Lecture 11 Fisher Information

Since the data are random, the score function from lecture 4 is a random variable, and the Fisher information, given by the variance of the score, measures how much information the sample contains about the parameter. In lecture 13 we will use the Fisher information to obtain a lower bound for the variance of any unbiased estimator, and in lecture 17 we will see that the reciprocal of the Fisher information is the asymptotic variance of the maximum likelihood estimator. In this lecture we will define the Fisher information, prove two identities which allow us to compute the Fisher information, and we will compute the Fisher information for the standard families of models.

11.1 Setting and Assumptions

Throughout this lecture, X1,…,XnX_{1},\dots,X_{n} are i.i.d. with density or probability weights f⁢(x;θ)f(x;\theta), where the parameter space Θ⊆ℝ\Theta\subseteq\mathbb{R} is an open interval. As in lecture 4, the log-likelihood is the sum ℓ⁢(θ)=∑i=1nℓi⁢(θ)\ell(\theta)=\sum_{i=1}^{n}\ell_{i}(\theta) of the terms ℓi⁢(θ)=log⁡f⁢(Xi;θ)\ell_{i}(\theta)=\log f(X_{i};\theta) and the score of definition 4.4 is its derivative ℓ′⁢(θ)=∑iℓi′⁢(θ)\ell^{\prime}(\theta)=\sum_{i}\ell_{i}^{\prime}(\theta). In this section we study the distribution of the score under ℙθ\mathbb{P}_{\theta}, for a fixed value of θ\theta.

The results of this lecture are obtained by differentiating, with respect to θ\theta, the identity ∫f⁢(x;θ)⁢dx=1\int f(x;\theta)\,\mathrm{d}x=1, and we need this differentiation to be legitimate. We assume the following throughout. The support {x∣f⁢(x;θ)>0}\{x\mid f(x;\theta)>0\} of the distribution does not depend on θ\theta. For every xx the function θ↦f⁢(x;θ)\theta\mapsto f(x;\theta) is twice continuously differentiable on Θ\Theta. Finally, the integral of f⁢(x;θ)f(x;\theta) over the support may be differentiated under the integral sign twice, so that

equation (11.1) (11.1)
dkd⁢θk⁢∫f⁢(x;θ)⁢dx=∫∂k∂θk⁢f⁢(x;θ)⁢dxfor ⁢k=1,2,\frac{\mathrm{d}^{k}}{\mathrm{d}\theta^{k}}\int f(x;\theta)\,\mathrm{d}x=\int% \frac{\partial^{k}}{\partial\theta^{k}}f(x;\theta)\,\mathrm{d}x\qquad\text{for% }k=1,2,

where the integrals on the right-hand side exist, with sums in place of integrals for a discrete model. In addition we assume that the score of a single observation has finite variance, Varθ(ℓ1′⁢(θ))<∞\mathop{\mathrm{Var}}\nolimits_{\theta}\bigl{(}\ell_{1}^{\prime}(\theta)\bigr{% )}<\infty for all θ∈Θ\theta\in\Theta, so that the quantities defined below are finite. These four assumptions are referred to as the conditions of this section. The one-parameter models from tables 1.1 and 1.2, and the two-parameter models with one parameter held fixed, satisfy these conditions, with the exception of the uniform distribution (see section 11.7). The exponential families in the sense of definition 7.1 with twice continuously differentiable natural parameter η⁢(θ)\eta(\theta) taking values in the interior of the natural parameter space also satisfy these conditions, as explained after lemma 7.10 in lecture 7. These conditions are included in the full list of regularity conditions in definition 16.1 of lecture 16.

11.2 The Score Has Mean Zero

The first identity is that, if θ\theta is the correct parameter value, the data do not systematically deviate from this value: the score at the true parameter value is zero on average.

Lemma 11.1.

Let the situation be as described in section 11.1. Then

𝔼θ⁢(ℓi′⁢(θ))=0for ⁢i=1,…,n,and thus𝔼θ⁢(ℓ′⁢(θ))=0\mathbb{E}_{\theta}\bigl{(}\ell_{i}^{\prime}(\theta)\bigr{)}=0\qquad\text{for % }i=1,\dots,n,\qquad\text{and thus}\qquad\mathbb{E}_{\theta}\bigl{(}\ell^{% \prime}(\theta)\bigr{)}=0

for all θ∈Θ\theta\in\Theta.

Proof.

Using the chain rule, we find the derivative of ℓi⁢(θ)=log⁡f⁢(Xi;θ)\ell_{i}(\theta)=\log f(X_{i};\theta) as

equation (11.2) (11.2)
ℓi′⁢(θ)=∂∂θ⁢f⁢(Xi;θ)f⁢(Xi;θ).\ell_{i}^{\prime}(\theta)=\frac{\frac{\partial}{\partial\theta}f(X_{i};\theta)% }{f(X_{i};\theta)}.

