A Derivation of the Linear Fisher Information

In this post, we’ll slog through the derivation of the Fisher information for Gaussian distributed noise while highlighting the linear Fisher information, a component of the end expression that has been particularly important in the computational neuroscience literature.


Suppose we have some scalar parameter $x$ and we transform it to an $N$-dimensional representation $\mathbf{r}$ through a function $\mathbf{f}$:

\begin{align} \mathbf{r} &= \mathbf{f}(x) + \boldsymbol{\epsilon}. \end{align}

The $\boldsymbol{\epsilon}$ term indicates that the transformation is noisy. How easy is it to decode $x$ given $\mathbf{r}$? One way to assess an estimator $\hat{x}(\mathbf{r})$ is to calculate its variance $\text{Var}(\hat{x})$ over possible data points $\mathbf{r}$. If the variance is large, we can’t necessarily trust our decoder in a given instance.

The Fisher information $I_F(x)$ places a lower bound on the variance any estimator of $x$ can possess:

\begin{align} \text{Var}\left[\hat{x}(\mathbf{r})\right] \geq \frac{1}{I_F(x)}. \end{align}

Thus, if we have a small Fisher information, then even the best estimator for $x$ will have a large variance: decoding $x$ from $\mathbf{r}$ will always be hard.

Given a log-likelihood $\log P[\mathbf{r} \vert x]$, there is a mathematical expression for the Fisher information:

\begin{align} I_F(x) &= \mathbb{E}_{\mathbf{r}\vert x}\left[\left(\frac{d}{dx} \log P[\mathbf{r}\vert x]\right)^2\right]. \\
&= \int d\mathbf{r} P[\mathbf{r} \vert x] \left(\frac{d}{dx} \log P[\mathbf{r} \vert x]\right)^2 \end{align}

That is, the Fisher information is the average square of the score function (the derivative of the log-likelihood with respect to the parameter).

In computational neuroscience, equation (1) is a simple encoding scheme modeling neural activity: an incoming stimulus $x$ is transformed into a (noisy) neural representation $\mathbf{r}$. Neural systems probably want to decode the stimulus at some point, so the Fisher information is a quantity of interest. In equation (1), we can analytically determine the Fisher information when the noise $\boldsymbol{\epsilon}$ is Gaussian.

Deriving the Fisher Information

Assume $\boldsymbol{\epsilon}$ is drawn from an $N$-dimensional Gaussian distribution with zero mean and covariance $\boldsymbol{\Sigma}(x)$ (i.e., the covariance is potentially dependent on $x$). Then, the conditional distribution $P[\mathbf{r}\vert x]$ is a simple Gaussian distribution:

\begin{align} P[\mathbf{r}\vert x] &= \frac{1}{\sqrt{(2\pi)^N \det \boldsymbol{\Sigma}(x)}}\exp\left[-\frac{1}{2}(\mathbf{r}-\mathbf{f}(x))(x)^{-1} (\mathbf{r} - \mathbf{f}(x))\right]. \end{align}

First, let’s compute the log-likelihood:

\begin{align} \log P[\mathbf{r}\vert x] &= -\frac{N}{2}\log(2\pi) -\frac{1}{2} \log \det \boldsymbol{\Sigma} \notag \\
&-\frac{1}{2} \left(\mathbf{r} - \mathbf{f}(x)\right)^T \boldsymbol{\Sigma}(x)^{-1} \left(\mathbf{r} - \mathbf{f}(x)\right). \end{align}

We then need the derivative with respect to $x$:

\begin{align} \frac{d}{dx} \log P[\mathbf{r}\vert x] &= 0 - \frac{1}{2}\frac{1}{\det \boldsymbol{\Sigma}(x)} \frac{d}{dx} \det\boldsymbol{\Sigma}(x) \\
&- \frac{1}{2} \frac{d(\mathbf{r} - \mathbf{f}(x))^T}{dx} \boldsymbol{\Sigma}(x)^{-1}(\mathbf{r}-\mathbf{f}(x)) \\
&-\frac{1}{2} (\mathbf{r} - \mathbf{f}(x))^T \frac{d\boldsymbol{\Sigma}(x)^{-1}}{dx} (\mathbf{r}-\mathbf{f}(x)) \\
&-\frac{1}{2} (\mathbf{r} - \mathbf{f}(x))^T \boldsymbol{\Sigma}(x)^{-1} \frac{d(\mathbf{r}-\mathbf{f}(x))}{dx} \end{align}

