# An Introduction to Fisher Information

Posted on September 6, 2021

A common question among statisticians and data analysts is how accurately we can estimate the parameters of a distribution given some samples from it. Fisher information can help answer this question by quantifying the amount of information that the samples contain about the unknown parameters of the distribution.

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Suppose we have samples from a distribution where the form is known but the parameters are not. For example, we might know the distribution is a Gaussian, but we don’t know the value of the mean or variance. We want to know how difficult the parameters are to estimate given the samples. One way to answer this question is to estimate the amount of information that the samples contain about the parameters. The more information the samples contain about the parameters, the easier they are to estimate. Conversely, the less information the samples contain about the parameters, the harder they are to estimate.

Fisher information is one way to measure how much information the samples contain about the parameters. There are alternatives, but Fisher information is the most well known. Before we get to the formal definition, which takes some time to get familiar with, let’s motivate Fisher information with an example.

Let’s say we have a sample from a Gaussian distribution with a mean $\mu$ and variance $\sigma^2$. The variance $\sigma^2$ is known, and the goal is to estimate the mean, $\mu$. To understand how difficult this is, we would like to know how much information we can expect the sample to contain about $\mu$. Figure 1 shows three Gaussian distributions with three different variances. We expect that the mean is easiest to estimate for the Gaussian distribution with the smallest variance. A random sample is more likely to be close to the mean when the variance is small than when the variance is large. This implies that the information content should grow as $\sigma$ shrinks.

Let’s $\mathcal{I}_x(\mu)$ to represent the information content of a sample $x$ at the mean $\mu$. In the case of the Gaussian, we might expect the information content of the sample to be inversely proportional to the variance, $\mathcal{I}_x(\mu) \propto 1 / \sigma^2$.

Notice also that the information content is a function of $\mu$, the parameter we want to estimate. Different values of the parameter could be easier or harder to estimate. The Gaussian distribution shifts based on $\mu$ but is otherwise unchanged, so the information content should only depend on $\sigma$ and not on $\mu$.

Another important point is that $x$ is a random sample. We don’t want to specify a value for $x$. Instead, we’d like the information content of $x$ to consider all possible values for $x$ and their corresponding probabilities. A value for $x$ which might tell us a lot about the parameter but is exceedingly unlikely shouldn’t contribute much to the expected information content of the sample. Taking an expectation over $x$ is a natural way to account for this.

The Fisher information can be expressed in multiple ways, none of which are easy to interpret at a glance. Let’s start with one of these definitions and follow it with an explanation. Assume $x$ is a random variable sampled from a distribution $p(x \mid \theta)$ where $\theta$ is an unknown scalar parameter.

Definition (Fisher information). The Fisher information content of $x$ at $\theta$ is:

$$\mathcal{I}_x(\theta) = \mathbb{E} \left[ \left(\frac{d}{d \theta} \log p(x \mid \theta)\right)^2 \right],$$ where the expectation is taken over $x$.

The expectation is an integral if $x$ is continuous: $\mathcal{I}_x (\theta) = \int_x p(x \mid \theta) \left(\frac{d}{d \theta} \log p(x \mid \theta)\right)^2 \, dx,$ and a sum if $x$ is discrete. The distribution $p(x \mid \theta)$ when viewed as a function of $\theta$ is the likelihood function, and $\log p(x \mid \theta)$ is the log-likelihood. To simplify notation, let’s use $\ell(\theta \mid x)$ to represent the log-likelihood of the parameter $\theta$ given the sample $x$. The derivative of the log-likelihood with respect to the parameter, $\ell^\prime(\theta \mid x)$, is called the score function. Using this terminology, the Fisher information is the expected value of the square of the score function: $\mathcal{I}_x(\theta) = \mathbb{E} \left[ \ell^\prime(\theta \mid x) ^2 \right].$

The Fisher information attempts to quantify the sensitivity of the random variable $x$ to the value of the parameter $\theta$. If small changes in $\theta$ result in large changes in the likely values of $x$, then the samples we observe tell us a lot about $\theta$. In this case the Fisher information should be high. This idea agrees with our interpretation of the Gaussian distributions in figure 1. A low variance means we will see large changes in the observed $x$ with small changes in the mean. In this case the Fisher information of $x$ about the mean $\mu$ is large.

Figure 2 plots an example of a log-likelihood function, $\ell(\theta \mid x)$, for a single value of $x$. The curve highlighted by the dashed line is a region where the log-likelihood changes rapidly as $\theta$ changes. This likely corresponds to a region of high Fisher information. The part of the curve highlighted by the dotted line barely changes as a function of $\theta$. This likely corresponds to a region of low Fisher information. I say “likely” because the Fisher information is an expectation over all values of $x$ and figure 2 only shows the log-likelihood for a single value of $x$.

