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Gaussian Likelihood and Invariance · Tutorial 178 of 975

Gaussian Scores and Fisher Information

Learn to differentiate the Gaussian log-likelihood in mean and covariance directions, assemble its Fisher information, and verify the information identities with explicit calculations.

Intermediate 11 min read

What You'll Learn

  • Derive the Gaussian score for the mean vector and covariance matrix
  • Express covariance information using symmetric perturbation directions
  • Identify when mean and covariance scores are uncorrelated
  • Distinguish Fisher information from sample-specific observed curvature
  • Verify the score-variance and expected-Hessian identities in a scalar Gaussian model

From Gaussian Likelihood to Score

Likelihood-based inference depends on how the log-likelihood changes as its parameters move. Its first derivative, the score, records the local direction and rate of change; the score's covariance gives Fisher information, which measures the precision available in the model. In a Gaussian model, these quantities can be derived explicitly for both the mean and covariance parameters.

Let \(X_1,\ldots,X_n\) be independent observations from \(N_p(\mu,\Sigma)\), where \(\mu\in\mathbb R^p\) and \(\Sigma\) is a positive-definite \(p\times p\) covariance matrix. Write \(y_i=X_i-\mu\). The log-likelihood, up to no omitted parameter-dependent terms, is

$$ \ell(\mu,\Sigma) =-\frac{np}{2}\log(2\pi)-\frac n2\log|\Sigma| -\frac12\sum_{i=1}^n y_i^\mathsf{T}\Sigma^{-1}y_i. $$

The score for a vector parameter is its column vector of partial derivatives. Because \(\Sigma\) is symmetric, its derivatives are most cleanly specified by symmetric directions: a perturbation \(H\) is any symmetric matrix, and the directional derivative is the derivative along \(\Sigma+tH\) at \(t=0\). This avoids treating the two copies of each off-diagonal covariance entry as separate parameters.

Definition: For a parameter vector \(\theta\), the score is \(U(\theta)=\nabla_\theta\ell(\theta)\). Its Fisher information is \(I(\theta)=E_\theta[U(\theta)U(\theta)^\mathsf{T}]\), when the expectation exists. In a covariance direction \(H\), the corresponding score is the directional derivative of \(\ell\) along \(\Sigma+tH\).

Mean and Covariance Scores

Differentiating the quadratic term with respect to \(\mu\) gives the mean score. For the covariance score, use the differential identities \(d\log|\Sigma|=\operatorname{tr}(\Sigma^{-1}d\Sigma)\) and \(d\Sigma^{-1}=-\Sigma^{-1}(d\Sigma)\Sigma^{-1}\). Both apply because \(\Sigma\) is positive definite.

Result: For the independent Gaussian sample,
$$ U_\mu =\frac{\partial\ell}{\partial\mu} =\Sigma^{-1}\sum_{i=1}^n(X_i-\mu). $$

For a symmetric covariance perturbation \(H\), the directional score is

$$ U_\Sigma[H] =\frac12\sum_{i=1}^n \operatorname{tr}\!\left[ \left\{\Sigma^{-1}y_i y_i^\mathsf{T}\Sigma^{-1} -\Sigma^{-1}\right\}H \right]. $$

Equivalently, under the trace pairing \(\operatorname{tr}(GH)\), the symmetric matrix representing the covariance score is

$$ G_\Sigma =\frac12\sum_{i=1}^n \left\{\Sigma^{-1}y_i y_i^\mathsf{T}\Sigma^{-1} -\Sigma^{-1}\right\}. $$

To check the covariance derivative, the log-determinant contributes \(-\tfrac n2\operatorname{tr}(\Sigma^{-1}H)\). The quadratic term contributes \(\tfrac12\sum_i y_i^\mathsf{T}\Sigma^{-1}H\Sigma^{-1}y_i\), which equals \(\tfrac12\sum_i\operatorname{tr}(\Sigma^{-1}y_i y_i^\mathsf{T}\Sigma^{-1}H)\). Adding the terms gives the stated directional score.

Worked Example: Evaluating a Covariance Direction Score

Consider one observation with \(p=2\), \(\mu=(0,0)^\mathsf{T}\), \(\Sigma=\begin{pmatrix}1&0\\0&4\end{pmatrix}\), and observed \(x=(1,2)^\mathsf{T}\). Find the covariance score matrix and its directional score for changing the off-diagonal covariance.

Here \(y=x\) and \(\Sigma^{-1}=\begin{pmatrix}1&0\\0&1/4\end{pmatrix}\). First,

$$ \Sigma^{-1}yy^\mathsf{T}\Sigma^{-1} = \begin{pmatrix}1&0\\0&1/4\end{pmatrix} \begin{pmatrix}1&2\\2&4\end{pmatrix} \begin{pmatrix}1&0\\0&1/4\end{pmatrix} = \begin{pmatrix}1&1/2\\1/2&1/4\end{pmatrix}. $$

Subtracting \(\Sigma^{-1}\) and multiplying by \(1/2\) gives

$$ G_\Sigma =\frac12 \begin{pmatrix}0&1/2\\1/2&0\end{pmatrix} = \begin{pmatrix}0&1/4\\1/4&0\end{pmatrix}. $$

An off-diagonal covariance change has direction \(H=\begin{pmatrix}0&1\\1&0\end{pmatrix}\), since both symmetric entries change together. Its score is \(\operatorname{tr}(G_\Sigma H)=1/4+1/4=1/2\). This factor of two is important: the matrix entry \(G_{12}\) alone is not the derivative with respect to the single off-diagonal covariance parameter.

