Lecture 14 Multiparameter Fisher Information

In lectures 11 and 13 the parameter was a single number, but most models we actually fit, e.g. the normal distribution with unknown mean and variance, have several parameters. In this lecture we will extend the concepts of Fisher information and the Cramér–Rao bound to a vector parameter θ=(θ1,…,θk)\theta=(\theta_{1},\dots,\theta_{k}): the score will be a vector, the information will be a k×kk\times k matrix, and the bound will be an inequality of covariance matrices. The new feature here is that we will need to invert the matrix, and the inverse will give the cost of estimating one parameter in the presence of the other parameters.

14.1 The Score Vector and the Information Matrix

Setting the stage as in lecture 11, we now assume that the parameter space is an open set Θ⊆ℝk\Theta\subseteq\mathbb{R}^{k}. The data X1,…,XnX_{1},\dots,X_{n} are again i.i.d. with density or probability weights f⁢(x;θ)f(x;\theta) and the log-likelihood function is ℓ⁢(θ)=∑i=1nℓi⁢(θ)\ell(\theta)=\sum_{i=1}^{n}\ell_{i}(\theta), where ℓi⁢(θ)=log⁡f⁢(Xi;θ)\ell_{i}(\theta)=\log f(X_{i};\theta), but θ\theta is now a vector. Derivatives with respect to θ\theta are now partial derivatives, denoted by ∂j\partial_{j} for the derivative with respect to θj\theta_{j}. We assume the conditions of section 11.1 with partial derivatives in place of d/d⁢θ\mathrm{d}/\mathrm{d}\theta, including the mixed derivatives ∂j∂l\partial_{j}\partial_{l}, and we assume that each coordinate of the score has finite variance under ℙθ\mathbb{P}_{\theta}.

Definition 14.1.

The score vector of the sample is given by the gradient of the log-likelihood function

∇ℓ⁢(θ)=(∂1ℓ⁢(θ),…,∂kℓ⁢(θ)),\nabla\ell(\theta)=\bigl{(}\partial_{1}\ell(\theta),\dots,\partial_{k}\ell(% \theta)\bigr{)},

as a column vector in ℝk\mathbb{R}^{k}.

For k=1k=1 the score vector is the score ℓ′⁢(θ)\ell^{\prime}(\theta) from definition 4.4. As in lecture 11, we consider the score vector as a random vector under ℙθ\mathbb{P}_{\theta}. The covariance matrix Cov(Y)\mathop{\mathrm{Cov}}\nolimits(Y) of a random vector Y=(Y1,…,Yk)Y=(Y_{1},\dots,Y_{k}) is the k×kk\times k matrix with entries Cov(Yj,Yl)\mathop{\mathrm{Cov}}\nolimits(Y_{j},Y_{l}), and its diagonal contains the variances.

Definition 14.2.

Given the conditions of section 11.1, the Fisher information matrix of the sample about θ\theta is given by

ℐ︀n⁢(θ)=Covθ(∇ℓ⁢(θ)),i.e.ℐ︀n⁢(θ)j⁢l=Covθ(∂jℓ⁢(θ),∂lℓ⁢(θ)).\mathcal{I}_{n}(\theta)=\mathop{\mathrm{Cov}}\nolimits_{\theta}\bigl{(}\nabla% \ell(\theta)\bigr{)},\qquad\text{{i.e.}}\qquad\mathcal{I}_{n}(\theta)_{jl}=% \mathop{\mathrm{Cov}}\nolimits_{\theta}\bigl{(}\partial_{j}\ell(\theta),% \partial_{l}\ell(\theta)\bigr{)}.

The information matrix for a single observation is denoted by ℐ︀⁢(θ)=ℐ︀1⁢(θ)\mathcal{I}(\theta)=\mathcal{I}_{1}(\theta).

The diagonal element ℐ︀n⁢(θ)j⁢j\mathcal{I}_{n}(\theta)_{jj} is the Fisher information about θj\theta_{j} in the sense of definition 11.2 with all other coordinates fixed to their true values; the off-diagonal elements are new and measure the strength of correlations between the components of the score. Both identities from lecture 11 carry over to this situation, separately for each coordinate.

Proposition 14.3.

