Probability Bayesian CDF greater than true CDF for Gaussian random variable A Gaussian random variable $X \sim \mathcal{N}(\mu,\sigma^2)$ has known variance $\sigma^2$ but unknown mean $\mu$. The cumulative distribution function of $X$ is denoted $F(x) = P(X < x)$. This question concerns the prediction of $F(x)$ based on $n$ samples of $X$ using the Bayesian approach. The conjugate prior for $\mu$ is Gaussian and posterior predictive $F_{B}(x)$ is also Gaussian. I am interested in the probability that the Bayesian prediction $F_B(x)$ is greater than $F(x)$ for a given value of $x$. Numerical simulation seems to suggest the following relationship:
$$P(F_B(x) > F(x)) = F(x)$$
Do you know of any literature that considers this problem or similar? Can you suggest why this might be the case?
 A: If 
\begin{align*}
X_0 &\sim \mathcal{N}(\mu,\sigma^2/n)\\ 
\mu &\sim\mathcal{N}(0,\sigma^2)\\
X_1 &\sim \mathcal{N}(\mu,\sigma^2)\\
\end{align*}
then
$$\mu|x_0\sim\mathcal{N}(nx_0/(n+1),\sigma^2/\{n+1\})$$
and
$$X_1|x_0\sim \mathcal{N}\left(\frac{n}{n+1}x_0,\frac{n+2}{n+1}\sigma^2\right)$$
Therefore
$$F_B(x)=\Phi\left( \frac{x-\frac{n}{n+1}x_0}{\sqrt{\frac{n+2}{n+1}\sigma}}\right)$$
Since$$F(x)=\Phi\left( \frac{x-\mu}{\sigma} \right)$$
we get
\begin{align*}
\mathbb{P}_\mu(F_B(x)>F(x)) &= \mathbb{P}_\mu\left(
\Phi\left( \frac{x-\frac{n}{n+1}X_0}{\sqrt{\frac{n+2}{n+1}}\sigma}\right)>\Phi\left( \frac{x-\mu}{\sigma} \right)\right)\\
&= \mathbb{P}_\mu\left(\frac{x-\frac{n}{n+1}X_0}{\sqrt{\frac{n+2}{n+1}}\sigma}>\frac{x-\mu}{\sigma} \right)\\
&= \mathbb{P}_\mu\left(x-\frac{n}{n+1}X_0>\sqrt{\frac{n+2}{n+1}}(x-\mu)\right)\\
&= \mathbb{P}_\mu\left(x-\sqrt{\frac{n+2}{n+1}}(x-\mu)>\frac{n}{n+1}X_0\right)\\
&= \mathbb{P}_\mu\left(X_0<\frac{n+1}{n}\left\{1-\sqrt{\frac{n+2}{n+1}}\right\}x-\sqrt{\frac{(n+2)(n+1)}{n}}\mu\right)\\
\end{align*}
