Interpretation of Bayesian 95% prediction interval

Question

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 $i = 1,\ldots, n$.

Assume a noninformative prior $p(\beta) \propto \text{constant}$, 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 $\hat{\beta} = (\sum_{i=1}^n y_ix_i)/(\sum_{i=1}^n x_i^2).$

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 expectation and variance $$ E[y_{n+1}|x_{n+1},\mathbf{y}] = \hat{\beta}x_{n+1},\quad {\rm var}[y_{n+1}|x_{n+1},\mathbf{y}] = \frac{9[x_{n+1}^2 + \sum_{i=1}^n x_i^2]}{\sum_{i=1}^n x_i^2}. $$ Thus the posterior probability density function for $y_{n+1}$, conditional on $x_{n+1}$, is \begin{align} 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}} \\ & \times \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\} \end{align}

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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?

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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 $b = 1.96v+m$. i.e., we can select many other possibilities of $a$ and $b$ which yield 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?

Answer 1

The interval accommodates all of the uncertainty in the problem. In your description of the problem, the only things you don't know are $\beta$ and $u$. The prediction interval that you derived accommodates the uncertainty in both of these. So there is no uncertainty left for the interval to "fail to accommodate".