Assume the following bivariate regression model:
$y_i = \beta x_i + u_i$ where $u_i$ is i.i.d $N(0, \sigma^2 = 9)$ for $i = 1, 2, ..., n$.
Assume a noninformative prior of the form:
$p(\beta) \propto constant$, then it can be shown that the posterior pdf for $\beta$ is:
$p(\beta|\mathbf{y}) = (18\pi)^{-\frac{1}{2}}\left(\sum_{i=1}^n x_i^2\right)^{\frac{1}{2}} \exp\left[-\frac{1}{18}\sum_{i=1}^n x_i^2 (\beta - \hat{\beta})^2\right]$
where $\displaystyle{\hat{\beta} = \frac{\sum_{i=1}^n y_ix_i}{\sum_{i=1}^n x_i^2}}$
Now consider the value of $y$ with a given future value of $x$, $x_{n+1}$:
$y_{n+1} = \beta x_{n+1} + u_{n+1}$ where $u_{n+1}$ is i.i.d $N(0, \sigma^2 = 9)$ , then we can show that:
$p(y_{n+1}|x_{n+1},\mathbf{y}) = \int_{\beta} p(y_{n+1}|x_{n+1}, \beta, \mathbf{y}) p(\beta|\mathbf{y})d\beta$ is a normal density with:
$E[y_{n+1}|x_{n+1},\mathbf{y}] = \hat{\beta}x_{n+1}$
and
$\displaystyle{var[y_{n+1}|x_{n+1},\mathbf{y}] = \frac{9[x_{n+1}^2 + \sum_{i=1}^n x_i^2]}{\sum_{i=1}^n x_i^2}}$
Thus the posterior probability density function for $y_{n+1}$, conditional on $x_{n+1}$, is given by: $$p(y_{n+1}|x_{n+1},\mathbf{y}) = \left(\frac{18\pi\left[x_{n+1}^2 + \displaystyle{\sum_{i=1}^n x_i^2}\right]}{ \displaystyle{\sum_{i=1}^n x_i^2}}\right)^{-\frac{1}{2}} \exp\left\{-\frac{\displaystyle{\sum_{i=1}^n x_i^2}}{18\left(x_{n+1}^2 + \displaystyle{\sum_{i=1}^n x_i^2}\right)}\left(y_{n+1}-\hat{\beta}x_{n+1}\right)^2\right\}$$
Now the question is: Specify a 95% prediction interval for $y_{n+1}$ and carefully interpret it. What aspect(s) of the data generating process does the interval fail to accommodate our uncertainty about?
I'm not exactly too sure on how to answer the question but here's my attempt:
So essentially we need to find some $a$ and $b$ such that $P(a < y_{n+1} < b) = \int_{a}^b p(y_{n+1}|x_{n+1}, \mathbf{y}) dy_{n+1} = 95\%$
Now we know that $y_{n+1}|x_{n+1}, \mathbf{y} \sim N(m, v^2)$ where $m = E[y_{n+1}|x_{n+1},\mathbf{y}]$ and $v^2 = var[y_{n+1}|x_{n+1},\mathbf{y}]$, Hence: $$\frac{y_{n+1}-m}{v} \sim N(0,1)$$ $$P(-1.96 < \frac{y_{n+1}-m}{v} < 1.96) = 95\%$$ $$P(-1.96v+m < y_{n+1} < 1.96v+m) = 95\%$$
Now because we are conditioning on $x_{n+1}$ and looking at the expression for $v$ and $m$, we see that both $v$ and $m$ are known values. So we can take $a = -1.96v+m$ and $b = 1.96v+m$.
Now I know that this $a$ and $b$ are not unique, ie, we can select many other possibilities of $a$ and $b$ which yields a probability of $95\%$... but how does this relate to answering the part of the question that asks what aspects of the data generating process this interval fails to accommodate?
Any help would be appreciated, thanks!