### Gaussian Distribution

As mentioned earlier, the log-likelihood in figure 2 is for a single value of $x$. To compute the Fisher information, we need to consider the log-likelihood for many values of $x$. Let’s illustrate this with the example of a Gaussian distribution.

As before, let’s say we have a random sample, $x$, from a Gaussian distribution for which we would like to compute the Fisher information at the unknown mean, $\mu$. Figure 3 shows from left to right the construction of the term inside the expectation in equation 1 used to compute the Fisher information. The probability distribution is a zero-mean, unit-variance Gaussian distribution ($\mu = 0$ and $\sigma = 1$). Figure 3c shows the derivative with respect to $\mu$ of the log-likelihood but as a function of $x$. Figure 3d shows the square of this derivative as a function of $x$. The Fisher information is computed by taking the expectation over $x$ of the curve in figure 3d. In other words, we multiply the curve in figure 3a with the curve in figure 3d and integrate the result.

The log of the Gaussian distribution is: $\ell(\mu \mid x, \sigma) = \log p(x \mid \mu, \sigma) = -\left(\log (\sqrt{2\pi} \sigma) + \frac{1}{2\sigma^2}(x - \mu)^2\right),$ and its derivative with respect to $\mu$, the score function, is: $\ell^\prime(\mu \mid x, \sigma) = \frac{d}{d\mu} \log p(x \mid \mu, \sigma) = \frac{1}{\sigma^2}(x - \mu).$ This derivative is shown in figure 3c but as a function of $x$. To visualize this derivative, we can plot the log-likelihood. The likelihood function of a Gaussian is also a Gaussian since the function is symmetric in $x$ and $\mu$. Figure 4a shows the log-likelihood for three values of $x$. For each value of $x$, a tangent to the slope of the curve at $\mu=0$ (i.e. the derivative plotted in figure 3c) is shown. The derivative of the log-likelihood with respect to $\mu$ but as a function of $x$ is reproduced in figure 4b. The slopes of the tangents in figure 4a correspond to the values of the points in figure 4b.

The Fisher information of the Gaussian at $\mu$ is the expected value of the squared score function: $\mathcal{I}_x(\mu) = \mathbb{E}\left[\left(\frac{1}{\sigma^2}(x - \mu)\right)^2\right] = \frac{1}{\sigma^4} \mathbb{E}\left[(x - \mu)^2\right] = \frac{\sigma^2}{\sigma^4} = \frac{1}{\sigma^2}.$ As expected, the Fisher information is inversely proportional to the variance.

### Bernoulli Distribution

Let’s use the Bernoulli distribution as another example. The Bernoulli distribution is that of a biased coin which has probability $\theta$ of turning up heads (or $1$) and probability $1-\theta$ of turning up tails (or $0$). We should expect that the more biased the coin, the easier it is to identify the bias from an observation of the coin toss. As an extreme example, if $\theta$ is $1$ or $0$, then a single coin toss will tell us the value of $\theta$. The Fisher information of the sample $x$ (the result of the coin toss) will be higher the closer $\theta$ is to either $1$ or $0$.

The Bernoulli distribution $p(x \mid \theta)$ is plotted as a function of the parameter $\theta$ in figure 5a. The two values of the distribution as a function of $\theta$ are $p(x=1 \mid \theta) = \theta$ and $p(x=0 \mid \theta) = 1 -\theta$. The log-likelihood for each value of $x$ is plotted as a function of $\theta$ in figure 5b, and the score function is plotted in figure 5c. The derivatives are: $\frac{d}{d \theta} \log p(x=1 \mid \theta) = \frac{1}{\theta} \quad \textrm{and} \quad \frac{d}{d \theta} \log p(x=0 \mid \theta) = \frac{1}{\theta - 1}.$ To get the Fisher information, shown in figure 5d, we take the expectation over $x$ of the squared derivatives: $\mathcal{I}_x(\theta) = \theta \frac{1}{\theta^2} + (1-\theta) \frac{1}{(\theta-1)^2} = \frac{1}{\theta} + \frac{1}{1 - \theta} = \frac{1}{\theta (1 - \theta)}.$ The Fisher information in figure 5d has the shape we expect. As $\theta$ approaches $0$ or $1$, the Fisher information grows rapidly. Just as in the Gaussian distribution, the Fisher information is inversely proportional to the variance of the Bernoulli distribution which is $\textrm{Var}(x) = \theta (1-\theta)$. The smaller the variance, the more we expect the sample of $x$ to tell us about the parameter $\theta$ and hence the higher the Fisher information.

