### A Quantum of the Fisher Information DerivationTweet

This post elaborates a derivation of Equation (2.20) on page 53 of Bayesian Data Analysis Third Edition.

$\pmb{\text{Figure 1}}$: Sir Ronald Aylmer Fisher (17 February 1890 - 29 July 1962). One of his many great contributions to Statistics is Fisher Information. Image taken from Wikipedia, some rights reserved.

Concretely, we want to show the derivation $J(\theta)$, the Fisher Information, from

$$$J(\theta) = \text{E}\left( \left( \frac{d \log \Pr(y \mid \theta )}{d\theta} \right)^2 \, \middle| \, \theta \right) \tag{1}\label{eq:start-point}$$$

to

$$$J(\theta) = - \text{E}\left( \frac{d^2 \log \Pr(y \mid \theta)}{d\theta^2} \, \middle| \, \theta \right). \tag{2}\label{eq:end-point}$$$

The idea of this derivation comes from a lecture note by John Duchi from Stanford Statistics class. The difference between this post and the lecture note is that the lecture note deals with multi-variables which employs second derivatives for multi-values (Hessian matrix); on the other hand, this post deals with a single variable and employs a second derivative for just one value.

Let’s start with computing $$$\text{E} \left( \frac{d \log \Pr(y \mid \theta)}{d\theta} \, \middle| \, \theta \right)$$$.

\require{cancel} \begin{align} \text{E}\left( \frac{d \log \Pr(y \mid \theta)}{d\theta} \, \middle| \, \theta \right) &= \int \frac{d \log \Pr(y \mid \theta)}{d\theta} \Pr(y \mid \theta) d\theta && \text{definition of expectation} \tag{3}\label{eq:dlog-1} \\ &= \int \frac{d \Pr(y \mid \theta)}{d\theta} \frac{1}{\Pr(y \mid \theta)} \; \Pr(y \mid \theta) d\theta && \text{derivation of }\frac{d \log \Pr(y \mid \theta)}{d\theta} \tag{4}\label{eq:dlog-2} \\ &= \int \frac{d \Pr(y \mid \theta)}{d\theta} \frac{1}{\cancel{\Pr(y \mid \theta)}} \; \cancel{\Pr(y \mid \theta)} d\theta \tag{5}\label{eq:dlog-3} \\ &= \int \frac{d \Pr(y \mid \theta)}{d\theta} d\theta \tag{6}\label{eq:dlog-4} \\ &= \frac{d}{d\theta} \int \Pr(y \mid \theta) d\theta && \text{exchange }\frac{d}{d\theta} \text{ and } \int \tag{7}\label{eq:dlog-5} \\ &= \frac{d}{d\theta} \underbrace{\int \Pr(y \mid \theta) d\theta}_{1} && \text{property of a pdf}\tag{8}\label{eq:dlog-6} \\ &= \frac{d}{d\theta} (1) \tag{9}\label{eq:dlog-7} \\ &= 0. \tag{10}\label{eq:dlog-8} \end{align}

Consider Equation \eqref{eq:dlog-5}, we shall utilize this exchangeability between integral and differentiation again later.

Equation \eqref{eq:dlog-2} states that

$$$\frac{d \log \Pr(y \mid \theta)}{d\theta} = \underbrace{\frac{1}{\Pr(y \mid \theta)}}_{u} \underbrace{\frac{d \Pr(y \mid \theta)}{d\theta}}_{v}. \tag{11}\label{eq:first-order-derivation}$$$

Therefore,

\begin{align} \frac{d^2 \log \Pr(y \mid \theta)}{d\theta^2} &= \underbrace{- \frac{1}{\Pr( y \mid \theta )^2} \frac{d \Pr(y \mid \theta)}{d\theta}}_{u^{\prime}} \underbrace{\frac{d \Pr(y \mid \theta)}{d\theta}}_{v} + \underbrace{\frac{1}{\Pr(y \mid \theta)}}_{u} \underbrace{\frac{d^2 \Pr(y \mid \theta)}{d\theta^2}}_{v^{\prime}} && \text{based on } u^{\prime} v + u v^{\prime} \tag{12}\label{eq:second-order-1} \\ &= \frac{d^2 \Pr(y \mid \theta)}{d\theta^2} \frac{1}{\Pr(y \mid \theta)} - \left( \frac{d\Pr(y \mid \theta)}{d\theta} \frac{1}{\Pr(y \mid \theta)} \right) \left( \frac{d\Pr(y \mid \theta)}{d\theta} \frac{1}{\Pr(y \mid \theta)} \right) && \text{just rearranging} \tag{13}\label{eq:second-order-2} \\ &= \frac{d^2 \Pr(y \mid \theta)}{d\theta^2} \frac{1}{\Pr(y \mid \theta)} - \left( \frac{d \log \Pr( y \mid \theta)}{d\theta} \right) \left( \frac{d \log \Pr( y \mid \theta)}{d\theta} \right) && \text{based on Equation }\eqref{eq:first-order-derivation} \tag{14}\label{eq:second-order-3} \\ &= \frac{d^2 \Pr(y \mid \theta)}{d\theta^2} \frac{1}{\Pr(y \mid \theta)} - \left( \frac{d \log \Pr( y \mid \theta)}{d\theta} \right)^{2} \tag{15}\label{eq:second-order-4} \end{align}