Fisher Information for Mean and Covariance

The covariance information is naturally a bilinear form on symmetric directions. For directions \(H\) and \(K\), its value is the expected product of the two directional scores. The next result gives all blocks of the information for the Gaussian model.

Theorem: For \(n\) independent observations from \(N_p(\mu,\Sigma)\), with \(\Sigma\) positive definite, the information for the mean is \(I_{\mu\mu}=n\Sigma^{-1}\). The mean and covariance information blocks are zero. For symmetric covariance directions \(H,K\),
$$ I_{\Sigma\Sigma}[H,K] =E[U_\Sigma[H]U_\Sigma[K]] =\frac n2\operatorname{tr}(\Sigma^{-1}H\Sigma^{-1}K). $$

If the covariance is represented by its \(p(p+1)/2\) distinct entries, this formula gives the information matrix in those coordinates. For example, the coordinate for a diagonal entry uses direction \(E_{jj}\), while an off-diagonal coordinate \(\sigma_{jk}\), \(j<k\), uses \(E_{jk}+E_{kj}\).

Proof. Set \(y_i=X_i-\mu\). The mean score is \(\Sigma^{-1}\sum_i y_i\). Independence and \(E[y_i]=0\), \(\operatorname{Var}(y_i)=\Sigma\) give

$$ E[U_\mu U_\mu^\mathsf{T}] =\Sigma^{-1}\left(\sum_{i=1}^n\operatorname{Var}(y_i)\right)\Sigma^{-1} =n\Sigma^{-1}. $$

For one observation, its covariance directional score is \(\tfrac12\{y^\mathsf{T}Ay-\operatorname{tr}(A\Sigma)\}\), where \(A=\Sigma^{-1}H\Sigma^{-1}\). Likewise, direction \(K\) uses \(B=\Sigma^{-1}K\Sigma^{-1}\). Write \(y=\Sigma^{1/2}z\), where \(z\sim N_p(0,I_p)\), and set \(C=\Sigma^{1/2}A\Sigma^{1/2}\), \(D=\Sigma^{1/2}B\Sigma^{1/2}\). Then the centered quadratic terms are \(z^\mathsf{T}Cz-\operatorname{tr}(C)\) and \(z^\mathsf{T}Dz-\operatorname{tr}(D)\).

For independent standard normal coordinates, \(E[z_a z_b z_c z_d]\) is the sum of the three products of Kronecker deltas corresponding to the pairings \((a,b)(c,d)\), \((a,c)(b,d)\), and \((a,d)(b,c)\). This follows from independence: a product has expectation zero if an index occurs an odd number of times; if all indices occur in pairs, the needed moments are \(E[z_j^2]=1\) and \(E[z_j^4]=3\). Expanding the two quadratic forms with this identity yields

$$ \operatorname{Cov}(z^\mathsf{T}Cz,z^\mathsf{T}Dz) =2\operatorname{tr}(CD). $$

Thus the covariance-direction information from one observation is \(\tfrac12\operatorname{tr}(CD)=\tfrac12\operatorname{tr}(\Sigma^{-1}H\Sigma^{-1}K)\); independence makes the \(n\)-observation information \(n\) times this value. Finally, the product of a mean score and a covariance score has expectation zero: after centering, each term is a product of three components of a centered Gaussian vector, and every such third moment is zero. Cross-observation terms vanish by independence and zero means. This proves all stated blocks. \(\square\)

The information is therefore block diagonal between mean and covariance under this parameterization, but covariance entries need not be mutually orthogonal. The trace formula handles their relationships without requiring a long coordinate-by-coordinate expansion.

Worked Example: Information for a Two-Variable Covariance

For one observation, take \(\Sigma=\operatorname{diag}(1,4)\). Use the distinct covariance coordinates \((\sigma_{11},\sigma_{12},\sigma_{22})\). Their direction matrices are \(H_{11}=E_{11}\), \(H_{12}=E_{12}+E_{21}\), and \(H_{22}=E_{22}\). Compute the diagonal information entries.

With \(A=\Sigma^{-1}=\operatorname{diag}(1,1/4)\), each entry is \(\tfrac12\operatorname{tr}(AH_aAH_b)\). For the first coordinate, \(AH_{11}AH_{11}=E_{11}\), so its information is \(1/2\). For the last coordinate, \(AH_{22}AH_{22}=(1/16)E_{22}\), giving information \(1/32\).