### Properties of Fisher Information

The Fisher information has several properties which make it easier to work with. I’ll mention two of the more salient ones here — the chain rule and the post-processing inequality.

Chain rule. Analogous to the chain rule of probability, the Fisher information obeys an additive chain rule: $\mathcal{I}_{x, y}(\theta) = \mathcal{I}_{x \mid y}(\theta) + \mathcal{I}_y(\theta).$ The conditional Fisher information is defined as: $\mathcal{I}_{x \mid y}(\theta) = \int_{y^\prime} p(y^\prime) \, \mathcal{I}_{x \mid y=y^\prime}(\theta) \, d y^\prime = \int_{y^\prime} p(y^\prime) \, \mathbb{E}\left[\left(\frac{d}{d \theta} \log p(x \mid y=y^\prime, \theta)\right)^2 \right] \, d y^\prime.$ When the samples $x$ and $y$ are independent, the chain rule simplifies to: $\mathcal{I}_{x, y}(\theta) = \mathcal{I}_{x}(\theta) + \mathcal{I}_y(\theta).$ So if we have $x_1, \ldots, x_n$ independent and identically distributed samples, the Fisher information for all $n$ samples simplifies to $n$ times the Fisher information of a single sample: $\mathcal{I}_{x_1, \ldots, x_n}(\theta) = \sum_{i=1}^n \mathcal{I}_{x_i}(\theta) = n \, \mathcal{I}_{x_1}(\theta).$

Post-processing. The Fisher information obeys a data processing inequality. The Fisher information of $x$ at $\theta$ cannot be increased by applying any function to $x$. If $f(x)$ is an arbitrary function of $x$, then: $\mathcal{I}_{f(x)}(\theta) \le \mathcal{I}_{x}(\theta).$ The inequality holds with equality when $f(x)$ is a sufficient statistic for $\theta$. A statistic is sufficient for $\theta$ if $\theta$ does not change the conditional probability of $x$ given the statistic: $p(x~\mid~f(x),~\theta)~ =~p(x~\mid~f(x)).$

### Alternate Definitions

Fisher information can be expressed in two other equivalent forms. The first form is: $\mathcal{I}_x(\theta) = -\mathbb{E} \left[\ell^{\prime\prime}(\theta \mid x) \right],$ and the second form is: $\mathcal{I}_x(\theta) = \textrm{Var}\left(\ell^\prime (\theta \mid x) \right).$ A reason to know about these alternate definitions is that in some cases they can be easier to compute than the version in equation 1. To show these are equivalent to the definition in equation 1, we need a couple of observations.

Observation 1. This observation is sometimes called the log-derivative trick in reinforcement learning algorithms. Using the chain rule of differentiation: $\ell^\prime(\theta \mid x) = \frac{d}{d\theta} \log p(x \mid \theta) = \frac{1}{p(x \mid \theta)} \frac{d}{d \theta} p(x \mid \theta).$

Observation 2. The expected value of the score function is zero. That is $\mathbb{E}\left[\ell^\prime(\theta \mid x)\right] = 0$. We can show this using the log-derivative trick from observation 1: $\begin{equation*} \begin{split} \mathbb{E}\left[\ell^\prime(\theta \mid x)\right] &= \mathbb{E}\left[\frac{d}{d\theta} \log p(x \mid \theta) \right] \cr &= \mathbb{E}\left[\frac{1}{p(x\mid \theta)}\frac{d}{d\theta} p(x \mid \theta) \right] \cr &= \int_x p(x \mid \theta) \frac{1}{p(x\mid \theta)}\frac{d}{d\theta} p(x \mid \theta) \, dx \cr &= \int_x \frac{d}{d\theta} p(x \mid \theta) \, dx \cr &= \frac{d}{d\theta} \int_x p(x \mid \theta) \, dx = \frac{d}{d\theta} 1 = 0. \end{split} \end{equation*}$ In the last step above (and in the rest of this tutorial) we assume the derivative and integral can be exchanged.

