling Uncertainty

A quadratic shape alone does not prove that an estimator has a normal sampling distribution. That conclusion needs regularity conditions on the data-generating model and a large-sample argument. The following standard result makes the connection precise in an independent, identically distributed setting.

Theorem: Let \(Y_1,\ldots,Y_n\) be independent and identically distributed under a scalar parameter \(\theta_0\), which is an interior point of the parameter space. Write \(U_n(\theta)=\ell_n'(\theta)\) and \(j_n(\theta)=-\ell_n''(\theta)\). Suppose: (1) a consistent estimator \(\hat{\theta}\) is an interior score root, so \(U_n(\hat{\theta})=0\) with probability tending to one; (2) the score satisfies \(U_n(\theta_0)/\sqrt n\xrightarrow{d}N(0,I(\theta_0))\), where \(0<I(\theta_0)<\infty\); and (3) for every random \(\tilde{\theta}\) between \(\hat{\theta}\) and \(\theta_0\), \(j_n(\tilde{\theta})/n\xrightarrow{p}I(\theta_0)\). Then \(\sqrt n(\hat{\theta}-\theta_0)\xrightarrow{d}N(0,1/I(\theta_0))\). Under these conditions, \(j_n(\hat{\theta})^{-1}\) is a consistent estimate of the leading asymptotic variance \(1/\{nI(\theta_0)\}\).

Proof. On the event that the score root exists, the mean value theorem applied to \(U_n\) gives a value \(\tilde{\theta}\) between \(\hat{\theta}\) and \(\theta_0\) such that

$$ 0=U_n(\hat{\theta}) =U_n(\theta_0)+U_n'(\tilde{\theta})(\hat{\theta}-\theta_0) =U_n(\theta_0)-j_n(\tilde{\theta})(\hat{\theta}-\theta_0). $$

Rearranging and multiplying by \(\sqrt n\),

$$ \sqrt n(\hat{\theta}-\theta_0) = \left\{\frac{j_n(\tilde{\theta})}{n}\right\}^{-1} \frac{U_n(\theta_0)}{\sqrt n}. $$

By assumption, the first factor converges in probability to \(1/I(\theta_0)\), and the second converges in distribution to \(N(0,I(\theta_0))\). Slutsky's theorem gives the stated limit, with variance \(I(\theta_0)^{-2}I(\theta_0)=1/I(\theta_0)\). The curvature condition also applies at the consistent estimate \(\hat{\theta}\), so \(j_n(\hat{\theta})/n\xrightarrow{p}I(\theta_0)\). Taking reciprocals gives \(j_n(\hat{\theta})^{-1}\) asymptotic to \(1/\{nI(\theta_0)\}\), as claimed. \(\square\)

This result explains why the inverse observed information at the estimate is commonly used as an approximate variance estimate. Replacing observed information with expected information gives the related approximation \(1/I_n(\hat{\theta})\), where \(I_n\) is the total expected information for the sample. The two can differ in a particular sample, even when both are useful under regular large-sample conditions.

Worked Example: Curvature in a Poisson Sample

Suppose six independent counts have total \(18\), and each count has a Poisson distribution with mean \(\lambda\). Find the estimate of \(\lambda\) and the curvature-based standard error.

Writing the observed counts as \(y_1,\ldots,y_6\), the log likelihood is

$$ \ell(\lambda)=18\log\lambda-6\lambda-\sum_{i=1}^{6}\log(y_i!), \qquad \lambda>0. $$

The score is \(18/\lambda-6\), so its zero is \(\hat{\lambda}=3\). Differentiating again gives

$$ j(\lambda)=-\ell''(\lambda)=\frac{18}{\lambda^2}, \qquad j(3)=\frac{18}{9}=2. $$

Therefore the approximate variance is \(1/2\), and the standard error is \(1/\sqrt{2}\approx0.7071\). The expected information is \(6/\lambda\), which is also \(2\) at \(\lambda=3\). For illustration, a normal-approximation interval using \(1.96\) standard errors is \(3\pm1.96(0.7071)\), or approximately \((1.614,4.386)\). Its approximate nature matters: a symmetric interval does not automatically respect every parameter constraint or achieve accurate coverage in a small sample.

Worked Example: An Exactly Quadratic Normal Likelihood

Let four independent observations be \(2,4,5,5\), each normally distributed with unknown mean \(\mu\) and known variance \(4\). Find the estimate and show how the curvature gives its exact sampling standard error.

The sample mean is \(\hat{\mu}=(2+4+5+5)/4=4\). Apart from constants, the log likelihood is \(-\sum_{i=1}^4(y_i-\mu)^2/8\). At \(\mu=4\), the residuals are \(-2,0,1,1\), whose sum is zero and whose squared sum is \(6\). For \(\mu=4+\delta\),

$$ \sum_{i=1}^{4}(y_i-\mu)^2 =\sum_{i=1}^{4}\{(y_i-4)-\delta\}^2 =6+4\delta^2, $$

because the cross term is \(-2\delta\sum_i(y_i-4)=0\). Consequently, \(\ell(4+\delta)-\ell(4)=-4\delta^2/8=-\delta^2/2\). The log likelihood is exactly quadratic, with \(j(\mu)=4/4=1\) and standard error \(1\). In this model, \(\bar Y\sim N(\mu,4/4)\), so that standard error is also exact for the sampling distribution, not merely a local approximation.

Using and Interpreting Approximate Standard Errors

When the normal approximation for \(\hat{\theta}\) is adequate, a common approximate confidence interval is

$$ \hat{\theta}\ \pm\ z_{1-\alpha/2}\frac{1}{\sqrt{j(\hat{\theta})}}, $$

where \(z_{1-\alpha/2}\) is the corresponding standard normal quantile. The formula combines the estimate with its local curvature-based standard error. It does not guarantee good coverage in a small sample, at a boundary, or when the likelihood is strongly asymmetric.

A frequent mistake is to interpret a large value of \(j(\hat{\theta})\) as a large standard error because it looks like a measure of “uncertainty.” The direction is the opposite: standard error is proportional to the inverse square root of information. Another mistake is to treat \(1/j(\hat{\theta})\) as the exact variance of \(\hat{\theta}\). The quadratic approximation concerns the local likelihood shape; the theorem shows a sampling-distribution connection only under its stated large-sample assumptions.

Curvature is less informative when the maximum is not interior, the second derivative is zero or positive, or the likelihood is notably non-quadratic around its maximum. In those cases, the local symmetric normal approximation may be poor. A parameter constraint can also make a symmetric interval extend outside the permissible parameter space. These are reasons to inspect the likelihood shape and the model conditions rather than applying the standard-error formula mechanically.

Key takeaway: Near a regular interior maximum, log-likelihood curvature \(j(\hat{\theta})\) sets the local precision scale: approximate variance is \(1/j(\hat{\theta})\) and approximate standard error is \(1/\sqrt{j(\hat{\theta})}\). The approximation is most useful when the likelihood is smooth and locally quadratic and the estimator has a suitable large-sample normal approximation.

Check Your Understanding

Answer each question by connecting the likelihood shape to the uncertainty scale.

  1. What is observed information, and how does it differ from expected Fisher information?
  2. If the observed information at an estimate is \(25\), what are the curvature-based approximate variance and standard error?
  3. Why does a larger positive curvature at the maximum correspond to a smaller standard error?
  4. In the Poisson example, why do observed and expected information agree at the maximum-likelihood estimate?
  5. Give two reasons why a curvature-based symmetric normal interval might be unreliable.
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