First, we can evaluate the derivative of a determinant using Jacobi’s formula, which states that

\begin{align} \frac{d}{dx} \det \mathbf{A}(x) &= \det \mathbf{A}(x) \text{ Tr}\left[\mathbf{A}(x)^{-1} \mathbf{A}’(x)\right] \end{align}

where $\mathbf{A}’(x) = \frac{d}{dx}\mathbf{A}(x)$. Thus,

\begin{align} \frac{d}{dx} \log P[\mathbf{r} \vert x] &= -\frac{1}{2}\text{Tr}\left[\boldsymbol{\Sigma}(x)^{-1} \boldsymbol{\Sigma}’(x)\right] +\frac{1}{2} \mathbf{f}’(x)^T \boldsymbol{\Sigma}(x)^{-1} (\mathbf{r}-\mathbf{f}(x)) \\
& \qquad \qquad - \frac{1}{2} (\mathbf{r} - \mathbf{f}(x))^T \boldsymbol{\Sigma}’(x)^{-1} (\mathbf{r}-\mathbf{f}(x)) \\
& \qquad \qquad +\frac{1}{2}(\mathbf{r} - \mathbf{f}(x))^T \boldsymbol{\Sigma}(x)^{-1} \mathbf{f}’(x). \end{align}

In the above expression, the second and fourth terms are equal since they’re both scalars and transposes of each other. As for the derivative of the matrix inverse, we note that

\begin{align} \frac{d}{dx} \boldsymbol{\Sigma}(x)^{-1} &= -\boldsymbol{\Sigma}(x)^{-1} \frac{d\boldsymbol{\Sigma}(x)}{dx} \boldsymbol{\Sigma}(x)^{-1} \end{align}

which can be derived by differentiating the definition of the matrix inverse $\boldsymbol{\Sigma}(x) \boldsymbol{\Sigma}(x)^{-1} = \mathbf{I}$. Thus, we have

\begin{align} \frac{d}{dx} \log P[\mathbf{r}\vert x] &= -\frac{1}{2}\text{Tr}\left[\boldsymbol{\Sigma}(x)^{-1} \boldsymbol{\Sigma}’(x)\right] + \mathbf{f}’(x)^T \boldsymbol{\Sigma}(x)^{-1} (\mathbf{r} - \mathbf{f}(x)) \\
& \qquad + \frac{1}{2} (\mathbf{r} - \mathbf{f}(x))^T\boldsymbol{\Sigma}(x)^{-1} \boldsymbol{\Sigma}’(x) \boldsymbol{\Sigma}(x)^{-1}(\mathbf{r} - \mathbf{f}(x))^T \\
&= -\frac{1}{2}\text{Tr}\left[\boldsymbol{\Sigma}^{-1}\boldsymbol{\Sigma}’\right] + \mathbf{f}’^T \boldsymbol{\Sigma}^{-1} (\mathbf{r} - \mathbf{f}) \notag \\
& \qquad + \frac{1}{2} (\mathbf{r} - \mathbf{f})^T\boldsymbol{\Sigma}^{-1} \boldsymbol{\Sigma}’ \boldsymbol{\Sigma}^{-1}(\mathbf{r} - \mathbf{f}). \end{align}

In the last line, we removed the dependence on $x$ to save space.