To show $\mathcal{I}_x(\theta) = -\mathbb{E} \left[\ell^{\prime\prime}(\theta \mid x)\right]$, we start by expanding the $\ell^{\prime\prime}(\theta \mid x)$ using observation 1 and then apply the product rule of differentiation: \begin{align*} \ell^{\prime \prime}(\theta \mid x) &= \frac{d}{d\theta} \frac{d}{d\theta} \log p(x \mid \theta) \cr &= \frac{d}{d \theta} \left( \frac{1}{p(x \mid \theta) } \frac{d}{d \theta} p(x \mid \theta) \right) \cr &= -\frac{1}{p(x \mid \theta)^2 } \left(\frac{d}{d \theta} p(x \mid \theta)\right)^2 + \frac{1}{p(x \mid \theta) } \frac{d^2}{d \theta^2} p(x \mid \theta) \cr &= -\ell^\prime(\theta \mid x)^2 + \frac{1}{p(x \mid \theta) } \frac{d^2}{d \theta^2} p(x \mid \theta), \end{align*} where we use observation 1 again in the last step. The next step is to take the expectation over $x$: $\mathbb{E} \left[\ell^{\prime \prime}(\theta \mid x) \right] = -\mathbb{E} \left[\ell^\prime(\theta \mid x)^2 \right] + \mathbb{E} \left[ \frac{1}{p(x \mid \theta) } \frac{d^2}{d \theta^2} p(x \mid \theta) \right].$ The first term is on the right is the negative of the Fisher information. The second term on the right is zero: \begin{align*} \mathbb{E} \left[ \frac{1}{p(x \mid \theta) } \frac{d^2}{d \theta^2} p(x \mid \theta) \right] &= \int_x p(x \mid \theta) \frac{1}{p(x \mid \theta) } \frac{d^2}{d \theta^2} p(x \mid \theta)\, dx \cr &= \int_x \frac{d^2}{d \theta^2} p(x \mid \theta) \, dx \cr &= \frac{d^2}{d \theta^2} \int_x p(x \mid \theta) \, dx = \frac{d^2}{d \theta^2} 1 = 0. \end{align*} Thus we have the result that $\mathcal{I}_x(\theta) = -\mathbb{E} \left[\ell^{\prime \prime}(\theta \mid x) \right]$.

The statement $\mathcal{I}_x(\theta) = \textrm{Var}\left(\ell^\prime (\theta \mid x) \right)$, follows directly from the fact that the expected value of the score function is $0$ (observation 2): $\textrm{Var}\left(\ell^\prime(\theta \mid x)\right) = \mathbb{E}\left[\ell^\prime(\theta \mid x)^2 \right] - \mathbb{E}\left[ \ell^\prime(\theta \mid x) \right]^2 = \mathbb{E}\left[ \ell^\prime(\theta \mid x)^2 \right] = \mathcal{I}_x(\theta).$

### Multivariate Fisher Information

The definition of Fisher information can be extended to include multiple parameters. In this case, the parameters of the distribution are now a $d$-dimensional vector, $\theta \in \mathbb{R}^d$. The multivariate first-order generalization of the Fisher information is: $\mathcal{I}_x(\theta) = \mathbb{E}\left[\nabla_\theta \ell(\theta \mid x) \nabla_\theta \ell(\theta \mid x)^\top\right],$ where $\nabla_\theta$ is the gradient operator which produces the vector of first derivatives of $\ell(\theta \mid x)$ with respect to $\theta$. Note in this case the Fisher information is a symmetric matrix in $\mathbb{R}^{d \times d}$. The diagonal entries of the matrix $\mathcal{I}_x(\theta)_{ii}$ have the usual interpretation. The higher these entries are, the more information $x$ contains about the $i$-th parameter, $\theta_i$. The off-diagonal entries are somewhat more subtle to interpret. Roughly speaking, if $\mathcal{I}_x(\theta)_{ij}$ is high then $x$ contains more information about the relationship between $\theta_i$ and $\theta_j$.

The multivariate second-order expression for the Fisher information is also a natural extension of the scalar version: $\mathcal{I}_x(\theta) = -\mathbb{E}\left[ \nabla^2_\theta \ell(\theta \mid x) \right],$ where $\nabla^2_\theta$ is the operator which produces the matrix of second derivatives of the log-likelihood with respect to $\theta$.

### Cramér-Rao Bound

An estimator for a parameter of a distribution is a function which takes as input the sample and returns an estimate for the parameter. Let’s use $\hat{\theta}(x)$ to represent an estimator for the parameter $\theta$. The estimator $\hat{\theta}(x)$ is unbiased if its expected value is equal to the parameter $\theta$: $\mathbb{E}\left[\hat{\theta}(x) \right] = \theta.$ The Cramér-Rao bound is an inequality which relates the variance of an estimator of a parameter $\theta$ to the Fisher information of a sample $x$ at $\theta$. If $x$ contains less information about $\theta$, then we expect $\theta$ to be harder to estimate given $x$. The Cramér-Rao bound makes this statement precise. In the simplest case, if $\hat{\theta}(x)$ is an unbiased estimator of $\theta$ given $x$, the Cramér-Rao bound states: $\textrm{Var}\left(\hat{\theta}(x)\right) \ge \frac{1}{\mathcal{I}_x(\theta)}.$ One way to think of the Cramér-Rao bound is as a two-player game say between Alice and Bob:

1. Alice chooses the parameter $\theta$.
2. Alice samples $x \sim p(x \mid \theta)$ and sends $x$ to Bob.
3. Bob estimates $\hat{\theta}(x)$.

