Important issues with Exercise 12.11.2

[Pipe-Vash]

As correctly showed in the beginning of part b), $\mu \lvert X^n \sim Normal\left(\overline{X} , \frac{\sigma^2}{n}\right)$, with $\sigma=1$. In part d), what it is expected is the posterior of $\theta= e^{\mu}, i.e., we want \mu \lvert X^n. It is important to notice what in the posterior analysis we treat the parameters as random values and the data as constants, so the solutions presented in part d) is completely incorrect. The correct procedure is the following:

• $\theta$'s CDF:

$$\begin{eqnarray} F_{\Theta}\left(\theta \vert X^n \right) &=& \mathbb{P}\left(\Theta \leq \theta \vert X^n \right) = \mathbb{P}\left(e^{\mu} \leq \theta \lvert X^n \right) = \mathbb{P}\left(\mu \leq \log(\theta) \vert X^n \right) \\ &=& \mathbb{P}\left(\left(\mu-\overline{X}\right) \frac{\sqrt{n}}{\sigma} \leq \left(\log(\theta)-\overline{X}\right) \frac{\sqrt{n}}{\sigma} \bigg\vert X^n \right) \\ &\approx& \mathbb{P}\left( Z \leq \left(\log(\theta)-\overline{X}\right) \frac{\sqrt{n}}{\sigma} \bigg\vert X^n \right)= \Phi\left[\left(\log(\theta)-\overline{X}\right) \frac{\sqrt{n}}{\sigma}\right] \end{eqnarray}$$

• $\theta$'s PDF comes out by taking the derivative respect to $\theta$:
  $f_\Theta(\theta)= \phi\left[\left(\log(\theta)-\overline{X}\right) \frac{\sqrt{n}}{\sigma}\right] \frac{\sqrt{n}}{\sigma} \frac{1}{\theta}$

• Notice that this solutions depends on the sample size as it is expected for a posteriori, which applies both to $\mu \lvert X^n$ and $\theta\lvert X^n$

• I also suggest to respect the symbol definitions in the preface in order to avoid confusions