Statement

The Cramér–Rao bound is stated in this section for several increasingly general cases, beginning with the case in which the parameter is a scalar and its estimator is unbiased. All versions of the bound require certain regularity conditions, which hold for most well-behaved distributions. These conditions are listed later in this section.

Scalar unbiased case

Suppose \theta is an unknown deterministic parameter that is to be estimated from n independent observations (measurements) of x, each from a distribution according to some probability density function f(x;\theta). The variance of any unbiased estimator \hat{\theta} of \theta is then bounded{{cite book | first=Frank | last= Nielsen | title= Connected at Infinity II | chapter= Cramér–Rao Lower Bound and Information Geometry | series= Texts and Readings in Mathematics | publisher= Hindustan Book Agency, Gurgaon page=18-37| doi= 10.1007/978-93-86279-56-9_2 | arxiv= 1301.3578 | isbn= 978-93-80250-51-9 | s2cid= 16759683 }} by the reciprocal of the Fisher information I(\theta): :\operatorname{var}(\hat{\theta}) \geq \frac{1}{I(\theta)} where the Fisher information I(\theta) is defined by : I(\theta) = n \operatorname{E}_{X;\theta} \left[ \left( \frac{\partial \ell(X;\theta)}{\partial\theta} \right)^2 \right]

and \ell(x;\theta)=\log (f(x;\theta)) is the natural logarithm of the likelihood function for a single sample x and \operatorname{E}_{x;\theta} denotes the expected value with respect to the density f(x;\theta) of X. If not indicated, in what follows, the expectation is taken with respect to X.

If \ell(x;\theta) is twice differentiable and certain regularity conditions hold, then the Fisher information can also be defined as follows:

: I(\theta) = -n \operatorname{E}_{X;\theta}\left[ \frac{\partial^2 \ell(X;\theta)}{\partial\theta^2} \right]

The efficiency of an unbiased estimator \hat{\theta} measures how close this estimator's variance comes to this lower bound; estimator efficiency is defined as

:e(\hat{\theta}) = \frac{I(\theta)^{-1}}{\operatorname{var}(\hat{\theta})}

or the minimum possible variance for an unbiased estimator divided by its actual variance. The Cramér–Rao lower bound thus gives :e(\hat{\theta}) \le 1.

General scalar case

A more general form of the bound can be obtained by considering a biased estimator T(X), whose expectation is not \theta but a function of this parameter, say, \psi(\theta). Hence E{T(X)} - \theta = \psi(\theta) - \theta is not generally equal to 0. In this case, the bound is given by : \operatorname{var}(T) \geq \frac{[\psi'(\theta)]^2}{I(\theta)} where \psi'(\theta) is the derivative of \psi(\theta) (by \theta), and I(\theta) is the Fisher information defined above.

Bound on the variance of biased estimators

Apart from being a bound on estimators of functions of the parameter, this approach can be used to derive a bound on the variance of biased estimators with a given bias, as follows. Consider an estimator \hat{\theta} with bias b(\theta) = E{\hat{\theta}} - \theta, and let \psi(\theta) = b(\theta) + \theta. By the result above, any unbiased estimator whose expectation is \psi(\theta) has variance greater than or equal to (\psi'(\theta))^2/I(\theta). Thus, any estimator \hat{\theta} whose bias is given by a function b(\theta) satisfies : \operatorname{var} \left(\hat{\theta}\right) \geq \frac{[1+b'(\theta)]^2}{I(\theta)}. The unbiased version of the bound is a special case of this result, with b(\theta)=0.

It's trivial to have a small variance − an "estimator" that is constant has a variance of zero. But from the above equation, we find that the mean squared error of a biased estimator is bounded by

:\operatorname{E}\left((\hat{\theta}-\theta)^2\right)\geq\frac{[1+b'(\theta)]^2}{I(\theta)}+b(\theta)^2,

using the standard decomposition of the MSE. Note, however, that if 1+b'(\theta) this bound might be less than the unbiased Cramér–Rao bound 1/I(\theta). For instance, in the example of estimating variance below, 1+b'(\theta)= \frac{n}{n+2} .

Multivariate case

Extending the Cramér–Rao bound to multiple parameters, define a parameter column vector :\boldsymbol{\theta} = \left[ \theta_1, \theta_2, \dots, \theta_d \right]^T \in \mathbb{R}^d with probability density function f(x; \boldsymbol{\theta}) which satisfies the two regularity conditions below.