We’ll need to square this expression to calculate the Fisher information. This is annoying, but let’s be organized:

\begin{align} \left(\frac{d}{dx} \log P[\mathbf{r}\vert x]\right)^2 &= \frac{1}{4} \text{Tr}\left[\boldsymbol{\Sigma}^{-1}\boldsymbol{\Sigma}’\right]^2 \notag \\
&+ \left[\mathbf{f}’^T \boldsymbol{\Sigma}^{-1} (\mathbf{r} - \mathbf{f})\right]^2 \notag \\
& + \frac{1}{4} \left[(\mathbf{r} - \mathbf{f})^T\boldsymbol{\Sigma}^{-1} \boldsymbol{\Sigma}’ \boldsymbol{\Sigma}^{-1}(\mathbf{r} - \mathbf{f})\right]^2 \notag \\
& -\text{Tr}\left[\boldsymbol{\Sigma}^{-1}\boldsymbol{\Sigma}’\right] \cdot \mathbf{f}’^T \boldsymbol{\Sigma}^{-1} (\mathbf{r} - \mathbf{f})\notag \\
& -\frac{1}{2}\text{Tr}\left[\boldsymbol{\Sigma}^{-1}\boldsymbol{\Sigma}’\right] \cdot (\mathbf{r} - \mathbf{f})^T\boldsymbol{\Sigma}^{-1} \boldsymbol{\Sigma}’ \boldsymbol{\Sigma}^{-1}(\mathbf{r} - \mathbf{f}) \notag \\
& + \mathbf{f}’^T \boldsymbol{\Sigma}^{-1} (\mathbf{r} - \mathbf{f}) \cdot (\mathbf{r} - \mathbf{f})^T\boldsymbol{\Sigma}^{-1} \boldsymbol{\Sigma}’ \boldsymbol{\Sigma}^{-1}(\mathbf{r} - \mathbf{f}) \\
&= E_1 + E_2 + E_3 + E_4 + E_5 + E_6. \label{eqn:addends} \end{align}

The Fisher information is the expectation of this expression over $P[\mathbf{r}\vert x]$. We’ve split up our expression into six addends, which we’ve listed in equation \eqref{eqn:addends}. Furthermore, terms like $\boldsymbol{\Sigma}(x)$ and $\mathbf{f}(x)$ have no $\mathbf{r}$ dependence and therefore are constants. That makes the expectation of the first term easy:

\begin{align} E_1 = \mathbb{E}_{\mathbf{r}\vert x}\left[\frac{1}{4} \text{Tr}\left[\boldsymbol{\Sigma}^{-1}\boldsymbol{\Sigma}’\right]^2\right] &= \frac{1}{4} \text{Tr}\left[\boldsymbol{\Sigma}^{-1}\boldsymbol{\Sigma}’\right]^2 \end{align}

Next, note the that addends with an odd number of $(\mathbf{r}-\mathbf{f})$ terms will vanish, since we are taking an expectation over a Gaussian. For example, the expectation of the fourth term is \begin{align} E_4 &= \mathbb{E}_{\mathbf{r}\vert x}\left[-\text{Tr}\left[\boldsymbol{\Sigma}^{-1}\boldsymbol{\Sigma}’\right] \mathbf{f}’^T \boldsymbol{\Sigma}^{-1} (\mathbf{r} - \mathbf{f})\right] \\
&= -\text{Tr}\left[\boldsymbol{\Sigma}^{-1}\boldsymbol{\Sigma}’\right] \mathbf{f}’^T \boldsymbol{\Sigma}^{-1} \mathbb{E}\left[(\mathbf{r} - \mathbf{f})\right] \\
&= 0 \end{align}

since $\mathbb{E}\left[\mathbf{r}\right] = \mathbf{f}$ by definition. Similarly, the expectation of the sixth term $E_6$ also vanishes, since it contains three instances of $(\mathbf{r}-\mathbf{f})$.