For the off-diagonal coordinate, \(AH_{12}=\begin{pmatrix}0&1\\1/4&0\end{pmatrix}\), so \((AH_{12})^2=\operatorname{diag}(1/4,1/4)\). Its trace is \(1/2\), and its information is therefore \(1/4\). The cross-information entries are zero: products involving a diagonal direction and the off-diagonal direction have zero trace, and \(H_{11}\) and \(H_{22}\) have disjoint diagonal support. The covariance information matrix for these coordinates is consequently \(\operatorname{diag}(1/2,1/4,1/32)\).

Information Identities in a Scalar Gaussian Model

For a scalar parameter, two common expressions for information are the variance of the score and the negative expected second derivative of the log-likelihood. In regular models they agree; the Gaussian example makes the calculation and the distinction from sample-specific curvature explicit. Let \(X_1,\ldots,X_n\) be independent \(N(\mu,v)\) observations, where \(v>0\), and write \(S_2=\sum_{i=1}^n(X_i-\mu)^2\). The log-likelihood is

$$ \ell(\mu,v) =-\frac n2\log(2\pi)-\frac n2\log v-\frac{S_2}{2v}. $$

Worked Example: Checking Both Information Identities

Differentiate the log-likelihood with respect to \(\mu\) and \(v\):

$$ U_\mu=\frac{\sum_i(X_i-\mu)}{v}, \qquad U_v=-\frac{n}{2v}+\frac{S_2}{2v^2}. $$

Since \(\sum_i(X_i-\mu)\sim N(0,nv)\), \(E[U_\mu]=0\) and \(\operatorname{Var}(U_\mu)=n/v\). Also, \(S_2/v\) has a chi-square distribution with \(n\) degrees of freedom, so \(E[S_2]=nv\) and \(\operatorname{Var}(S_2)=2nv^2\). It follows that \(E[U_v]=0\) and

$$ \operatorname{Var}(U_v) =\frac{\operatorname{Var}(S_2)}{4v^4} =\frac{2nv^2}{4v^4} =\frac{n}{2v^2}. $$

The cross-information is zero. Indeed, the covariance between \(U_\mu\) and \(U_v\) is a constant times a sum of centered third moments and is therefore zero. Thus the score-variance information matrix is diagonal, with entries \(n/v\) and \(n/(2v^2)\).

Now differentiate again:

$$ \ell_{\mu\mu}=-\frac n v,\qquad \ell_{\mu v}=-\frac{\sum_i(X_i-\mu)}{v^2},\qquad \ell_{vv}=\frac{n}{2v^2}-\frac{S_2}{v^3}. $$

Taking expectations gives \(E[\ell_{\mu\mu}]=-n/v\), \(E[\ell_{\mu v}]=0\), and

$$ E[\ell_{vv}] =\frac{n}{2v^2}-\frac{nv}{v^3} =-\frac{n}{2v^2}. $$

Therefore \(-E[\nabla^2\ell]\) equals the score-variance information matrix, confirming both identities in this scalar model.

For a numerical check, take \(n=4\), \(\mu=0\), \(v=2\), and observations \(0,1,-1,\sqrt{3}\). Then \(S_2=0^2+1^2+(-1)^2+(\sqrt3)^2=5\), and the observed second derivative is

$$ \ell_{vv} =\frac{4}{2(2)^2}-\frac{5}{2^3} =\frac12-\frac58 =-\frac18. $$

The observed negative second derivative is \(1/8\), whereas the Fisher information for \(v\) is \(n/(2v^2)=1/2\). They differ because Fisher information is an expectation over possible samples, not the negative curvature at every particular observed data set.

Why the Distinction Matters

The score identifies how the likelihood responds locally to a parameter change, while Fisher information summarizes the score's variability under the model. The mean-covariance cross block being zero is a precise statement about this Gaussian parameterization; it does not say that sample estimates are independent in every finite-sample setting or that all covariance coordinates are uncorrelated.

A frequent error is to call the negative observed Hessian “the Fisher information” without taking an expectation. In the numerical calculation above, the observed negative curvature for \(v\) is \(1/8\), while the Fisher information is \(1/2\). The correction is to distinguish the data-dependent Hessian from its model expectation. A second common error is to differentiate with respect to one off-diagonal matrix entry while forgetting that a symmetric covariance perturbation changes both entries; using the direction \(E_{jk}+E_{kj}\) makes the coordinate derivative unambiguous.

Key takeaway: For a Gaussian sample, the mean score is linear in centered observations and the covariance score is centered quadratic. Their information blocks are \(n\Sigma^{-1}\), zero cross-information, and the covariance-direction form \(\tfrac n2\operatorname{tr}(\Sigma^{-1}H\Sigma^{-1}K)\). Keep expected information distinct from observed curvature.

Check Your Understanding

Use the definitions and calculations above to answer these questions.

  1. What is the mean score for \(n\) independent observations from \(N_p(\mu,\Sigma)\), and which sample quantity does it depend on?
  2. Why does an off-diagonal covariance coordinate use the direction \(E_{jk}+E_{kj}\) rather than \(E_{jk}\) alone?
  3. For symmetric directions \(H\) and \(K\), write the Fisher information bilinear form for the covariance parameter.
  4. In the scalar model, why is the covariance between the mean and variance scores zero?
  5. For the numerical scalar example, compare the observed negative second derivative with the Fisher information and explain why they differ.
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