Since XiX_{i} lies in the support with probability one, the denominator is positive and the ratio is well defined. Cancelling the density against the denominator we find

𝔼θ⁢(ℓi′⁢(θ))\displaystyle\mathbb{E}_{\theta}\bigl{(}\ell_{i}^{\prime}(\theta)\bigr{)}
=∫∂∂θ⁢f⁢(x;θ)f⁢(x;θ)⁢f⁢(x;θ)⁢dx=∫∂∂θ⁢f⁢(x;θ)⁢dx\displaystyle=\int\frac{\frac{\partial}{\partial\theta}f(x;\theta)}{f(x;\theta% )}\,f(x;\theta)\,\mathrm{d}x=\int\frac{\partial}{\partial\theta}f(x;\theta)\,% \mathrm{d}x
=dd⁢θ⁢∫f⁢(x;θ)⁢dx=dd⁢θ⁢ 1=0,\displaystyle=\frac{\mathrm{d}}{\mathrm{d}\theta}\int f(x;\theta)\,\mathrm{d}x% =\frac{\mathrm{d}}{\mathrm{d}\theta}\,1=0,

where the integral is over the support of the distribution, which does not depend on θ\theta (with a sum in place of the integral for a discrete model), and the third equality comes from the case k=1k=1 of (11.1). Summing over ii gives the statement for the whole sample. This completes the proof. ∎

If the parameter value is wrong, the score is not mean zero: for the exponential model from example 4.5, the score ℓ′⁢(θ)=n/θ−∑iXi\ell^{\prime}(\theta)=n/\theta-\sum_{i}X_{i} has expectation 𝔼θ0⁢(ℓ′⁢(θ))=n/θ−n/θ0\mathbb{E}_{\theta_{0}}\bigl{(}\ell^{\prime}(\theta)\bigr{)}=n/\theta-n/\theta% _{0} if the true parameter value is θ0\theta_{0}, positive for θ<θ0\theta<\theta_{0} and negative for θ>θ0\theta>\theta_{0}. The score on average points towards the correct parameter value from both sides. In lecture 16 we will use this behaviour to prove that the maximum likelihood estimator is consistent.

11.3 Fisher Information

The next question we could ask is about the fluctuations of the score around zero. If the score is typically large in magnitude, this indicates that the data allow us to distinguish well between nearby parameter values. In contrast, if the score is typically close to zero, this indicates that the log-likelihood is nearly flat at the truth. The variance of the score is a measure for how informative the sample is.

Definition 11.2.

Under the conditions of section 11.1, the Fisher information of the sample X1,…,XnX_{1},\dots,X_{n} about θ\theta is given by

ℐ︀n⁢(θ)=Varθ(ℓ′⁢(θ)),\mathcal{I}_{n}(\theta)=\mathop{\mathrm{Var}}\nolimits_{\theta}\bigl{(}\ell^{% \prime}(\theta)\bigr{)},

where the variance of the score is taken at θ\theta. The Fisher information of a single observation is denoted by ℐ︀⁢(θ)=ℐ︀1⁢(θ)\mathcal{I}(\theta)=\mathcal{I}_{1}(\theta) and is given by the variance of ℓ1′⁢(θ)\ell_{1}^{\prime}(\theta).

The subscript in ℐ︀n\mathcal{I}_{n} is the sample size. Where we compare two different parametrisations of the same model, we write ℐ︀θ\mathcal{I}_{\theta}, ℐ︀η\mathcal{I}_{\eta} or ℐ︀ψ\mathcal{I}_{\psi} for the information of a single observation about the parameter named in the subscript.

The variance of the score is often awkward to compute directly, and the second expression of the following lemma is nearly always the quickest route to the information.

Lemma 11.3.

Suppose we are in the situation of section 11.1. Then

ℐ︀n⁢(θ)=𝔼θ⁢(ℓ′⁢(θ)2)=−𝔼θ⁢(ℓ′′⁢(θ))\mathcal{I}_{n}(\theta)=\mathbb{E}_{\theta}\bigl{(}\ell^{\prime}(\theta)^{2}% \bigr{)}=-\mathbb{E}_{\theta}\bigl{(}\ell^{\prime\prime}(\theta)\bigr{)}

for all θ∈Θ\theta\in\Theta.

Proof.

The first equality follows from the fact that ℓ′⁢(θ)\ell^{\prime}(\theta) has mean zero by lemma 11.1, and thus the variance equals the second moment. The second equality is obtained by taking the derivative of the ratio (11.2) using the quotient rule:

ℓi′′⁢(θ)=∂2∂θ2⁢f⁢(Xi;θ)f⁢(Xi;θ)−(∂∂θ⁢f⁢(Xi;θ)f⁢(Xi;θ))2=∂2∂θ2⁢f⁢(Xi;θ)f⁢(Xi;θ)−ℓi′⁢(θ)2.\ell_{i}^{\prime\prime}(\theta)=\frac{\frac{\partial^{2}}{\partial\theta^{2}}f% (X_{i};\theta)}{f(X_{i};\theta)}-\Bigl{(}\frac{\frac{\partial}{\partial\theta}% f(X_{i};\theta)}{f(X_{i};\theta)}\Bigr{)}^{2}=\frac{\frac{\partial^{2}}{% \partial\theta^{2}}f(X_{i};\theta)}{f(X_{i};\theta)}-\ell_{i}^{\prime}(\theta)% ^{2}.

