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 : the score will be a vector, the information will be a 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 . The data are again i.i.d. with density or probability weights and the log-likelihood function is , where , but is now a vector. Derivatives with respect to are now partial derivatives, denoted by for the derivative with respect to . We assume the conditions of section 11.1 with partial derivatives in place of , including the mixed derivatives , and we assume that each coordinate of the score has finite variance under .
The score vector of the sample is given by the gradient of the log-likelihood function
as a column vector in .
For the score vector is the score from definition 4.4. As in lecture 11, we consider the score vector as a random vector under . The covariance matrix of a random vector is the matrix with entries , and its diagonal contains the variances.
Given the conditions of section 11.1, the Fisher information matrix of the sample about is given by
The information matrix for a single observation is denoted by .
The diagonal element is the Fisher information about 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.
With the assumptions of section 11.1 in place, the score has mean , and we have
Thus, 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 . In outline, the ratio has mean zero since the density in the numerator and denominator cancels. Using the quotient rule we find 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 is positive semi-definite, if for all , and positive definite, if for all ; a positive definite matrix has a positive definite inverse.
Suppose the conditions from section 11.1 hold. Then the following statements hold.
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The matrix is symmetric and we have for all . In particular, is positive semi-definite.
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Information is additive: .
For the first statement, the symmetry of the matrix follows from the fact that for all and . For , the variance of the linear combination is given by
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 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 in some direction, for example for the model . From now on we assume that 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, is the same as . For symmetric matrices and we use the second form as the definition: we think of a matrix as larger than , if is positive definite, and as at least as large as , if is positive semi-definite. Thus is at least as large as exactly when for every ; unlike for numbers, two matrices may not be comparable at all.
Suppose we have the conditions of section 11.1 and that the matrix is positive definite. Let be an unbiased estimator for , so that for all and all , where the variances are finite and each satisfies (13.1) in every component of . Then the matrix
is positive semi-definite for all . In particular we have
for and for all .
The argument is the proof of theorem 13.2, applied to linear combinations. If we hold all coordinates of other than fixed, lemma 13.1 applies to the one-parameter model with score and gives , which is for and otherwise. Thus, by bilinearity, we have for all . Using the Cauchy–Schwarz inequality, lemma A.8, and the first statement of proposition 14.4, we find
Choosing , both and equal , which is positive for , and dividing by this value gives . This is the claim, since , and for the diagonal case we have , with the th unit vector. This completes the proof. ∎
The same argument gives a bound for a function of the parameter. Let be an unbiased estimator for with finite variance which satisfies (13.1) in every component of . Then lemma 13.1 gives , and the proof above with replaced by and replaced by shows
For this is theorem 13.2. The case 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.
Let be i.i.d. where , both unknown. From example 5.4 we know that the log-likelihood function is
and the first partial derivatives are given by
Taking second derivatives, where we treat as a single variable, we find the three second derivatives as
To compute the expectations, we use the fact that and . The mixed derivative has expectation zero. The last term has expectation . Thus, by proposition 14.3, we find the information matrix and its inverse to be
The upper left entry is the information about the mean, where the variance is known from example 11.7. The sample mean, with variance , achieves the bound for . For the variance the bound is . The UMVUE from example 10.8 has variance , as shown in the computation of section 2.3, and thus does not achieve the bound. Since is the UMVUE, no unbiased estimator for achieves the bound when 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 only, and that the remaining components are nuisance parameters, i.e. unknown but not of interest. If the values of the nuisance parameters were known, would be the only parameter and the Cramér–Rao bound for would be . Since the values are unknown, the bound is by proposition 14.5. The following lemma compares the two bounds.
For every we have for . For , write
Then we have
is the correlation between the two components of the score vector, and .
For the first statement, we apply the Cauchy–Schwarz inequality from lemma A.8 to the two random variables and . By the first statement of proposition 14.4, these have variances and and covariance , and thus we find . For the inverse matrix is
where since the matrix is positive definite. Dividing numerator and denominator of by gives the claimed formula. Since and are the variances of the two components of the score vector and is their covariance, is their correlation, and gives . This completes the proof. ∎
Thus, not knowing the other parameters can never decrease the bound. For the bound for with unknown is times the bound with known, and the two bounds agree if and only if . This case has a name.
The two components of a parameter are called orthogonal, if for all .
By the additivity of information, the condition does not depend on the sample size. For orthogonal parameters, not knowing does not increase the bound for , and vice versa. We consider one example of each case.
For the normal distribution with , the information matrix (14.1) is diagonal, and thus and are orthogonal. The bound for is whether or not is known, and the bound for is whether or not is known. Equal bounds do not mean that not knowing the nuisance parameter is harmless, though: exercise 13.5 shows that the bound for is attained when is known, while example 14.6 shows that it is not attained when is unknown.
Orthogonality is a special feature of the normal distribution, not the rule, as the following example shows.
Consider the gamma distribution with shape and rate both unknown, as in exercise 7.1. The log-likelihood of one observation has second derivatives , and with respect to twice, to and , and to twice, where denotes the trigamma function, the second derivative of . Since these are non-random, the information matrix of one observation is
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
For we use, without proof, the series . With , this gives : the bound for with unknown is about four and a half times the bound with 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 .
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For a vector parameter the score is the gradient , and the Fisher information matrix is its covariance matrix, equal to minus the expected Hessian of the log-likelihood.
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The information matrix is symmetric and positive semi-definite, and adds over independent observations, .
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By the Cramér–Rao inequality, for an unbiased estimator for the matrix is positive semi-definite; in particular .
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For with both parameters unknown, with inverse .
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The diagonal of the inverse dominates the reciprocal of the diagonal, with equality for orthogonal parameters: nuisance parameters inflate the bound by in the two-parameter case, which is one for the normal distribution and about for the gamma distribution with shape .
Let be i.i.d. and consider the parametrisation , where the standard deviation is used.
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Write down the log-likelihood and compute its first partial derivatives.
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By taking derivatives a second time and taking expectations, show that the information matrix for is .
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Write down the Cramér–Rao bound for an unbiased estimator for . Using Jensen’s inequality, lemma A.7, show that the sample standard deviation is not unbiased for and thus cannot be compared to directly.
Consider the gamma distribution with shape and rate . The log-likelihood of observing one value is from exercise 7.1, and the information matrix is given in section 14.4.
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Compute the first and second partial derivatives of and check the matrix.
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Assume that the shape parameter is known to be . In this case, is the only parameter. From the matrix derive the Fisher information about . Why is this a diagonal element of and not of the inverse matrix?
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Now assume that both parameters are unknown. Show that the Cramér–Rao bound for an unbiased estimator for from a sample of size is given by
and determine the ratio of this bound to the bound for known shape parameter .
Let be i.i.d. with distribution and, independently of these, let be i.i.d. with distribution , where the common variance is known and is unknown. The score of all observations is the sum of the scores of the two samples, and the results of this lecture carry over with replaced by the covariance matrix of this score.
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Write down the log-likelihood of all observations and show that the information matrix is .
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Using proposition 14.5, find the Cramér–Rao bound for an unbiased estimator for the difference . Show that achieves this bound.
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Assume now that is also unknown, i.e. when we have . Show that the information matrix is still diagonal and discuss the implications for the estimation of when is unknown.