The Cramér-Rao bound says that on average the squared difference between Bob’s estimate and the true value of the parameter will be greater than $1 / \mathcal{I}_x(\theta)$.

The proof of the Cramér-Rao bound is only a few lines. First, since the expected value of the score function is zero (observation 2), we have: $\textrm{Cov}\left(\hat{\theta}(x), \ell^\prime(\theta \mid x) \right) = \mathbb{E} \left[\hat{\theta}(x)\ell^\prime(\theta \mid x) \right] - \mathbb{E} \left[\hat{\theta}(x)\right]\mathbb{E}\left[\ell^\prime(\theta \mid x) \right] = \mathbb{E} \left[\hat{\theta}(x) \ell^\prime(\theta \mid x) \right].$ Using the log-derivative trick again (observation 1): \begin{align*} \mathbb{E} \left[\hat{\theta}(x) \ell^\prime(\theta \mid x) \right] &= \int_x p(x\mid \theta) \hat{\theta}(x) \frac{\partial}{\partial \theta} \log p(x \mid \theta) d\,x \cr &= \int_x p(x\mid \theta) \hat{\theta}(x) \frac{1}{p(x\mid \theta)} \frac{\partial}{\partial \theta} p(x \mid \theta) d\,x \cr &= \frac{\partial}{\partial \theta} \int_x \hat{\theta}(x) p(x \mid \theta) d\,x \cr &= \frac{\partial}{\partial \theta} \mathbb{E} \left[\hat{\theta}(x)\right] \cr &= \frac{\partial}{\partial \theta} \theta = 1, \end{align*} where in the second-to-last step we use the fact that the estimator is unbiased. This tells us that the covariance is one: $\textrm{Cov}\left(\hat{\theta}(x), \ell^\prime(\theta \mid x) \right) = 1.$ Combining this with the Cauchy-Schwarz inequality we have: $\textrm{Var}\left(\hat{\theta}(x)\right) \textrm{Var}\left(\ell^\prime(\theta \mid x) \right) \ge \textrm{Cov}\left(\hat{\theta}(x), \ell^\prime(\theta \mid x)\right)^2 = 1,$ hence: $\textrm{Var}\left(\hat{\theta}(x)\right) \ge \frac{1}{\textrm{Var}\left(\ell^\prime(\theta \mid x) \right)} = \frac{1}{\mathcal{I}_x(\theta)}.$

### More Uses of Fisher Information

The Fisher information has applications beyond quantifying the difficulty in estimating parameters of a distribution given samples from it. I’ll briefly discuss two such applications: natural gradient descent and data privacy.

Natural gradient descent. Fisher information is used to compute the natural gradient used in numerical optimization. Natural gradient descent1 is not commonly used directly in large machine-learning problems due to computational difficulties, but it motivates some more commonly used optimization methods.

One way to view standard gradient descent is that it searches for the best update within a small region around the current parameters. This region is defined by the standard Euclidean distance to the existing parameters. Natural gradient descent is the same idea, but instead of defining the region with Euclidean distance it defines the region using the Kullback-Liebler (KL) divergence. In this case the KL divergence is used to measure the distance between the likelihood function at the current parameters and the likelihood function at the updated parameters.

Natural gradient descent looks similar to Newton’s method. It replaces the inverse Hessian with the inverse of the expected Hessian, which is the same as the inverse of the Fisher information matrix. The update is: $\mathcal{I}_(\theta)^{-1} \nabla_\theta \mathcal{L}(\theta),$ where $\mathcal{L}$ is the likelihood function.

Data privacy. One relatively recent use for Fisher information (which is an area I am working on2) is to use Fisher information as a tool for data privacy. We can invert the role of the samples and the parameters and measure the Fisher information of the parameters at the samples. The parameters in this case could be a machine-learning model and the samples are data from different individuals on which the model was trained. The Fisher information of the model about the data then quantifies the privacy loss for different individuals when revealing the model.

### Footnotes

1. See for example Shun-ichi Amari, Natural Gradient Works Efficiently in Learning, Neural Computation, 1998. (link)

2. Our recent research on this is detailed in Hannun, et al., Measuring Data Leakage in Machine-Learning Models with Fisher Information, Uncertainty in Artificial Intelligence, 2021. (link)