Similar to the proof of lemma 11.1, the expectation of the first term vanishes, using the case k=2k=2 of (11.1):

𝔼θ⁢(∂2∂θ2⁢f⁢(Xi;θ)f⁢(Xi;θ))=∫∂2∂θ2⁢f⁢(x;θ)⁢dx=d2d⁢θ2⁢∫f⁢(x;θ)⁢dx=0.\mathbb{E}_{\theta}\Bigl{(}\frac{\frac{\partial^{2}}{\partial\theta^{2}}f(X_{i% };\theta)}{f(X_{i};\theta)}\Bigr{)}=\int\frac{\partial^{2}}{\partial\theta^{2}% }f(x;\theta)\,\mathrm{d}x=\frac{\mathrm{d}^{2}}{\mathrm{d}\theta^{2}}\int f(x;% \theta)\,\mathrm{d}x=0.

Since the second moment of ℓi′⁢(θ)\ell_{i}^{\prime}(\theta) is finite by the conditions of section 11.1, we can take expectations term by term and get 𝔼θ⁢(ℓi′′⁢(θ))=−𝔼θ⁢(ℓi′⁢(θ)2)\mathbb{E}_{\theta}\bigl{(}\ell_{i}^{\prime\prime}(\theta)\bigr{)}=-\mathbb{E}% _{\theta}\bigl{(}\ell_{i}^{\prime}(\theta)^{2}\bigr{)} for all ii. The terms ℓ1′⁢(θ),…,ℓn′⁢(θ)\ell_{1}^{\prime}(\theta),\dots,\ell_{n}^{\prime}(\theta) are independent of each other, since they are functions of different observations, and thus the variance of the sum is the sum of the variances. Since each term has mean zero, its variance is its second moment, and we find

Varθ(ℓ′⁢(θ))\displaystyle\mathop{\mathrm{Var}}\nolimits_{\theta}\bigl{(}\ell^{\prime}(% \theta)\bigr{)}
=∑i=1nVarθ(ℓi′⁢(θ))=∑i=1n𝔼θ⁢(ℓi′⁢(θ)2)\displaystyle=\sum_{i=1}^{n}\mathop{\mathrm{Var}}\nolimits_{\theta}\bigl{(}% \ell_{i}^{\prime}(\theta)\bigr{)}=\sum_{i=1}^{n}\mathbb{E}_{\theta}\bigl{(}% \ell_{i}^{\prime}(\theta)^{2}\bigr{)}
=−∑i=1n𝔼θ⁢(ℓi′′⁢(θ))=−𝔼θ⁢(ℓ′′⁢(θ)),\displaystyle=-\sum_{i=1}^{n}\mathbb{E}_{\theta}\bigl{(}\ell_{i}^{\prime\prime% }(\theta)\bigr{)}=-\mathbb{E}_{\theta}\bigl{(}\ell^{\prime\prime}(\theta)\bigr% {)},

where we used ℓ′′⁢(θ)=∑iℓi′′⁢(θ)\ell^{\prime\prime}(\theta)=\sum_{i}\ell_{i}^{\prime\prime}(\theta) in the last step. This completes the proof. ∎

The second expression gives the name of the quantity: it is the expected curvature of the log-likelihood at the true parameter value. Large information means that the log-likelihood has a sharp maximum, well-defined by the data. Small information means the log-likelihood is flat. Note that the identity holds in expectation only: for a given sample the curvature −ℓ′′⁢(θ)-\ell^{\prime\prime}(\theta) is random and in general differs from ℐ︀n⁢(θ)\mathcal{I}_{n}(\theta). In lecture 17 this idea will be turned into a theorem: the approximate variance of the maximum likelihood estimator is 1/ℐ︀n⁢(θ)1/\mathcal{I}_{n}(\theta) for large sample size, and ℐ︀n⁢(θ)\mathcal{I}_{n}(\theta) can be replaced there by the curvature −ℓ′′⁢(θ^)-\ell^{\prime\prime}(\hat{\theta}) at the maximum likelihood estimate, which can be computed from the data, numerically if need be (lecture 18).

The proof also shows how the information of the sample relates to the information of one observation: each observation contributes the same amount, and nn observations carry nn times the information of one.

Proposition 11.4.

With the same assumptions as in section 11.1 we have

ℐ︀n⁢(θ)=n⁢ℐ︀⁢(θ)\mathcal{I}_{n}(\theta)=n\,\mathcal{I}(\theta)

for all θ∈Θ\theta\in\Theta.

Proof.