With the assumptions of section 11.1 in place, the score has mean 𝔼θ⁢(∇ℓ⁢(θ))=0\mathbb{E}_{\theta}\bigl{(}\nabla\ell(\theta)\bigr{)}=0, and we have

ℐ︀n⁢(θ)j⁢l=𝔼θ⁢(∂jℓ⁢(θ)⁢∂lℓ⁢(θ))=−𝔼θ⁢(∂j∂lℓ⁢(θ))for ⁢j,l=1,…,k.\mathcal{I}_{n}(\theta)_{jl}=\mathbb{E}_{\theta}\bigl{(}\partial_{j}\ell(% \theta)\,\partial_{l}\ell(\theta)\bigr{)}=-\mathbb{E}_{\theta}\bigl{(}\partial% _{j}\partial_{l}\ell(\theta)\bigr{)}\qquad\text{for }j,l=1,\dots,k.

Thus, ℐ︀n⁢(θ)\mathcal{I}_{n}(\theta) is the negative of the expected Hessian matrix of the log-likelihood.

The proof of this result is similar to the proof of lemma 11.1 and lemma 11.3, but with partial derivatives instead of d/d⁢θ\mathrm{d}/\mathrm{d}\theta. In outline, the ratio ∂jf/f\partial_{j}f/f has mean zero since the density in the numerator and denominator cancels. Using the quotient rule we find ∂j∂lℓi=∂j∂lf⁢(Xi;θ)/f⁢(Xi;θ)−∂jℓi⁢∂lℓi\partial_{j}\partial_{l}\ell_{i}=\partial_{j}\partial_{l}f(X_{i};\theta)/f(X_{% i};\theta)-\partial_{j}\ell_{i}\,\partial_{l}\ell_{i} where the first term has mean zero. Finally, the independence of the observations allows us to carry over the resulting identity from one observation to the sample. As for the scalar case, the second expression is often the more practical one.

14.2 Properties of the Information Matrix

Two properties of the scalar information from lecture 11 have matrix counterparts. Recall that a symmetric matrix AA is positive semi-definite, if a⊤⁢A⁢a≥0a^{\top}Aa\geq 0 for all a∈ℝka\in\mathbb{R}^{k}, and positive definite, if a⊤⁢A⁢a>0a^{\top}Aa>0 for all a≠0a\neq 0; a positive definite matrix has a positive definite inverse.

Proposition 14.4.

Suppose the conditions from section 11.1 hold. Then the following statements hold.

  1. 1.
    ​

    The matrix ℐ︀n⁢(θ)\mathcal{I}_{n}(\theta) is symmetric and we have a⊤⁢ℐ︀n⁢(θ)⁢a=Varθ(a⊤⁢∇ℓ⁢(θ))≥0a^{\top}\mathcal{I}_{n}(\theta)a=\mathop{\mathrm{Var}}\nolimits_{\theta}\bigl{% (}a^{\top}\nabla\ell(\theta)\bigr{)}\geq 0 for all a∈ℝka\in\mathbb{R}^{k}. In particular, ℐ︀n⁢(θ)\mathcal{I}_{n}(\theta) is positive semi-definite.

  2. 2.
    ​

    Information is additive: ℐ︀n⁢(θ)=n⁢ℐ︀⁢(θ)\mathcal{I}_{n}(\theta)=n\,\mathcal{I}(\theta).

Proof.

For the first statement, the symmetry of the matrix follows from the fact that Cov(Yj,Yl)=Cov(Yl,Yj)\mathop{\mathrm{Cov}}\nolimits(Y_{j},Y_{l})=\mathop{\mathrm{Cov}}\nolimits(Y_{% l},Y_{j}) for all jj and ll. For a∈ℝka\in\mathbb{R}^{k}, the variance of the linear combination a⊤⁢∇ℓ⁢(θ)a^{\top}\nabla\ell(\theta) is given by

Varθ(∑j=1kaj⁢∂jℓ⁢(θ))=∑j=1k∑l=1kaj⁢al⁢ℐ︀n⁢(θ)j⁢l=a⊤⁢ℐ︀n⁢(θ)⁢a,\mathop{\mathrm{Var}}\nolimits_{\theta}\Bigl{(}\sum_{j=1}^{k}a_{j}\,\partial_{% j}\ell(\theta)\Bigr{)}=\sum_{j=1}^{k}\sum_{l=1}^{k}a_{j}a_{l}\,\mathcal{I}_{n}% (\theta)_{jl}=a^{\top}\mathcal{I}_{n}(\theta)a,

and thus is non-negative. The second statement is proved by using the fact that the score vector of the sample is the sum of nn i.i.d. score vectors of the individual observations and that the covariance matrix of the sum of independent random vectors equals the sum of the covariance matrices. This completes the proof. ∎

Strict positive definiteness is violated in cases where the log-likelihood does not respond to changes of θ\theta in some direction, for example for the model N⁢(θ1+θ2,1)N(\theta_{1}+\theta_{2},1). From now on we assume that ℐ︀n⁢(θ)\mathcal{I}_{n}(\theta) is positive definite so that the inverse exists.

