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, are i.i.d. with density or probability weights , where the parameter space is an open interval. As in lecture 4, the log-likelihood is the sum of the terms and the score of definition 4.4 is its derivative . In this section we study the distribution of the score under , for a fixed value of .
The results of this lecture are obtained by differentiating, with respect to , the identity , and we need this differentiation to be legitimate. We assume the following throughout. The support of the distribution does not depend on . For every the function is twice continuously differentiable on . Finally, the integral of over the support may be differentiated under the integral sign twice, so that
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, for all , 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 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 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.
Using the chain rule, we find the derivative of as
Since 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
where the integral is over the support of the distribution, which does not depend on (with a sum in place of the integral for a discrete model), and the third equality comes from the case of (11.1). Summing over 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 has expectation if the true parameter value is , positive for and negative for . 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.
Under the conditions of section 11.1, the Fisher information of the sample about is given by
where the variance of the score is taken at . The Fisher information of a single observation is denoted by and is given by the variance of .
The subscript in is the sample size. Where we compare two different parametrisations of the same model, we write , or 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.
The first equality follows from the fact that 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:
Similar to the proof of lemma 11.1, the expectation of the first term vanishes, using the case of (11.1):
Since the second moment of is finite by the conditions of section 11.1, we can take expectations term by term and get for all . The terms 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
where we used 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 is random and in general differs from . In lecture 17 this idea will be turned into a theorem: the approximate variance of the maximum likelihood estimator is for large sample size, and can be replaced there by the curvature 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 observations carry times the information of one.
The first equality in the last display of the proof of lemma 11.3 writes as the sum . Since the observations are identically distributed, each of the terms equals . 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 .
For the Bernoulli distribution with success probability we have and thus
Using we find
Thus, the information of the sample is . Exercise 11.1 checks these identities for this model directly.
For the Poisson distribution with mean we have and thus
In the case of the normal distribution with unknown mean and known variance , the parameter is and . Thus we have
Since the second derivative does not depend on the data, we do not need to consider the expectation. Note that by lemma 2.2, and in lecture 13 we will see that no unbiased estimator can have a smaller variance.
For the exponential distribution with rate , we have and thus
Again, the second derivative is non-random. Section 11.6 shows how the information changes if we describe the model by the mean 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 from definition 7.9.
Let be an exponential family in its natural parametrisation, where ranges over an open interval inside the natural parameter space . Then the Fisher information of a single observation about is given by
On the other hand, if the family is written as as in definition 7.1, where is twice continuously differentiable and takes values in the interior of , then the information about is given by
The assumptions on are those under which the conditions of section 11.1 hold, as discussed there, so that the information is defined.
In the natural parametrisation we have and thus and . By lemma 7.10 we have and and thus, since the second derivative is non-random, we find . For the general parametrisation we have , since both quantities are determined by the requirement that the density must integrate to one, and thus, using the chain rule, we find
The factor is a constant and thus the variance of the score is
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 and by example 7.2, so that 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 in proposition 11.9 is an instance of the general rule.
Let where is a twice continuously differentiable bijection with for all . Let denote the Fisher information for a single observation when the model is parametrised by and let be the information when the model is parametrised by . Then,
With the new parametrisation, the density is and the log-likelihood for a single observation is . Since is continuous and never vanishes on the interval , it has the same sign throughout, so that is strictly monotonic and is twice continuously differentiable. Using the chain rule, the conditions of section 11.1 for the parametrisation by carry over to the parametrisation by ; we omit the verification. The derivative of the inverse function is for . This gives
Since is fixed for fixed , is a constant and taking variances under gives , as claimed. ∎
For the exponential distribution, reparametrising from the rate to the mean changes the information from example 11.8 to ; 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.
Let be uniformly distributed on , i.e. for . The support of the distribution depends on , but we can still compute the derivatives on the support to see whether the identities hold. For we have and thus and . The score is a negative constant and thus we have , in contradiction to lemma 11.1. The variance is zero, whereas and : 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 differ by the term 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 , decreases like , whereas the lower bound of lecture 13, where it applies, does not allow the variance of an unbiased estimator to decrease faster than ; and the maximum likelihood estimator is not approximately normal for large , in contrast to lecture 17.
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Under the conditions of section 11.1 (a support free of , 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 is the variance of the score, and it equals and , the expected curvature of the log-likelihood at the truth.
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Information from independent observations can be combined, ; for an exponential family we have ; the information depends on the parametrisation, and under we get .
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We have learned the following standard values: for the Bernoulli distribution, for the Poisson distribution, for the mean of the normal distribution and 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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Let be i.i.d. Bernoulli with success probability , and let be the score of a single observation from example 11.5.
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Verify lemma 11.1 for this model by computing directly from the two possible values of .
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Compute directly, and check that it agrees with the value of found in example 11.5.
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Compare with the variance of the estimator for .
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Now let be a single observation from the binomial distribution with known number of trials and success probability . Show that the information of about is , and explain the relation to the Bernoulli case using proposition 11.4.
Let be i.i.d. with density for , where , the gamma distribution with known shape and scale from exercise 5.2, i.e. in the rate parametrisation from table A.2.
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Compute the score of a single observation and verify that it has mean zero.
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Show that , once via and once via the variance of the score, using .
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In exercise 5.2 we found that the MLE is unbiased with . Compare this with .
Let be i.i.d. from the geometric distribution with parameter , i.e. for , where and , as in exercises 4.2 and 5.1.
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Compute the score of a single observation and verify that it has mean zero.
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Show that and write down .
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Verify the value from the previous part by taking the variance of the score.
Let be i.i.d. Poisson with mean .