The first equality in the last display of the proof of lemma 11.3 writes ℐ︀n⁢(θ)\mathcal{I}_{n}(\theta) as the sum ∑iVarθ(ℓi′⁢(θ))\sum_{i}\mathop{\mathrm{Var}}\nolimits_{\theta}\bigl{(}\ell_{i}^{\prime}(% \theta)\bigr{)}. Since the observations are identically distributed, each of the nn terms equals Varθ(ℓ1′⁢(θ))=ℐ︀⁢(θ)\mathop{\mathrm{Var}}\nolimits_{\theta}\bigl{(}\ell_{1}^{\prime}(\theta)\bigr{% )}=\mathcal{I}(\theta). This completes the proof. ∎

11.4 Examples

We now compute the information for four of the standard families of tables 1.1 and 1.2; the exercises cover the binomial, gamma and geometric distributions. By proposition 11.4 it suffices to work with a single observation X=X1X=X_{1}.

Example 11.5.

For the Bernoulli distribution with success probability θ∈(0,1)\theta\in(0,1) we have ℓ1⁢(θ)=X⁢log⁡θ+(1−X)⁢log⁡(1−θ)\ell_{1}(\theta)=X\log\theta+(1-X)\log(1-\theta) and thus

ℓ1′⁢(θ)=Xθ−1−X1−θ,ℓ1′′⁢(θ)=−Xθ2−1−X(1−θ)2.\ell_{1}^{\prime}(\theta)=\frac{X}{\theta}-\frac{1-X}{1-\theta},\qquad\ell_{1}% ^{\prime\prime}(\theta)=-\frac{X}{\theta^{2}}-\frac{1-X}{(1-\theta)^{2}}.

Using 𝔼θ⁢(X)=θ\mathbb{E}_{\theta}(X)=\theta we find

ℐ︀⁢(θ)=−𝔼θ⁢(ℓ1′′⁢(θ))=θθ2+1−θ(1−θ)2=1θ⁢(1−θ).\mathcal{I}(\theta)=-\mathbb{E}_{\theta}\bigl{(}\ell_{1}^{\prime\prime}(\theta% )\bigr{)}=\frac{\theta}{\theta^{2}}+\frac{1-\theta}{(1-\theta)^{2}}=\frac{1}{% \theta(1-\theta)}.

Thus, the information of the sample is ℐ︀n⁢(θ)=n/(θ⁢(1−θ))\mathcal{I}_{n}(\theta)=n/\bigl{(}\theta(1-\theta)\bigr{)}. Exercise 11.1 checks these identities for this model directly.

Example 11.6.

For the Poisson distribution with mean θ>0\theta>0 we have ℓ1⁢(θ)=X⁢log⁡θ−θ−log⁡X!\ell_{1}(\theta)=X\log\theta-\theta-\log X! and thus

ℓ1′⁢(θ)=Xθ−1,ℓ1′′⁢(θ)=−Xθ2,ℐ︀⁢(θ)=𝔼θ⁢(X)θ2=1θ.\ell_{1}^{\prime}(\theta)=\frac{X}{\theta}-1,\qquad\ell_{1}^{\prime\prime}(% \theta)=-\frac{X}{\theta^{2}},\qquad\mathcal{I}(\theta)=\frac{\mathbb{E}_{% \theta}(X)}{\theta^{2}}=\frac{1}{\theta}.
Example 11.7.

In the case of the normal distribution with unknown mean μ\mu and known variance σ2\sigma^{2}, the parameter is μ\mu and ℓ1⁢(μ)=−12⁢log⁡(2⁢π⁢σ2)−(X−μ)2/(2⁢σ2)\ell_{1}(\mu)=-\tfrac{1}{2}\log(2\pi\sigma^{2})-(X-\mu)^{2}/(2\sigma^{2}). Thus we have

ℓ1′⁢(μ)=X−μσ2,ℓ1′′⁢(μ)=−1σ2,ℐ︀⁢(μ)=1σ2.\ell_{1}^{\prime}(\mu)=\frac{X-\mu}{\sigma^{2}},\qquad\ell_{1}^{\prime\prime}(% \mu)=-\frac{1}{\sigma^{2}},\qquad\mathcal{I}(\mu)=\frac{1}{\sigma^{2}}.

Since the second derivative does not depend on the data, we do not need to consider the expectation. Note that Varμ(X¯)=σ2/n=1/ℐ︀n⁢(μ)\mathop{\mathrm{Var}}\nolimits_{\mu}(\bar{X})=\sigma^{2}/n=1/\mathcal{I}_{n}(\mu) by lemma 2.2, and in lecture 13 we will see that no unbiased estimator can have a smaller variance.

Example 11.8.

For the exponential distribution with rate θ>0\theta>0, we have ℓ1⁢(θ)=log⁡θ−θ⁢X\ell_{1}(\theta)=\log\theta-\theta X and thus

ℓ1′⁢(θ)=1θ−X,ℓ1′′⁢(θ)=−1θ2,ℐ︀⁢(θ)=1θ2.\ell_{1}^{\prime}(\theta)=\frac{1}{\theta}-X,\qquad\ell_{1}^{\prime\prime}(% \theta)=-\frac{1}{\theta^{2}},\qquad\mathcal{I}(\theta)=\frac{1}{\theta^{2}}.

Again, the second derivative is non-random. Section 11.6 shows how the information changes if we describe the model by the mean 1/θ1/\theta instead of the rate.