14.3 The Cramér–Rao Bound for Vector Parameters

Theorem 13.2 bounds the variance of an unbiased estimator for a scalar parameter from below. For a vector parameter the natural object to bound is the covariance matrix, and the natural reading of “at least” between symmetric matrices is that the difference is positive semi-definite.

For real numbers, a>ba>b is the same as a−b>0a-b>0. For symmetric matrices AA and BB we use the second form as the definition: we think of a matrix AA as larger than BB, if A−BA-B is positive definite, and as at least as large as BB, if A−BA-B is positive semi-definite. Thus AA is at least as large as BB exactly when a⊤⁢A⁢a≥a⊤⁢B⁢aa^{\top}Aa\geq a^{\top}Ba for every a∈ℝka\in\mathbb{R}^{k}; unlike for numbers, two matrices may not be comparable at all.

Proposition 14.5.

Suppose we have the conditions of section 11.1 and that the matrix ℐ︀n⁢(θ)\mathcal{I}_{n}(\theta) is positive definite. Let T=(T1,…,Tk)T=(T_{1},\dots,T_{k}) be an unbiased estimator for θ\theta, so that 𝔼θ⁢(Tj)=θj\mathbb{E}_{\theta}(T_{j})=\theta_{j} for all jj and all θ\theta, where the variances are finite and each TjT_{j} satisfies (13.1) in every component of θ\theta. Then the matrix

Covθ(T)−ℐ︀n⁢(θ)−1\mathop{\mathrm{Cov}}\nolimits_{\theta}(T)-\mathcal{I}_{n}(\theta)^{-1}

is positive semi-definite for all θ∈Θ\theta\in\Theta. In particular we have

Varθ(Tj)≥(ℐ︀n⁢(θ)−1)j⁢jandVarθ(a⊤⁢T)≥a⊤⁢ℐ︀n⁢(θ)−1⁢a\mathop{\mathrm{Var}}\nolimits_{\theta}(T_{j})\geq\bigl{(}\mathcal{I}_{n}(% \theta)^{-1}\bigr{)}_{jj}\qquad\text{and}\qquad\mathop{\mathrm{Var}}\nolimits_% {\theta}(a^{\top}T)\geq a^{\top}\mathcal{I}_{n}(\theta)^{-1}a

for j=1,…,kj=1,\dots,k and for all a∈ℝka\in\mathbb{R}^{k}.

Proof.

The argument is the proof of theorem 13.2, applied to linear combinations. If we hold all coordinates of θ\theta other than θl\theta_{l} fixed, lemma 13.1 applies to the one-parameter model with score ∂lℓ⁢(θ)\partial_{l}\ell(\theta) and gives Covθ(Tj,∂lℓ⁢(θ))=∂lθj\mathop{\mathrm{Cov}}\nolimits_{\theta}\bigl{(}T_{j},\partial_{l}\ell(\theta)% \bigr{)}=\partial_{l}\theta_{j}, which is 11 for l=jl=j and 0 otherwise. Thus, by bilinearity, we have Covθ(a⊤⁢T,b⊤⁢∇ℓ⁢(θ))=a⊤⁢b\mathop{\mathrm{Cov}}\nolimits_{\theta}\bigl{(}a^{\top}T,b^{\top}\nabla\ell(% \theta)\bigr{)}=a^{\top}b for all a,b∈ℝka,b\in\mathbb{R}^{k}. Using the Cauchy–Schwarz inequality, lemma A.8, and the first statement of proposition 14.4, we find

(a⊤⁢b)2≤(a⊤⁢Covθ(T)⁢a)⁢(b⊤⁢ℐ︀n⁢(θ)⁢b).(a^{\top}b)^{2}\leq\bigl{(}a^{\top}\mathop{\mathrm{Cov}}\nolimits_{\theta}(T)% \,a\bigr{)}\bigl{(}b^{\top}\mathcal{I}_{n}(\theta)\,b\bigr{)}.

