Exercise 13.8 - Deriving E step for GSM prior

Answers

The hints contains typoes and are unnecessary. By definition, we are to calculate:

\begin{align} 𝔼 [\frac{1}{τ_{j}^{2}} | w_{j}] = & \int \frac{1}{τ_{j}^{2}} p (τ_{j}^{2} | w_{j}) d τ_{j}^{2} = \int \frac{1}{τ_{j}^{2}} \frac{p (w_{j} | τ_{j}^{2}) p (τ_{j}^{2})}{p (w_{j})} d τ_{j}^{2} \\ = & \frac{1}{p (w_{j})} \int \frac{1}{τ_{j}^{2}} 𝒩 (w_{j} | 0, τ_{j}^{2}) p (τ_{j}^{2}) d τ_{j}^{2} \\ = & \int \frac{1}{\sqrt{2 π}} {(τ_{j}^{2})}^{- 1.5} \exp (- \frac{w_{j}^{2}}{2 τ_{j}^{2}}) p (τ_{j}^{2}) d τ_{j}^{2} . \end{align}

While we also have:

p^{'} (w_{j}) = \int \frac{1}{\sqrt{2 π}} {(τ_{j}^{2})}^{- 1.5} \exp (- \frac{w_{j}^{2}}{2 τ_{j}^{2}}) p (τ_{j}^{2}) \cdot w_{j} d τ_{j}^{2} .

Plugging both side into (13.197) finishes the proof.

solour_lfq

2021-03-24 13:42