11.5 Exponential Families

All four models from section 11.4 are exponential families. For such models, in the natural parametrisation, the information is the second derivative of the cumulant function KK from definition 7.9.

Proposition 11.9.

Let f⁢(x;η)=h⁢(x)⁢exp⁡(η⁢T⁢(x)−K⁢(η))f(x;\eta)=h(x)\exp\bigl{(}\eta T(x)-K(\eta)\bigr{)} be an exponential family in its natural parametrisation, where η\eta ranges over an open interval inside the natural parameter space 𝒩︀\mathcal{N}. Then the Fisher information of a single observation about η\eta is given by

ℐ︀η⁢(η)=K′′⁢(η)=Varη(T⁢(X)).\mathcal{I}_{\eta}(\eta)=K^{\prime\prime}(\eta)=\mathop{\mathrm{Var}}\nolimits% _{\eta}\bigl{(}T(X)\bigr{)}.

On the other hand, if the family is written as f⁢(x;θ)=h⁢(x)⁢exp⁡(η⁢(θ)⁢T⁢(x)−B⁢(θ))f(x;\theta)=h(x)\exp\bigl{(}\eta(\theta)T(x)-B(\theta)\bigr{)} as in definition 7.1, where η\eta is twice continuously differentiable and takes values in the interior of 𝒩︀\mathcal{N}, then the information about θ\theta is given by

ℐ︀θ⁢(θ)=η′⁢(θ)2⁢K′′⁢(η⁢(θ))=η′⁢(θ)2⁢Varθ(T⁢(X)).\mathcal{I}_{\theta}(\theta)=\eta^{\prime}(\theta)^{2}\,K^{\prime\prime}\bigl{% (}\eta(\theta)\bigr{)}=\eta^{\prime}(\theta)^{2}\,\mathop{\mathrm{Var}}% \nolimits_{\theta}\bigl{(}T(X)\bigr{)}.

The assumptions on η\eta are those under which the conditions of section 11.1 hold, as discussed there, so that the information is defined.

Proof.

In the natural parametrisation we have ℓ1⁢(η)=log⁡h⁢(X)+η⁢T⁢(X)−K⁢(η)\ell_{1}(\eta)=\log h(X)+\eta T(X)-K(\eta) and thus ℓ1′⁢(η)=T⁢(X)−K′⁢(η)\ell_{1}^{\prime}(\eta)=T(X)-K^{\prime}(\eta) and ℓ1′′⁢(η)=−K′′⁢(η)\ell_{1}^{\prime\prime}(\eta)=-K^{\prime\prime}(\eta). By lemma 7.10 we have 𝔼η⁢(T⁢(X))=K′⁢(η)\mathbb{E}_{\eta}\bigl{(}T(X)\bigr{)}=K^{\prime}(\eta) and Varη(T⁢(X))=K′′⁢(η)\mathop{\mathrm{Var}}\nolimits_{\eta}\bigl{(}T(X)\bigr{)}=K^{\prime\prime}(\eta) and thus, since the second derivative is non-random, we find ℐ︀η⁢(η)=K′′⁢(η)=Varη(T⁢(X))\mathcal{I}_{\eta}(\eta)=K^{\prime\prime}(\eta)=\mathop{\mathrm{Var}}\nolimits% _{\eta}\bigl{(}T(X)\bigr{)}. For the general parametrisation we have B⁢(θ)=K⁢(η⁢(θ))B(\theta)=K\bigl{(}\eta(\theta)\bigr{)}, since both quantities are determined by the requirement that the density must integrate to one, and thus, using the chain rule, we find

ℓ1′⁢(θ)=η′⁢(θ)⁢(T⁢(X)−K′⁢(η⁢(θ))).\ell_{1}^{\prime}(\theta)=\eta^{\prime}(\theta)\,\Bigl{(}T(X)-K^{\prime}\bigl{% (}\eta(\theta)\bigr{)}\Bigr{)}.

The factor η′⁢(θ)\eta^{\prime}(\theta) is a constant and thus the variance of the score is

ℐ︀θ⁢(θ)=η′⁢(θ)2⁢Varθ(T⁢(X))=η′⁢(θ)2⁢K′′⁢(η⁢(θ)),\mathcal{I}_{\theta}(\theta)=\eta^{\prime}(\theta)^{2}\,\mathop{\mathrm{Var}}% \nolimits_{\theta}\bigl{(}T(X)\bigr{)}=\eta^{\prime}(\theta)^{2}\,K^{\prime% \prime}\bigl{(}\eta(\theta)\bigr{)},

as claimed. ∎

The proposition turns the computation of the information into that of a variance, which for the standard families can be read off tables A.1 and A.2. As a check, the Bernoulli distribution has T⁢(x)=xT(x)=x and η⁢(θ)=log⁡(θ/(1−θ))\eta(\theta)=\log\bigl{(}\theta/(1-\theta)\bigr{)} by example 7.2, so that η′⁢(θ)2⁢Varθ(X)=θ⁢(1−θ)/(θ⁢(1−θ))2\eta^{\prime}(\theta)^{2}\mathop{\mathrm{Var}}\nolimits_{\theta}(X)=\theta(1-% \theta)/\bigl{(}\theta(1-\theta)\bigr{)}^{2} recovers example 11.5. Exercises 11.2 and 11.4 use the proposition in the same way.