Choosing b=ℐ︀n⁢(θ)−1⁢ab=\mathcal{I}_{n}(\theta)^{-1}a, both a⊤⁢ba^{\top}b and b⊤⁢ℐ︀n⁢(θ)⁢bb^{\top}\mathcal{I}_{n}(\theta)b equal a⊤⁢ℐ︀n⁢(θ)−1⁢aa^{\top}\mathcal{I}_{n}(\theta)^{-1}a, which is positive for a≠0a\neq 0, and dividing by this value gives a⊤⁢Covθ(T)⁢a≥a⊤⁢ℐ︀n⁢(θ)−1⁢aa^{\top}\mathop{\mathrm{Cov}}\nolimits_{\theta}(T)\,a\geq a^{\top}\mathcal{I}_% {n}(\theta)^{-1}a. This is the claim, since a⊤⁢Covθ(T)⁢a=Varθ(a⊤⁢T)a^{\top}\mathop{\mathrm{Cov}}\nolimits_{\theta}(T)\,a=\mathop{\mathrm{Var}}% \nolimits_{\theta}(a^{\top}T), and for the diagonal case we have a=eja=e_{j}, with the jjth unit vector. This completes the proof. ∎

The same argument gives a bound for a function g⁢(θ)g(\theta) of the parameter. Let WW be an unbiased estimator for g⁢(θ)g(\theta) with finite variance which satisfies (13.1) in every component of θ\theta. Then lemma 13.1 gives Covθ(W,b⊤⁢∇ℓ⁢(θ))=b⊤⁢∇g⁢(θ)\mathop{\mathrm{Cov}}\nolimits_{\theta}\bigl{(}W,b^{\top}\nabla\ell(\theta)% \bigr{)}=b^{\top}\nabla g(\theta), and the proof above with a⊤⁢Ta^{\top}T replaced by WW and aa replaced by ∇g⁢(θ)\nabla g(\theta) shows

Varθ(W)≥∇g⁢(θ)⊤⁢ℐ︀n⁢(θ)−1⁢∇g⁢(θ).\mathop{\mathrm{Var}}\nolimits_{\theta}(W)\geq\nabla g(\theta)^{\top}\mathcal{% I}_{n}(\theta)^{-1}\nabla g(\theta).

For k=1k=1 this is theorem 13.2. The case g⁢(θ)=a⊤⁢θg(\theta)=a^{\top}\theta is used in exercise 14.3.

As an example, we apply the bound to the most important case, the normal distribution with both parameters unknown.

Example 14.6.

Let X1,…,XnX_{1},\dots,X_{n} be i.i.d. N⁢(μ,σ2)N(\mu,\sigma^{2}) where θ=(μ,σ2)\theta=(\mu,\sigma^{2}), both unknown. From example 5.4 we know that the log-likelihood function is

ℓ⁢(μ,σ2)=−n2⁢log⁡(2⁢π)−n2⁢log⁡σ2−12⁢σ2⁢∑i=1n(Xi−μ)2,\ell(\mu,\sigma^{2})=-\frac{n}{2}\log(2\pi)-\frac{n}{2}\log\sigma^{2}-\frac{1}% {2\sigma^{2}}\sum_{i=1}^{n}(X_{i}-\mu)^{2},

and the first partial derivatives are given by

∂ℓ∂μ=1σ2⁢∑i=1n(Xi−μ),∂ℓ∂σ2=−n2⁢σ2+12⁢σ4⁢∑i=1n(Xi−μ)2.\frac{\partial\ell}{\partial\mu}=\frac{1}{\sigma^{2}}\sum_{i=1}^{n}(X_{i}-\mu)% ,\qquad\frac{\partial\ell}{\partial\sigma^{2}}=-\frac{n}{2\sigma^{2}}+\frac{1}% {2\sigma^{4}}\sum_{i=1}^{n}(X_{i}-\mu)^{2}.