The Fisher information matrix is a d \times d matrix with element I_{m, k} defined as : I_{m, k} = \operatorname{E} \left[ \frac{\partial }{\partial \theta_m} \log f\left(x; \boldsymbol{\theta}\right) \frac{\partial }{\partial \theta_k} \log f\left(x; \boldsymbol{\theta}\right) \right] = -\operatorname{E} \left[ \frac{\partial ^2}{\partial \theta_m , \partial \theta_k} \log f\left(x; \boldsymbol{\theta}\right) \right].

Let \boldsymbol{T}(X) be an estimator of any vector function of parameters, \boldsymbol{T}(X) = (T_1(X), \ldots, T_d(X))^T, and denote its expectation vector \operatorname{E}[\boldsymbol{T}(X)] by \boldsymbol{\psi}(\boldsymbol{\theta}). The Cramér–Rao bound then states that the covariance matrix of \boldsymbol{T}(X) satisfies : I\left(\boldsymbol{\theta}\right) \geq \phi(\theta)^T \operatorname{cov}{\boldsymbol{\theta}}\left(\boldsymbol{T}(X)\right)^{-1}\phi(\theta) , : \operatorname{cov}{\boldsymbol{\theta}}\left(\boldsymbol{T}(X)\right) \geq \phi(\theta) I\left(\boldsymbol{\theta}\right)^{-1} \phi(\theta)^T where

• The matrix inequality A \ge B is understood to mean that the matrix A-B is positive semidefinite, and
• \phi(\theta) := \partial \boldsymbol{\psi}(\boldsymbol{\theta})/\partial \boldsymbol{\theta} is the Jacobian matrix whose ij element is given by \partial \psi_i(\boldsymbol{\theta})/\partial \theta_j.

If \boldsymbol{T}(X) is an unbiased estimator of \boldsymbol{\theta} (i.e., \boldsymbol{\psi}\left(\boldsymbol{\theta}\right) = \boldsymbol{\theta}), then the Cramér–Rao bound reduces to : \operatorname{cov}_{\boldsymbol{\theta}}\left(\boldsymbol{T}(X)\right) \geq I\left(\boldsymbol{\theta}\right)^{-1}.

If it is inconvenient to compute the inverse of the Fisher information matrix, then one can simply take the reciprocal of the corresponding diagonal element to find a (possibly loose) lower bound.

: \operatorname{var}_{\boldsymbol{\theta}}(T_m(X))

\left[\operatorname{cov}{\boldsymbol{\theta}}\left(\boldsymbol{T}(X)\right)\right]{mm} \geq \left[I\left(\boldsymbol{\theta}\right)^{-1}\right]{mm} \geq \frac{1}{I\left(\boldsymbol{\theta}\right){mm}}.

Regularity conditions

The bound relies on two weak regularity conditions on the probability density function, f(x; \theta), and the estimator T(X):

• The Fisher information is always defined; equivalently, for all x such that f(x; \theta) 0, \frac{\partial}{\partial\theta} \log f(x;\theta) exists, and is finite.
• The operations of integration with respect to x and differentiation with respect to \theta can be interchanged in the expectation of T; that is, \frac{\partial}{\partial\theta} \left[ \int T(x) f(x;\theta) ,dx \right] = \int T(x) \left[ \frac{\partial}{\partial\theta} f(x;\theta) \right] ,dx whenever the right-hand side is finite. This condition can often be confirmed by using the fact that integration and differentiation can be swapped when either of the following cases hold:
  1. The function f(x;\theta) has bounded support in x, and the bounds do not depend on \theta;
  2. The function f(x;\theta) has infinite support, is continuously differentiable, and the integral converges uniformly for all \theta.

Proof

Proof for the general case based on the [[Chapman–Robbins bound]]

Proof based on.

First equation:

Let \delta be an infinitesimal, then for any v\in \R^n, plugging \theta' = \theta + \delta v in, we have (E_{\theta'}[T] - E_{\theta}[T]) = v^T \phi(\theta)\delta; \quad \chi^2(\mu_{\theta'} ; \mu_\theta) = v^T I(\theta) v \delta^2

Plugging this into multivariate Chapman–Robbins bound gives I(\theta) \geq \phi(\theta) \operatorname{Cov}_\theta[T]^{-1} \phi(\theta)^T.