From Equation \eqref{eq:second-order-4} we obtain

\begin{align} \frac{d^2 \log \Pr(y \mid \theta)}{d\theta^2} = \frac{d^2 \Pr(y \mid \theta)}{d\theta^2} \frac{1}{\Pr(y \mid \theta)} - \left( \frac{d \log \Pr( y \mid \theta)}{d\theta} \right)^{2} &\Longleftrightarrow \left( \frac{d \log \Pr( y \mid \theta)}{d\theta} \right)^{2} = - \frac{d^2 \log \Pr(y \mid \theta)}{d\theta^2} + \frac{d^2 \Pr(y \mid \theta)}{d\theta^2} \frac{1}{\Pr(y \mid \theta)} \tag{16}\label{eq:second-order-last} \end{align}

Now we are ready to calculate $$$\text{E}\left( \left( \frac{d \log \Pr(y \mid \theta )}{d\theta} \right)^2 \, \middle| \, \theta \right).$$$

\require{cancel} \begin{align} \text{E}\left( \left( \frac{d \log \Pr(y \mid \theta )}{d\theta} \right)^2 \, \middle| \, \theta \right) &= \int \left( \frac{d \log \Pr(y \mid \theta )}{d\theta} \right)^2 \Pr(y \mid \theta) d\theta && \text{by definition} \tag{17}\label{eq:final-showdown-1}\\ &= \int \left( - \frac{d^2 \log \Pr(y \mid \theta)}{d\theta^2} + \frac{d^2 \Pr(y \mid \theta)}{d\theta^2} \frac{1}{\Pr(y \mid \theta)} \right) \Pr(y \mid \theta) d\theta && \text{by Equation }\eqref{eq:second-order-last} \tag{18}\label{eq:final-showdown-2}\\ &= \int \left( - \frac{d^2 \log \Pr(y \mid \theta)}{d\theta^2} \right) \Pr(y \mid \theta) d\theta + \int \frac{d^2 \Pr(y \mid \theta)}{d\theta^2} \frac{1}{\Pr(y \mid \theta)} \Pr(y \mid \theta) d\theta && \text{by distributive} \tag{19}\label{eq:final-showdown-3}\\ &= \int \left( - \frac{d^2 \log \Pr(y \mid \theta)}{d\theta^2} \right) \Pr(y \mid \theta) d\theta + \int \frac{d^2 \Pr(y \mid \theta)}{d\theta^2} \frac{1}{\cancel{\Pr(y \mid \theta)}} \cancel{\Pr(y \mid \theta)} d\theta \tag{20}\label{eq:final-showdown-4}\\ &= \int \left( - \frac{d^2 \log \Pr(y \mid \theta)}{d\theta^2} \right) \Pr(y \mid \theta) d\theta + \int \frac{d^2 \Pr(y \mid \theta)}{d\theta^2} d\theta \tag{21}\label{eq:final-showdown-5}\\ &= \int \left( - \frac{d^2 \log \Pr(y \mid \theta)}{d\theta^2} \right) \Pr(y \mid \theta) d\theta + \frac{d^2}{d\theta^2} \left( \int \Pr(y \mid \theta) d\theta \right) && \text{by exchangeability again} \tag{22}\label{eq:final-showdown-6}\\ &= \int \left( - \frac{d^2 \log \Pr(y \mid \theta)}{d\theta^2} \right) \Pr(y \mid \theta) d\theta + \underbrace{\frac{d^2}{d\theta^2} \left( \int \Pr(y \mid \theta) d\theta \right)}_{0} && \text{similar to Equation }\eqref{eq:dlog-8} \tag{23}\label{eq:final-showdown-7}\\ &= \int \left( - \frac{d^2 \log \Pr(y \mid \theta)}{d\theta^2} \right) \Pr(y \mid \theta) d\theta \tag{24}\label{eq:final-showdown-8}\\ &= - \int \left( \frac{d^2 \log \Pr(y \mid \theta)}{d\theta^2} \right) \Pr(y \mid \theta) d\theta \tag{25}\label{eq:final-showdown-9}\\ &= - \text{E}\left( \frac{d^2 \log \Pr(y \mid \theta)}{d\theta^2} \, \middle| \, \theta \right). && \text{by definition} \tag{26}\label{eq:final-showdown-10} \end{align}

At last, we have finally shown that

$$$J(\theta) = \text{E}\left( \left( \frac{d \log \Pr(y \mid \theta )}{d\theta} \right)^2 \, \middle| \, \theta \right) = - \text{E}\left( \frac{d^2 \log \Pr(y \mid \theta)}{d\theta^2} \, \middle| \, \theta \right)$$$

as it is explained by Equation (2.20) on page 53 of the book.

Written on February 23, 2021