11.6 Reparametrisation

The information depends on which parameter we use to describe a model, and the factor η′⁢(θ)2\eta^{\prime}(\theta)^{2} in proposition 11.9 is an instance of the general rule.

Proposition 11.10.

Let ψ=g⁢(θ)\psi=g(\theta) where g:Θ→g⁢(Θ)g\colon\Theta\to g(\Theta) is a twice continuously differentiable bijection with g′⁢(θ)≠0g^{\prime}(\theta)\neq 0 for all θ∈Θ\theta\in\Theta. Let ℐ︀θ\mathcal{I}_{\theta} denote the Fisher information for a single observation when the model is parametrised by θ\theta and let ℐ︀ψ\mathcal{I}_{\psi} be the information when the model is parametrised by ψ\psi. Then,

ℐ︀ψ⁢(ψ)=ℐ︀θ⁢(θ)g′⁢(θ)2for ⁢ψ=g⁢(θ).\mathcal{I}_{\psi}(\psi)=\frac{\mathcal{I}_{\theta}(\theta)}{g^{\prime}(\theta% )^{2}}\qquad\text{for }\psi=g(\theta).
Proof.

With the new parametrisation, the density is f~⁢(x;ψ)=f⁢(x;g−1⁢(ψ))\tilde{f}(x;\psi)=f\bigl{(}x;g^{-1}(\psi)\bigr{)} and the log-likelihood for a single observation is ℓ~1⁢(ψ)=ℓ1⁢(g−1⁢(ψ))\tilde{\ell}_{1}(\psi)=\ell_{1}\bigl{(}g^{-1}(\psi)\bigr{)}. Since g′g^{\prime} is continuous and never vanishes on the interval Θ\Theta, it has the same sign throughout, so that gg is strictly monotonic and g−1g^{-1} is twice continuously differentiable. Using the chain rule, the conditions of section 11.1 for the parametrisation by θ\theta carry over to the parametrisation by ψ\psi; we omit the verification. The derivative of the inverse function is (g−1)′⁢(ψ)=1/g′⁢(θ)(g^{-1})^{\prime}(\psi)=1/g^{\prime}(\theta) for θ=g−1⁢(ψ)\theta=g^{-1}(\psi). This gives

ℓ~1′⁢(ψ)=ℓ1′⁢(θ)⁢(g−1)′⁢(ψ)=ℓ1′⁢(θ)g′⁢(θ).\tilde{\ell}_{1}^{\prime}(\psi)=\ell_{1}^{\prime}(\theta)\,(g^{-1})^{\prime}(% \psi)=\frac{\ell_{1}^{\prime}(\theta)}{g^{\prime}(\theta)}.

Since θ\theta is fixed for fixed ψ\psi, 1/g′⁢(θ)1/g^{\prime}(\theta) is a constant and taking variances under ℙθ\mathbb{P}_{\theta} gives ℐ︀ψ⁢(ψ)=ℐ︀θ⁢(θ)/g′⁢(θ)2\mathcal{I}_{\psi}(\psi)=\mathcal{I}_{\theta}(\theta)/g^{\prime}(\theta)^{2}, as claimed. ∎

For the exponential distribution, reparametrising from the rate θ\theta to the mean μ=1/θ\mu=1/\theta changes the information 1/θ21/\theta^{2} from example 11.8 to ℐ︀μ⁢(μ)=1/μ2\mathcal{I}_{\mu}(\mu)=1/\mu^{2}; see exercise 11.5. Thus, the information is not a property of the model in isolation, but depends on the parametrisation. In lecture 26 we will use the transformation rule of the proposition to construct a prior distribution, the Jeffreys prior, which does not depend on the parametrisation chosen.

11.7 When the Identities Fail

The uniform distribution from example 1.2 shows what can go wrong if the conditions from section 11.1 are violated.

Example 11.11.

Let XX be uniformly distributed on [0,θ][0,\theta], i.e. f⁢(x;θ)=1/θf(x;\theta)=1/\theta for 0≤x≤θ0\leq x\leq\theta. The support of the distribution depends on θ\theta, but we can still compute the derivatives on the support to see whether the identities hold. For 0<X<θ0<X<\theta we have ℓ1⁢(θ)=−log⁡θ\ell_{1}(\theta)=-\log\theta and thus ℓ1′⁢(θ)=−1/θ\ell_{1}^{\prime}(\theta)=-1/\theta and ℓ1′′⁢(θ)=1/θ2\ell_{1}^{\prime\prime}(\theta)=1/\theta^{2}. The score is a negative constant and thus we have 𝔼θ⁢(ℓ1′⁢(θ))=−1/θ≠0\mathbb{E}_{\theta}\bigl{(}\ell_{1}^{\prime}(\theta)\bigr{)}=-1/\theta\neq 0, in contradiction to lemma 11.1. The variance is zero, whereas 𝔼θ⁢(ℓ1′⁢(θ)2)=1/θ2\mathbb{E}_{\theta}\bigl{(}\ell_{1}^{\prime}(\theta)^{2}\bigr{)}=1/\theta^{2} and −𝔼θ⁢(ℓ1′′⁢(θ))=−1/θ2-\mathbb{E}_{\theta}\bigl{(}\ell_{1}^{\prime\prime}(\theta)\bigr{)}=-1/\theta^% {2}: the three expressions for lemma 11.3 take three different values, one of which is negative. The proof breaks down at the interchange between differentiation and integration: for this model, the two sides of (11.1) with k=1k=1 differ by the term f⁢(θ;θ)=1/θf(\theta;\theta)=1/\theta from the moving upper limit, which is missed by the differentiation under the integral sign. Exercise 11.6 can be used to work out the two computations.