Taking second derivatives, where we treat σ2\sigma^{2} as a single variable, we find the three second derivatives as

∂2ℓ∂μ2\displaystyle\frac{\partial^{2}\ell}{\partial\mu^{2}}
=−nσ2,∂2ℓ∂μ⁢∂σ2=−1σ4⁢∑i=1n(Xi−μ),\displaystyle=-\frac{n}{\sigma^{2}},\qquad\frac{\partial^{2}\ell}{\partial\mu% \,\partial\sigma^{2}}=-\frac{1}{\sigma^{4}}\sum_{i=1}^{n}(X_{i}-\mu),
∂2ℓ∂(σ2)2\displaystyle\frac{\partial^{2}\ell}{\partial(\sigma^{2})^{2}}
=n2⁢σ4−1σ6⁢∑i=1n(Xi−μ)2.\displaystyle=\frac{n}{2\sigma^{4}}-\frac{1}{\sigma^{6}}\sum_{i=1}^{n}(X_{i}-% \mu)^{2}.

To compute the expectations, we use the fact that 𝔼θ⁢(Xi−μ)=0\mathbb{E}_{\theta}(X_{i}-\mu)=0 and 𝔼θ⁢((Xi−μ)2)=σ2\mathbb{E}_{\theta}\bigl{(}(X_{i}-\mu)^{2}\bigr{)}=\sigma^{2}. The mixed derivative has expectation zero. The last term has expectation n/(2⁢σ4)−n⁢σ2/σ6=−n/(2⁢σ4)n/(2\sigma^{4})-n\sigma^{2}/\sigma^{6}=-n/(2\sigma^{4}). Thus, by proposition 14.3, we find the information matrix and its inverse to be

equation (14.1) (14.1)
ℐ︀n⁢(μ,σ2)=(nσ200n2⁢σ4)andℐ︀n⁢(μ,σ2)−1=(σ2n002⁢σ4n).\mathcal{I}_{n}(\mu,\sigma^{2})=\begin{pmatrix}\dfrac{n}{\sigma^{2}}&0\\[9.429% 14pt] 0&\dfrac{n}{2\sigma^{4}}\end{pmatrix}\qquad\text{and}\qquad\mathcal{I}_{n}(\mu% ,\sigma^{2})^{-1}=\begin{pmatrix}\dfrac{\sigma^{2}}{n}&0\\[9.42914pt] 0&\dfrac{2\sigma^{4}}{n}\end{pmatrix}.

The upper left entry is the information n/σ2n/\sigma^{2} about the mean, where the variance is known from example 11.7. The sample mean, with variance σ2/n\sigma^{2}/n, achieves the bound for μ\mu. For the variance the bound is 2⁢σ4/n2\sigma^{4}/n. The UMVUE S2S^{2} from example 10.8 has variance 2⁢σ4/(n−1)2\sigma^{4}/(n-1), as shown in the computation of section 2.3, and thus does not achieve the bound. Since S2S^{2} is the UMVUE, no unbiased estimator for σ2\sigma^{2} achieves the bound when μ\mu is unknown.

The information matrix for the normal distribution is diagonal. The following section explains the significance of this fact and what happens when this is not the case.

14.4 Nuisance Parameters

Assume that we are interested in θ1\theta_{1} only, and that the remaining components θ2,…,θk\theta_{2},\dots,\theta_{k} are nuisance parameters, i.e. unknown but not of interest. If the values of the nuisance parameters were known, θ1\theta_{1} would be the only parameter and the Cramér–Rao bound for θ1\theta_{1} would be 1/ℐ︀n⁢(θ)111/\mathcal{I}_{n}(\theta)_{11}. Since the values are unknown, the bound is (ℐ︀n⁢(θ)−1)11\bigl{(}\mathcal{I}_{n}(\theta)^{-1}\bigr{)}_{11} by proposition 14.5. The following lemma compares the two bounds.

Lemma 14.7.

For every θ∈Θ\theta\in\Theta we have (ℐ︀n⁢(θ)−1)j⁢j≥1/ℐ︀n⁢(θ)j⁢j\bigl{(}\mathcal{I}_{n}(\theta)^{-1}\bigr{)}_{jj}\geq 1/\mathcal{I}_{n}(\theta% )_{jj} for j=1,…,kj=1,\dots,k. For k=2k=2, write

ℐ︀n⁢(θ)=(abbc).\mathcal{I}_{n}(\theta)=\begin{pmatrix}a&b\\ b&c\end{pmatrix}.