Second equation:

It suffices to prove this for scalar case, with h(X) taking values in \R . Because for general T(X) , we can take any v \in \R^m, then defining h:= \sum_j v_j T_j, the scalar case gives \operatorname{Var}\theta[h] = v^T \operatorname{Cov}\theta[T]v \geq v^T \phi(\theta) I(\theta)^{-1}\phi(\theta)^T vThis holds for all v \in \R^m, so we can conclude\operatorname{Cov}\theta[T] \geq \phi(\theta) I(\theta)^{-1}\phi(\theta)^TThe scalar case states that \operatorname{Var}\theta[h] \geq \phi(\theta)^T I(\theta)^{-1} \phi(\theta) with \phi(\theta) := \nabla_\theta E_\theta[h].

Let \delta be an infinitesimal, then for any v\in \R^n, taking \theta' = \theta + \delta v in the single-variate Chapman–Robbins bound gives \operatorname{Var}_\theta[h] \geq \frac{\langle v, \phi(\theta)\rangle^2}{v^T I(\theta) v}.

By linear algebra, \sup_{v\neq 0} \frac{\langle w, v\rangle^2}{v^T M v} = w^T M^{-1}w for any positive-definite matrix M, thus we obtain \operatorname{Var}_\theta[h] \geq \phi(\theta)^T I(\theta)^{-1} \phi(\theta).

A standalone proof for the general scalar case

For the general scalar case:

Assume that T=t(X) is an estimator with expectation \psi(\theta) (based on the observations X), i.e. that \operatorname{E}(T) = \psi (\theta). The goal is to prove that, for all \theta, :\operatorname{var}(t(X)) \geq \frac{[\psi^\prime(\theta)]^2}{I(\theta)}.

Let X be a random variable with probability density function f(x; \theta). Here T = t(X) is a statistic, which is used as an estimator for \psi (\theta). Define V as the score:

:V = \frac{\partial}{\partial\theta} \ln f(X;\theta) = \frac{1}{f(X;\theta)}\frac{\partial}{\partial\theta}f(X;\theta)

where the chain rule is used in the final equality above. Then the expectation of V, written \operatorname{E}(V), is zero. This is because:

: \operatorname{E}(V) = \int f(x;\theta)\left[\frac{1}{f(x;\theta)}\frac{\partial }{\partial \theta} f(x;\theta)\right] , dx = \frac{\partial}{\partial\theta}\int f(x;\theta) , dx = 0

where the integral and partial derivative have been interchanged (justified by the second regularity condition).

If we consider the covariance \operatorname{cov}(V, T) of V and T, we have \operatorname{cov}(V, T) = \operatorname{E}(V T), because \operatorname{E}(V) = 0. Expanding this expression we have

: \begin{align} \operatorname{cov}(V,T) & = \operatorname{E} \left( T \cdot\left[\frac{1}{f(X;\theta)}\frac{\partial}{\partial\theta}f(X;\theta) \right] \right) \[6pt] & = \int t(x) \left[\frac{1}{f(x;\theta)} \frac{\partial}{\partial\theta} f(x;\theta) \right] f(x;\theta), dx \[6pt] & = \frac{\partial}{\partial\theta} \left[ \int t(x) f(x;\theta),dx \right] = \frac{\partial}{\partial\theta} E(T) = \psi^\prime(\theta) \end{align}

again because the integration and differentiation operations commute (second condition).

The Cauchy–Schwarz inequality shows that

: \sqrt{ \operatorname{var} (T) \operatorname{var} (V)} \geq \left| \operatorname{cov}(V,T) \right| = \left | \psi^\prime (\theta) \right |

therefore

: \operatorname{var} (T) \geq \frac{[\psi^\prime(\theta)]^2}{\operatorname{var} (V)} = \frac{[\psi^\prime(\theta)]^2}{I(\theta)} which proves the proposition.

Examples

Multivariate normal distribution

For the case of a d-variate normal distribution : \boldsymbol{x} \sim \mathcal{N}d \left( \boldsymbol{\mu}( \boldsymbol{\theta}) , {\boldsymbol C} ( \boldsymbol{\theta}) \right) the Fisher information matrix has elements{{cite book : I{m, k} = \frac{\partial \boldsymbol{\mu}^T}{\partial \theta_m} {\boldsymbol C}^{-1} \frac{\partial \boldsymbol{\mu}}{\partial \theta_k}

• \frac{1}{2} \operatorname{tr} \left( {\boldsymbol C}^{-1} \frac{\partial {\boldsymbol C}}{\partial \theta_m} {\boldsymbol C}^{-1} \frac{\partial {\boldsymbol C}}{\partial \theta_k} \right) where "tr" is the trace.