For this model the Fisher information is not defined, and the two results based on it do not apply. Their conclusions fail as well: by exercise 2.2 the mean squared error of the sample maximum, and with it the variance of the unbiased estimator (n+1)⁢maxi⁡Xi/n(n+1)\max_{i}X_{i}/n, decreases like 1/n21/n^{2}, whereas the lower bound of lecture 13, where it applies, does not allow the variance of an unbiased estimator to decrease faster than 1/n1/n; and the maximum likelihood estimator maxi⁡Xi\max_{i}X_{i} is not approximately normal for large nn, in contrast to lecture 17.

Summary.
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    Under the conditions of section 11.1 (a support free of θ\theta, twice continuously differentiable density, differentiation under the integral sign and a score with finite variance), the score at the true parameter value has mean zero.

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    The Fisher information ℐ︀n⁢(θ)\mathcal{I}_{n}(\theta) is the variance of the score, and it equals 𝔼θ⁢(ℓ′⁢(θ)2)\mathbb{E}_{\theta}\bigl{(}\ell^{\prime}(\theta)^{2}\bigr{)} and −𝔼θ⁢(ℓ′′⁢(θ))-\mathbb{E}_{\theta}\bigl{(}\ell^{\prime\prime}(\theta)\bigr{)}, the expected curvature of the log-likelihood at the truth.

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    Information from independent observations can be combined, ℐ︀n⁢(θ)=n⁢ℐ︀⁢(θ)\mathcal{I}_{n}(\theta)=n\mathcal{I}(\theta); for an exponential family we have ℐ︀θ⁢(θ)=η′⁢(θ)2⁢Varθ(T)\mathcal{I}_{\theta}(\theta)=\eta^{\prime}(\theta)^{2}\mathop{\mathrm{Var}}% \nolimits_{\theta}(T); the information depends on the parametrisation, and under ψ=g⁢(θ)\psi=g(\theta) we get ℐ︀ψ⁢(ψ)=ℐ︀θ⁢(θ)/g′⁢(θ)2\mathcal{I}_{\psi}(\psi)=\mathcal{I}_{\theta}(\theta)/g^{\prime}(\theta)^{2}.

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    We have learned the following standard values: 1/(θ⁢(1−θ))1/\bigl{(}\theta(1-\theta)\bigr{)} for the Bernoulli distribution, 1/θ1/\theta for the Poisson distribution, 1/σ21/\sigma^{2} for the mean of the normal distribution and 1/θ21/\theta^{2} for the rate parameter in the exponential distribution. For the uniform distribution these relations do not hold, since the support of the distribution moves with the parameter.

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    The information is the currency of the two results presented later: it gives a bound on the variance of unbiased estimators in lecture 13 and it determines the asymptotic variance of the MLE in lecture 17.

Exercise 11.1.

Let X1,…,XnX_{1},\dots,X_{n} be i.i.d. Bernoulli with success probability θ∈(0,1)\theta\in(0,1), and let ℓ1′⁢(θ)\ell_{1}^{\prime}(\theta) be the score of a single observation from example 11.5.

  1. 1.
    ​

    Verify lemma 11.1 for this model by computing 𝔼θ⁢(ℓ1′⁢(θ))\mathbb{E}_{\theta}\bigl{(}\ell_{1}^{\prime}(\theta)\bigr{)} directly from the two possible values of X1X_{1}.

  2. 2.
    ​

    Compute Varθ(ℓ1′⁢(θ))\mathop{\mathrm{Var}}\nolimits_{\theta}\bigl{(}\ell_{1}^{\prime}(\theta)\bigr{)} directly, and check that it agrees with the value of ℐ︀⁢(θ)\mathcal{I}(\theta) found in example 11.5.

  3. 3.
    ​

    Compare 1/ℐ︀n⁢(θ)1/\mathcal{I}_{n}(\theta) with the variance of the estimator X¯\bar{X} for θ\theta.

  4. 4.
    ​

    Now let YY be a single observation from the binomial distribution with known number of trials mm and success probability θ\theta. Show that the information of YY about θ\theta is m/(θ⁢(1−θ))m/\bigl{(}\theta(1-\theta)\bigr{)}, and explain the relation to the Bernoulli case using proposition 11.4.

Exercise 11.2.