The expectation of the second term is

\begin{align} E_2 &= \mathbb{E}_{\mathbf{r}\vert x}\left[\left[\mathbf{f}’^T \boldsymbol{\Sigma}^{-1} (\mathbf{r} - \mathbf{f})\right]^2\right] \\
&= \mathbb{E}\left[\mathbf{f}’^T \boldsymbol{\Sigma}^{-1} (\mathbf{r}-\mathbf{f})(\mathbf{r}-\mathbf{f})^T \boldsymbol{\Sigma}^{-1} \mathbf{f}’\right] \\
&= \mathbf{f}’^T \boldsymbol{\Sigma}^{-1} \mathbb{E}\left[(\mathbf{r}-\mathbf{f})(\mathbf{r}-\mathbf{f})^T\right] \boldsymbol{\Sigma}^{-1} \mathbf{f}’ \\
&= \mathbf{f}’^T \boldsymbol{\Sigma}^{-1} \boldsymbol{\Sigma} \boldsymbol{\Sigma}^{-1} \mathbf{f}’ \\
&= \mathbf{f}’^T \boldsymbol{\Sigma}^{-1} \mathbf{f}’, \end{align}

which, incidentally, is the linear Fisher information.

Next, the expectation of the fifth term is \begin{align} E_5 &= -\frac{1}{2} \text{Tr} \left[\boldsymbol{\Sigma}^{-1}\boldsymbol{\Sigma}’\right] \mathbb{E}\left[(\mathbf{r}-\mathbf{f})^T \boldsymbol{\Sigma}^{-1} \boldsymbol{\Sigma}’ \boldsymbol{\Sigma}^{-1} (\mathbf{r}-\mathbf{f}) \right]. \end{align}

Here, we evaluating over a quadratic form for which we’ll need to invoke the identity \begin{align} \mathbb{E}\left[\boldsymbol{\epsilon}^T \boldsymbol{\Lambda} \boldsymbol{\epsilon} \right] &= \text{Tr}\left[\boldsymbol{\Lambda} \text{Cov}(\boldsymbol{\epsilon})\right] + \mathbb{E}[\boldsymbol{\epsilon}]^T \boldsymbol{\Lambda} \mathbb{E}[\boldsymbol{\epsilon}] \end{align} where $\boldsymbol{\Lambda}$ is a matrix and $\boldsymbol{\epsilon}$ is a random vector. In our case, the expectation of $(\mathbf{r}-\mathbf{f})$ vanishes, so we’re only concerned with the first term. The expectation becomes \begin{align} E_5 &= -\frac{1}{2} \text{Tr}\left[\boldsymbol{\Sigma}^{-1}\boldsymbol{\Sigma}’\right] \cdot \text{Tr}\left[\boldsymbol{\Sigma}^{-1} \boldsymbol{\Sigma}’ \boldsymbol{\Sigma}^{-1} \boldsymbol{\Sigma}\right] \\
&= -\frac{1}{2} \text{Tr} \left[\boldsymbol{\Sigma}^{-1}\boldsymbol{\Sigma}’\right] \cdot \text{Tr}\left[\boldsymbol{\Sigma}^{-1} \boldsymbol{\Sigma}’\right] \\
&= -\frac{1}{2} \text{Tr} \left[\boldsymbol{\Sigma}^{-1}\boldsymbol{\Sigma}’\right]^2. \end{align}

We’ve left the trickiest for last. The expectation of the third term is effectively the expectation over the product of two quadratic forms:

\begin{align} E_3 &= \frac{1}{4} \mathbb{E}\left[(\mathbf{r} - \mathbf{f})^T\boldsymbol{\Sigma}^{-1} \boldsymbol{\Sigma}’ \boldsymbol{\Sigma}^{-1}(\mathbf{r} - \mathbf{f})(\mathbf{r} - \mathbf{f})^T\boldsymbol{\Sigma}^{-1} \boldsymbol{\Sigma}’ \boldsymbol{\Sigma}^{-1}(\mathbf{r} - \mathbf{f})\right] \end{align}

for which we’ll need the identity (see here).

