Given a normal distribution $P_A(x)$, I want to find another normal distribution $P_B(x)$ with maximal uncertainty that satisfies the following:
Any sample drawn from $P_B(x)$ has a probability larger than $1-\epsilon$ of this sample being within a certain confidence interval $[\mu_A-c_A, \mu_A+c_A]$ of $P_A(x)$.
I managed to get a solution for this problem in the following way:
Let:
$P_A \sim N(\mu_A, \sigma_A^2)$ with PDF $f(x) = \frac{1}{\sigma_A \sqrt{2\pi}}e^{-\frac{1}{2}(\frac{x-\mu_A}{\sigma_A})^2}$ and
$P_B \sim N(\mu_B, \sigma_B^2)$ with PDF $f(x) = \frac{1}{\sigma_B \sqrt{2\pi}}e^{-\frac{1}{2}(\frac{x-\mu_B}{\sigma_B})^2}$
I can create the inequality
$1-\epsilon \le \int_{\mu_A-c_A}^{\mu_A+c_A}\frac{1}{\sigma_B \sqrt{2\pi}}e^{-\frac{1}{2}(\frac{x-\mu_B}{\sigma_B})^2}\,dx$
Which (according to WolframAlpha) resolves into the following:
$1-\epsilon \le \frac{1}{2} \text{erf}(\frac{\mu_A+c_A-\mu_B}{\sqrt{2}\sigma_B})+\frac{1}{2} \text{erf}(\frac{-\mu_A+c_A+\mu_B}{\sqrt{2}\sigma_B})$
From logical thinking, I can come to the conclusion that the maximum uncertainty is given if $\mu_B = \mu_A$ since this centers the $P_B$ distribution in the center of the given interval. Thus the equation becomes
$1-\epsilon \le \frac{1}{2} \text{erf}(\frac{c_A}{\sqrt{2}\sigma_B})+\frac{1}{2} \text{erf}(\frac{c_A}{\sqrt{2}\sigma_B})$
Which can be simplified to
$1-\epsilon \le \text{erf}(\frac{c_A}{\sqrt{2}\sigma_B})$
This way, I can calculate the maximum $\sigma_B$ dependent on $\sigma_A$.
However, I cannot think of a proof that the uncertainty can be maximized iff $\mu_A = \mu_B$. It seems absolutely logical to me, but I'm having a hard time formulating it properly...
Notice that $\int_{\mu_A-c_A}^{\mu_A+c_A}\frac{1}{\sigma_B \sqrt{2\pi}}e^{-\frac{1}{2}(\frac{x-\mu_B}{\sigma_B})^2}\,dx \leq \int_{\mu_A-c_A}^{\mu_A+c_A}\frac{1}{\sigma_B \sqrt{2\pi}}e^{-\frac{1}{2}(\frac{x-\mu_A}{\sigma_B})^2}\,dx$. Therefore, if $(\sigma_B,\mu_B)$ is a solution, $(\sigma_B,\mu_A)$ also is.