Taylor series of a convolution

Question

The derivation below is from Probability Theory: The Logic Of Science By E. T. Jaynes, chapter 7 "The Central Gaussian, Or Normal, Distribution", p.706 The Landon derivation. Text available online:

Given a specific $\epsilon$, the probability for the new noise voltage to have the value $v'$ would be just the previous probability that $v$ should have the value $(v' - \epsilon)$. Thus, by the product and sum rules of probability theory, the new probability distribution is the convolution

$(7.18)\quad f(v') = \int p(v'-\epsilon|\sigma)q(\epsilon)d\epsilon $

Expanding this in powers of the small quantity $\epsilon$ and dropping the prime, we have

$(7.19)\quad f(v) = p(v|\sigma) - {\partial p(v|\sigma)\over \partial v} \int \epsilon q(\epsilon)d\epsilon + \frac 12 {\partial^2 p(v|\sigma)\over \partial v^2} \int \epsilon^2 q(\epsilon)d\epsilon + \ldots$

Question: How exactly was 7.19 derived from 7.18? I'm assuming this is just the Taylor series, but maybe I'm mistaken. Anyways I would be grateful for someone filling in the details of how this derivation was accomplished.

Answer 1

It is a straight forward Taylor expansion. Let $\phi(v)=p(v\mid\sigma)$. Then $$ \phi(v-\epsilon)=\phi(v)+\phi'(v)\,(-\epsilon)+\frac12\,\phi''(v)\,(-\epsilon)^2+\dots=\phi(v)-\phi'(v)\,\epsilon+\frac12\,\phi''(v)\,\epsilon^2+\dots $$ Now multiply by $q(\epsilon)$ and integrate.

Answer 2

The Taylor series of a function f(x) around a critical point $x_0$ can be written as: $$f(x)= \Sigma_{n=0}^\infty \frac{1}{n!} \frac{d^n}{dx^n}f(x)|_{x=x_0} (x-x_0)^n $$ $$=f(x)|_{x=x_0}+\frac{d}{dx}f(x)|_{x=x_0}(x-x_0)+\frac{1}{2} \frac{d^2}{dx^2}f(x)|_{x=x_0}(x-x_0)^2+...$$

Therefore expanding p around$\epsilon$ leaves $$\quad f(v) = \int (p(v|\sigma) - {\partial p(v|\sigma)\over \partial v} \epsilon + \frac 12 {\partial^2 p(v|\sigma)\over \partial v^2} \epsilon^2 + \ldots) q(\epsilon) d \epsilon$$

$$ = p(v|\sigma) - {\partial p(v|\sigma)\over \partial v} \int \epsilon q(\epsilon)d\epsilon + \frac 12 {\partial^2 p(v|\sigma)\over \partial v^2} \int \epsilon^2 q(\epsilon)d\epsilon + \ldots$$