Then we have

(ℐ︀n⁢(θ)−1)11=1a⋅11−ρ2,where ⁢ρ=ba⁢c\bigl{(}\mathcal{I}_{n}(\theta)^{-1}\bigr{)}_{11}=\frac{1}{a}\cdot\frac{1}{1-% \rho^{2}},\qquad\text{where }\rho=\frac{b}{\sqrt{ac}}

is the correlation between the two components of the score vector, and 0≤ρ2<10\leq\rho^{2}<1.

Proof.

For the first statement, we apply the Cauchy–Schwarz inequality from lemma A.8 to the two random variables ej⊤⁢ℐ︀n⁢(θ)−1⁢∇ℓ⁢(θ)e_{j}^{\top}\mathcal{I}_{n}(\theta)^{-1}\nabla\ell(\theta) and ∂jℓ⁢(θ)\partial_{j}\ell(\theta). By the first statement of proposition 14.4, these have variances (ℐ︀n⁢(θ)−1)j⁢j\bigl{(}\mathcal{I}_{n}(\theta)^{-1}\bigr{)}_{jj} and ℐ︀n⁢(θ)j⁢j\mathcal{I}_{n}(\theta)_{jj} and covariance ej⊤⁢ℐ︀n⁢(θ)−1⁢ℐ︀n⁢(θ)⁢ej=1e_{j}^{\top}\mathcal{I}_{n}(\theta)^{-1}\mathcal{I}_{n}(\theta)e_{j}=1, and thus we find 1≤(ℐ︀n⁢(θ)−1)j⁢j⁢ℐ︀n⁢(θ)j⁢j1\leq\bigl{(}\mathcal{I}_{n}(\theta)^{-1}\bigr{)}_{jj}\,\mathcal{I}_{n}(\theta% )_{jj}. For k=2k=2 the inverse matrix is

ℐ︀n⁢(θ)−1=1a⁢c−b2⁢(c−b−ba),\mathcal{I}_{n}(\theta)^{-1}=\frac{1}{ac-b^{2}}\begin{pmatrix}c&-b\\ -b&a\end{pmatrix},

where a⁢c−b2>0ac-b^{2}>0 since the matrix is positive definite. Dividing numerator and denominator of c/(a⁢c−b2)c/(ac-b^{2}) by a⁢cac gives the claimed formula. Since aa and cc are the variances of the two components of the score vector and bb is their covariance, ρ\rho is their correlation, and a⁢c−b2>0ac-b^{2}>0 gives ρ2<1\rho^{2}<1. This completes the proof. ∎

Thus, not knowing the other parameters can never decrease the bound. For k=2k=2 the bound for θ1\theta_{1} with θ2\theta_{2} unknown is 1/(1−ρ2)1/(1-\rho^{2}) times the bound with θ2\theta_{2} known, and the two bounds agree if and only if b=0b=0. This case has a name.

Definition 14.8.

The two components of a parameter θ=(θ1,θ2)\theta=(\theta_{1},\theta_{2}) are called orthogonal, if ℐ︀⁢(θ)12=0\mathcal{I}(\theta)_{12}=0 for all θ∈Θ\theta\in\Theta.

By the additivity of information, the condition does not depend on the sample size. For orthogonal parameters, not knowing θ2\theta_{2} does not increase the bound for θ1\theta_{1}, and vice versa. We consider one example of each case.

Example 14.9.

For the normal distribution with θ=(μ,σ2)\theta=(\mu,\sigma^{2}), the information matrix (14.1) is diagonal, and thus μ\mu and σ2\sigma^{2} are orthogonal. The bound for μ\mu is σ2/n\sigma^{2}/n whether or not σ2\sigma^{2} is known, and the bound for σ2\sigma^{2} is 2⁢σ4/n2\sigma^{4}/n whether or not μ\mu is known. Equal bounds do not mean that not knowing the nuisance parameter is harmless, though: exercise 13.5 shows that the bound for σ2\sigma^{2} is attained when μ\mu is known, while example 14.6 shows that it is not attained when μ\mu is unknown.

Orthogonality is a special feature of the normal distribution, not the rule, as the following example shows.

Example 14.10.