Let X1,…,XnX_{1},\dots,X_{n} be i.i.d. with density f⁢(x;θ)=θ−2⁢x⁢e−x/θf(x;\theta)=\theta^{-2}x\,e^{-x/\theta} for x≥0x\geq 0, where θ>0\theta>0, the gamma distribution with known shape 22 and scale θ\theta from exercise 5.2, i.e. Gamma⁢(2,1/θ)\text{Gamma}(2,1/\theta) in the rate parametrisation from table A.2.

  1. 1.
    ​

    Compute the score ℓ1′⁢(θ)\ell_{1}^{\prime}(\theta) of a single observation and verify that it has mean zero.

  2. 2.
    ​

    Show that ℐ︀⁢(θ)=2/θ2\mathcal{I}(\theta)=2/\theta^{2}, once via −𝔼θ⁢(ℓ1′′⁢(θ))-\mathbb{E}_{\theta}\bigl{(}\ell_{1}^{\prime\prime}(\theta)\bigr{)} and once via the variance of the score, using Varθ(X1)=2⁢θ2\mathop{\mathrm{Var}}\nolimits_{\theta}(X_{1})=2\theta^{2}.

  3. 3.
    ​

    Write the density in the form of definition 7.1 with T⁢(x)=xT(x)=x, identify η⁢(θ)\eta(\theta), and recover ℐ︀⁢(θ)\mathcal{I}(\theta) from proposition 11.9.

  4. 4.
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    In exercise 5.2 we found that the MLE θ^=X¯/2\hat{\theta}=\bar{X}/2 is unbiased with MSE(θ^)=θ2/(2⁢n)\mathop{\mathrm{MSE}}\nolimits(\hat{\theta})=\theta^{2}/(2n). Compare this with 1/ℐ︀n⁢(θ)1/\mathcal{I}_{n}(\theta).

Exercise 11.3.

Let X1,…,XnX_{1},\dots,X_{n} be i.i.d. from the geometric distribution with parameter p∈(0,1)p\in(0,1), i.e. f⁢(x;p)=(1−p)x−1⁢pf(x;p)=(1-p)^{x-1}p for x∈ℕx\in\mathbb{N}, where 𝔼p⁢(X1)=1/p\mathbb{E}_{p}(X_{1})=1/p and Varp(X1)=(1−p)/p2\mathop{\mathrm{Var}}\nolimits_{p}(X_{1})=(1-p)/p^{2}, as in exercises 4.2 and 5.1.

  1. 1.
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    Compute the score ℓ1′⁢(p)\ell_{1}^{\prime}(p) of a single observation and verify that it has mean zero.

  2. 2.
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    Show that ℐ︀⁢(p)=1/(p2⁢(1−p))\mathcal{I}(p)=1/\bigl{(}p^{2}(1-p)\bigr{)} and write down ℐ︀n⁢(p)\mathcal{I}_{n}(p).

  3. 3.
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    Verify the value from the previous part by taking the variance of the score.

Exercise 11.4.

Let X1,…,XnX_{1},\dots,X_{n} be i.i.d. Poisson with mean θ>0\theta>0.

  1. 1.
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    From lecture 7 we know that the natural parameter is η=log⁡θ\eta=\log\theta and the cumulant function is K⁢(η)=eηK(\eta)=e^{\eta}. Using proposition 11.9, find the information ℐ︀η⁢(η)\mathcal{I}_{\eta}(\eta) about η\eta in terms of θ\theta.

  2. 2.
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    Recover ℐ︀θ⁢(θ)=1/θ\mathcal{I}_{\theta}(\theta)=1/\theta from the previous part using proposition 11.10.

  3. 3.
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    In exercise 5.7 we have estimated the probability p0=ℙθ⁢(X1=0)=e−θp_{0}=\mathbb{P}_{\theta}(X_{1}=0)=e^{-\theta}. Determine the information ℐ︀p0⁢(p0)\mathcal{I}_{p_{0}}(p_{0}) about p0p_{0} and describe its behaviour as θ→∞\theta\to\infty. Give a one-sentence answer as to why this behaviour is reasonable.

Exercise 11.5.

Let X1,…,XnX_{1},\dots,X_{n} be i.i.d. exponential with rate θ>0\theta>0 and let μ=1/θ\mu=1/\theta be the mean.

  1. 1.
    ​

    Using proposition 11.10 and example 11.8, verify the value ℐ︀μ⁢(μ)=1/μ2\mathcal{I}_{\mu}(\mu)=1/\mu^{2} from section 11.6.

  2. 2.
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    Show that the information of the sample about μ\mu is n/μ2n/\mu^{2} and compare the reciprocal of this quantity to the variance of the estimator X¯\bar{X} for μ\mu.

Exercise 11.6.

Let XX be uniformly distributed on the interval [0,θ][0,\theta] as described in example 11.11. Compute both sides of (11.1) for this model, using k=1k=1, and verify that the difference between the two quantities corresponds to the contribution f⁢(θ;θ)=1/θf(\theta;\theta)=1/\theta of the (moving) upper limit of the integral.