Predictive distribution with divergent integral I am having some trouble understanding the basic definition of a posterior predictive distribution and how to apply it to simple examples.
I have that if $$y \sim g(y\mid \theta, x)$$
then the predictive posterior distribution is
$$g(y\mid x)=\int g(y\mid\theta,x)\pi(\theta\mid x) \,d\theta$$
So I am trying to work through an example of this where
$x \sim \operatorname{Normal}(\theta, \sigma^2)$ and $y \sim \operatorname{Normal}(ex,\sigma^2)$ and $\pi(\theta,\sigma^2)=\frac{1}{\sigma^2}$
And I am not sure I understand all the notations correctly. I don't have any examples either of actually solving for this, but I would be interested in seeing any.
So I know that $$\pi(\theta\mid x)  \propto \pi(x\mid\theta) \pi(\theta)$$
so $$\pi(\theta,\sigma^2\mid x) \propto \exp\left(\frac{-1}{2\sigma^2}{(x-\theta)^2}\right) \frac{1}{\sigma^2}$$
But then I am not sure how to proceed.
I assume I am looking for $\pi(Y\mid X=x)$
Well  $P(Y\mid X=x,\theta , \sigma^2) = \exp(\frac{-1}{2\sigma^2}(y-ex)^2)$
so,
$$P(Y\mid X=x,\theta , \sigma^2) \pi(\theta,\sigma^2 \mid x)= \frac{1}{\sigma^2} \exp\left(\frac{-1}{2\sigma^2}[(y-ex)^2+(x-\theta)^2]\right)$$
Note that if we try to double integral to normalise over $\pi(x\mid\theta , \sigma^{2})\pi(\theta,\sigma^2)$ we get a divergent integral. So we cannot make equality.
 A: To sum up, you should know the prior distribution of $\sigma^2$. And, yes, you need double integral to integrate out both $\sigma^2, \theta$, unless $\sigma^2$ has been assumed to be a given constant.
Let's clarify what you are looking for. Your are looking for the posterior predictive, i.e. the distribution of $Y | X=x$. Your model is as follows. Given constant $e$,
$$
\begin{aligned}
X \mid \theta, \sigma^2 &\sim N(\theta,\sigma^2) \\[7pt]
Y \mid x, \theta, \sigma^2 &\sim N(ex,\sigma^2) \\[7pt]
π(θ \mid \sigma^2) &= 1/\sigma^2 I_\Theta \\[7pt]
\sigma^2 &\sim f_{\sigma^2}
\end{aligned}
$$
So you have 
$$
\begin{aligned}
\pi(\theta, \sigma^2 \mid x ) &\propto \pi(x \mid \theta, \sigma^2) \pi(\theta \mid \sigma^2) \pi(\sigma^2)\\[7pt]
&\propto \exp(-\frac{1}{2\sigma^2} (x-\theta)^2) 1/\sigma^2 \cdot f_{\sigma^2}
\end{aligned}
$$
Thus
$$
\begin{aligned}
P(Y \mid X=x) &= \int_{\sigma^2} \int_\theta P( Y, \theta, \sigma^2 \mid X=x) \\[7pt]
&= \int_{\sigma^2} \int_\theta \pi(Y \mid x, \theta, \sigma^2) \pi (\theta \mid \sigma^2) f_{\sigma^2} \\[7pt]
&= \int_{\sigma^2} \int_\theta N(ex,\sigma^2) 1/\sigma^2 f_{\sigma^2} \\[7pt]
\end{aligned}
$$