Consider the gamma distribution with shape α>0\alpha>0 and rate β>0\beta>0 both unknown, as in exercise 7.1. The log-likelihood ℓ1⁢(α,β)\ell_{1}(\alpha,\beta) of one observation has second derivatives −ψ′⁢(α)-\psi^{\prime}(\alpha), 1/β1/\beta and −α/β2-\alpha/\beta^{2} with respect to α\alpha twice, to α\alpha and β\beta, and to β\beta twice, where ψ′\psi^{\prime} denotes the trigamma function, the second derivative of log⁡Γ\log\Gamma. Since these are non-random, the information matrix of one observation is

ℐ︀⁢(α,β)=(ψ′⁢(α)−1β−1βαβ2);\mathcal{I}(\alpha,\beta)=\begin{pmatrix}\psi^{\prime}(\alpha)&-\dfrac{1}{% \beta}\\[9.42914pt] -\dfrac{1}{\beta}&\dfrac{\alpha}{\beta^{2}}\end{pmatrix};

exercise 14.2 checks this computation. The off-diagonal entry is non-zero, so the parameters are not orthogonal. One can show that this matrix is positive definite. By lemma 14.7 we have

ρ2=1α⁢ψ′⁢(α).\rho^{2}=\frac{1}{\alpha\psi^{\prime}(\alpha)}.

For α=2\alpha=2 we use, without proof, the series ψ′⁢(α)=∑m=0∞1/(α+m)2\psi^{\prime}(\alpha)=\sum_{m=0}^{\infty}1/(\alpha+m)^{2}. With ψ′⁢(2)=∑m=2∞1/m2=π2/6−1≈0.645\psi^{\prime}(2)=\sum_{m=2}^{\infty}1/m^{2}=\pi^{2}/6-1\approx 0.645, this gives 1/(1−ρ2)≈4.451/(1-\rho^{2})\approx 4.45: the bound for α\alpha with β\beta unknown is about four and a half times the bound with β\beta known.

The inverse information matrix will appear again in lecture 17, where we will see that for large samples the maximum likelihood estimator is approximately normal with covariance matrix ℐ︀n⁢(θ0)−1\mathcal{I}_{n}(\theta_{0})^{-1}.

Summary.
  • •
    ​

    For a vector parameter the score is the gradient ∇ℓ⁢(θ)\nabla\ell(\theta), and the Fisher information matrix ℐ︀n⁢(θ)\mathcal{I}_{n}(\theta) is its covariance matrix, equal to minus the expected Hessian of the log-likelihood.

  • •
    ​

    The information matrix is symmetric and positive semi-definite, and adds over independent observations, ℐ︀n=n⁢ℐ︀\mathcal{I}_{n}=n\mathcal{I}.

  • •
    ​

    By the Cramér–Rao inequality, for an unbiased estimator TT for θ\theta the matrix Covθ(T)−ℐ︀n⁢(θ)−1\mathop{\mathrm{Cov}}\nolimits_{\theta}(T)-\mathcal{I}_{n}(\theta)^{-1} is positive semi-definite; in particular Varθ(Tj)≥(ℐ︀n⁢(θ)−1)j⁢j\mathop{\mathrm{Var}}\nolimits_{\theta}(T_{j})\geq\bigl{(}\mathcal{I}_{n}(% \theta)^{-1}\bigr{)}_{jj}.

  • •
    ​

    For N⁢(μ,σ2)N(\mu,\sigma^{2}) with both parameters unknown, ℐ︀n=diag⁢(n/σ2,n/(2⁢σ4))\mathcal{I}_{n}=\mathrm{diag}\bigl{(}n/\sigma^{2},\,n/(2\sigma^{4})\bigr{)} with inverse diag⁢(σ2/n, 2⁢σ4/n)\mathrm{diag}(\sigma^{2}/n,\,2\sigma^{4}/n).

  • •
    ​

    The diagonal of the inverse dominates the reciprocal of the diagonal, with equality for orthogonal parameters: nuisance parameters inflate the bound by 1/(1−ρ2)1/(1-\rho^{2}) in the two-parameter case, which is one for the normal distribution and about 4.454.45 for the gamma distribution with shape 22.

Exercise 14.1.

Let X1,…,XnX_{1},\dots,X_{n} be i.i.d. N⁢(μ,σ2)N(\mu,\sigma^{2}) and consider the parametrisation φ=(μ,σ)\varphi=(\mu,\sigma), where the standard deviation is used.

  1. 1.
    ​

    Write down the log-likelihood ℓ⁢(μ,σ)\ell(\mu,\sigma) and compute its first partial derivatives.