\begin{align} \mathbb{E}\left[\boldsymbol{\epsilon}^T \mathbf{A} \boldsymbol{\epsilon}\boldsymbol{\epsilon}^T \mathbf{B} \boldsymbol{\epsilon}\right] &= \text{Tr}\left[\mathbf{A}\mathbf{C}\right]\text{Tr}\left[\mathbf{B}\mathbf{C}\right] + 2 \text{Tr}\left[\mathbf{A}\mathbf{C} \mathbf{B}\mathbf{C}\right] \end{align} where $\mathbf{C} = \text{Cov}(\boldsymbol{\epsilon})$. Applying the identity gives us our last addend:

\begin{align} E_3 &= \frac{1}{4} \text{Tr}\left[\boldsymbol{\Sigma}^{-1} \boldsymbol{\Sigma}’ \boldsymbol{\Sigma}^{-1}\boldsymbol{\Sigma}\right]^2 + \frac{1}{2}\text{Tr}\left[\left(\boldsymbol{\Sigma}^{-1} \boldsymbol{\Sigma}’ \boldsymbol{\Sigma}^{-1}\boldsymbol{\Sigma}\right)^2\right] \\
&= \frac{1}{4}\text{Tr}\left[\boldsymbol{\Sigma}^{-1} \boldsymbol{\Sigma}’\right]^2 + \frac{1}{2} \text{Tr}\left[\boldsymbol{\Sigma}^{-1} \boldsymbol{\Sigma}’ \boldsymbol{\Sigma}^{-1}\boldsymbol{\Sigma}’\right]. \end{align}

Proceeding to the final sum, we have:

\begin{align} I_F(s) &= \mathbb{E}_{\mathbf{r}\vert x} \left[\left(\frac{d}{dx} \log P[\mathbf{r}\vert x]\right)^2\right] \\
&= \left(\frac{1}{4} \text{Tr} \left[\boldsymbol{\Sigma}^{-1} \boldsymbol{\Sigma}’\right]^2\right) + \left(\mathbf{f}’^T \boldsymbol{\Sigma}^{-1} \mathbf{f}’\right) \\
&+ \left(\frac{1}{4}\text{Tr}\left[\boldsymbol{\Sigma}^{-1} \boldsymbol{\Sigma}’\right]^2 + \frac{1}{2} \text{Tr}\left[\boldsymbol{\Sigma}^{-1} \boldsymbol{\Sigma}’ \boldsymbol{\Sigma}^{-1}\boldsymbol{\Sigma}’\right]\right) + 0 \\
&-\left(\frac{1}{2}\text{Tr} \left[\boldsymbol{\Sigma}^{-1} \boldsymbol{\Sigma}’\right]^2\right) + 0 \\
&= \mathbf{f}’^T \boldsymbol{\Sigma}^{-1} \mathbf{f}’ + \frac{1}{2} \text{Tr}\left[\boldsymbol{\Sigma}^{-1} \boldsymbol{\Sigma}’ \boldsymbol{\Sigma}^{-1}\boldsymbol{\Sigma}’\right]. \end{align}

And we’re done!

The Linear Fisher Information

In the above expression, the second term (often dubbed the “trace” term) will vanish if the covariance matrix is stimulus-independent, leaving the first term:

\begin{align} I_F(s) &= \mathbf{f}’(s)^T \boldsymbol{\Sigma}^{-1}\mathbf{f}’(s) \end{align}

which is referred to as the linear Fisher information. The nice thing about this estimator (which acts as a lower bound to the Fisher information) is that it comes with its own linear decoder, known as the locally optimal linear estimator:

\begin{align} \hat{s} &= s_0 + \frac{\mathbf{f}’(s_0)^T \boldsymbol{\Sigma}^{-1}(s_0) \left[\mathbf{r} - \mathbf{f}(s_0)\right]}{\mathbf{f}’(s_0)^T \boldsymbol{\Sigma}^{-1}(s_0) \mathbf{f}’(s_0)}. \end{align}

Here, $s_0$ is some base stimulus value. Therefore, the above decoder allows us to discriminate stimuli near $s_0$ with variance equal to the linear Fisher information. Thus, linear Fisher information is nice for three reasons: first, that we even have a decoder in the first place (the Fisher information has no guarantees about the existence of a decoder); second, it’s linear (linearity is always nice); and third, it’s generally easy to calculate (if you stare at it long enough, the linear Fisher information basically becomes a signal-to-noise ratio).

Overall, it’s a useful tool for measuring the coding fidelity of different representations and as such has been vitally important in the computational neuroscience literature.