For example, let w[j] be a sample of n independent observations with unknown mean \theta and known variance \sigma^2 . :w[j] \sim \mathcal{N}_{d, n} \left(\theta {\boldsymbol 1}, \sigma^2 {\boldsymbol I} \right). Then the Fisher information is a scalar given by

: I(\theta)

\left(\frac{\partial\boldsymbol{\mu}(\theta)}{\partial\theta}\right)^T{\boldsymbol C}^{-1} \left(\frac{\partial\boldsymbol{\mu}(\theta)}{\partial\theta}\right) = \sum^{n}_{i=1} \frac{1}{\sigma^2} = \frac{n}{\sigma^2},

and so the Cramér–Rao bound is

: \operatorname{var}(\hat\theta) \geq \frac{\sigma^2}{n}.

Normal variance with known mean

Suppose X is a normally distributed random variable with known mean \mu and unknown variance \sigma^2. Consider the following statistic:

: T=\frac{\sum_{i=1}^n (X_i-\mu)^2}{n}.

Then T is unbiased for \sigma^2, as E(T)=\sigma^2. What is the variance of T?

: \operatorname{var}(T) = \operatorname{var}\left(\frac{\sum_{i=1}^n(X_i-\mu)^2}{n}\right) = \frac{\sum_{i=1}^n\operatorname{var}(X_i-\mu)^2}{n^2} = \frac{n\operatorname{var}(X-\mu)^2}{n^2}=\frac{1}{n} \left[ \operatorname{E}\left{(X-\mu)^4\right}-\left(\operatorname{E}{(X-\mu)^2}\right)^2 \right]

(the second equality follows directly from the definition of variance and the fact that X_i's are independent). The first term is the fourth moment about the mean and has value 3(\sigma^2)^2; the second is the square of the variance, or (\sigma^2)^2. Thus

:\operatorname{var}(T)=\frac{2(\sigma^2)^2}{n}.

Now, what is the Fisher information in the sample? Recall that the score V is defined as

: V=\frac{\partial}{\partial\sigma^2}\log\left[ L(\sigma^2,X)\right]

where L is the likelihood function. Thus in this case,

: \log\left[L(\sigma^2,X)\right]=\log\left[\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(X-\mu)^2 /{2\sigma^2}}\right] =-\log(\sqrt{2\pi\sigma^2})-\frac{(X-\mu)^2}{2\sigma^2}

: V=\frac{\partial}{\partial\sigma^2}\log \left[ L(\sigma^2,X) \right]=\frac{\partial}{\partial\sigma^2}\left[-\log(\sqrt{2\pi\sigma^2})-\frac{(X-\mu)^2}{2\sigma^2}\right] =-\frac{1}{2\sigma^2}+\frac{(X-\mu)^2}{2(\sigma^2)^2}

where the second equality is from elementary calculus. Thus, the information in a single observation is just minus the expectation of the derivative of V, or

: I =-\operatorname{E}\left(\frac{\partial V}{\partial\sigma^2}\right) =-\operatorname{E}\left(-\frac{(X-\mu)^2}{(\sigma^2)^3}+\frac{1}{2(\sigma^2)^2}\right) =\frac{\sigma^2}{(\sigma^2)^3}-\frac{1}{2(\sigma^2)^2} =\frac{1}{2(\sigma^2)^2}.

Thus the information in a sample of n independent observations is just n times this, or \frac{n}{2(\sigma^2)^2}.

The Cramér–Rao bound states that

: \operatorname{var}(T)\geq\frac{1}{I}.

In this case, the inequality is saturated (equality is achieved), showing that the estimator is efficient.

However, we can achieve a lower mean squared error using a biased estimator. The estimator

: T=\frac{\sum_{i=1}^n (X_i-\mu)^2}{n+2}.

obviously has a smaller variance, which is in fact

:\operatorname{var}(T)=\frac{2n(\sigma^2)^2}{(n+2)^2}.

Its bias is

:\left(1-\frac{n}{n+2}\right)\sigma^2=\frac{2\sigma^2}{n+2}

so its mean squared error is

:\operatorname{MSE}(T)=\left(\frac{2n}{(n+2)^2}+\frac{4}{(n+2)^2}\right)(\sigma^2)^2 =\frac{2(\sigma^2)^2}{n+2}

which is less than what unbiased estimators can achieve according to the Cramér–Rao bound.

When the mean is not known, the minimum mean squared error estimate of the variance of a sample from Gaussian distribution is achieved by dividing by n+1, rather than n-1 or n+2.

References and notes

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