  2. 2.
    ​

    By taking derivatives a second time and taking expectations, show that the information matrix for (μ,σ)(\mu,\sigma) is diag⁢(n/σ2, 2⁢n/σ2)\mathrm{diag}(n/\sigma^{2},\,2n/\sigma^{2}).

  3. 3.
    ​

    Write down the Cramér–Rao bound for an unbiased estimator for σ\sigma. Using Jensen’s inequality, lemma A.7, show that the sample standard deviation S=S2S=\sqrt{S^{2}} is not unbiased for σ\sigma and thus cannot be compared to Var(S)\mathop{\mathrm{Var}}\nolimits(S) directly.

Exercise 14.2.

Consider the gamma distribution with shape α>0\alpha>0 and rate β>0\beta>0. The log-likelihood of observing one value is ℓ1⁢(α,β)=α⁢log⁡β−log⁡Γ⁢(α)+(α−1)⁢log⁡X−β⁢X\ell_{1}(\alpha,\beta)=\alpha\log\beta-\log\Gamma(\alpha)+(\alpha-1)\log X-\beta X from exercise 7.1, and the information matrix ℐ︀⁢(α,β)\mathcal{I}(\alpha,\beta) is given in section 14.4.

  1. 1.
    ​

    Compute the first and second partial derivatives of ℓ1⁢(α,β)\ell_{1}(\alpha,\beta) and check the matrix.

  2. 2.
    ​

    Assume that the shape parameter is known to be α=2\alpha=2. In this case, β\beta is the only parameter. From the matrix derive the Fisher information about β\beta. Why is this a diagonal element of ℐ︀⁢(α,β)\mathcal{I}(\alpha,\beta) and not of the inverse matrix?

  3. 3.
    ​

    Exercise 11.3 considers parametrisation of the same model in terms of the scale parameter θ=1/β\theta=1/\beta. Use proposition 11.9 to derive the value ℐ︀⁢(θ)=2/θ2\mathcal{I}(\theta)=2/\theta^{2} found in that exercise.

  4. 4.
    ​

    Now assume that both parameters are unknown. Show that the Cramér–Rao bound for an unbiased estimator for β\beta from a sample of size nn is given by

    β2⁢ψ′⁢(α)n⁢(α⁢ψ′⁢(α)−1),\frac{\beta^{2}\psi^{\prime}(\alpha)}{n\bigl{(}\alpha\psi^{\prime}(\alpha)-1% \bigr{)}},

    and determine the ratio of this bound to the bound for known shape parameter α=2\alpha=2.

Exercise 14.3.

Let X1,…,XmX_{1},\dots,X_{m} be i.i.d. with distribution N⁢(μ1,σ2)N(\mu_{1},\sigma^{2}) and, independently of these, let Y1,…,YnY_{1},\dots,Y_{n} be i.i.d. with distribution N⁢(μ2,σ2)N(\mu_{2},\sigma^{2}), where the common variance σ2\sigma^{2} is known and θ=(μ1,μ2)\theta=(\mu_{1},\mu_{2}) is unknown. The score of all m+nm+n observations is the sum of the scores of the two samples, and the results of this lecture carry over with ℐ︀n⁢(θ)\mathcal{I}_{n}(\theta) replaced by the covariance matrix ℐ︀m+n⁢(θ)\mathcal{I}_{m+n}(\theta) of this score.

  1. 1.
    ​

    Write down the log-likelihood of all m+nm+n observations and show that the information matrix is ℐ︀m+n⁢(μ1,μ2)=diag⁢(m/σ2,n/σ2)\mathcal{I}_{m+n}(\mu_{1},\mu_{2})=\mathrm{diag}(m/\sigma^{2},\,n/\sigma^{2}).

  2. 2.
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    Using proposition 14.5, find the Cramér–Rao bound for an unbiased estimator for the difference μ1−μ2\mu_{1}-\mu_{2}. Show that X¯−Y¯\bar{X}-\bar{Y} achieves this bound.

  3. 3.
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    Assume now that σ2\sigma^{2} is also unknown, i.e. when we have θ=(μ1,μ2,σ2)\theta=(\mu_{1},\mu_{2},\sigma^{2}). Show that the 3×33\times 3 information matrix is still diagonal and discuss the implications for the estimation of μ1−μ2\mu_{1}-\mu_{2} when σ2\sigma^{2} is unknown.