Consider the bivariate normal distribution with standard deviation $\sigma$ and mean displaced to a point $x_0$ along the $x$-axis. In polar coordinates, the pdf is expressed as:
$$\rho(r,\theta)=\frac{r}{2\pi\sigma^2}\exp\left[-\frac{(r\cos{\theta}-x_0)^2+(r\sin{\theta})^2}{2\sigma^2}\right]$$
The expectation value for $\theta$ is then:
$$\overline{\theta}=\int_0^{2\pi}d\theta\int_0^\infty dr\ \theta\ \rho(r,\theta) $$
I am not aware of an analytical result to the above integral, but numerical calculation for $x_0\gt0$ and $\sigma\ll x_0$ gives $\overline{\theta}=\pi$. At first glance, this result seems counterintuitive, because it indicates that the mean of the pdf lies somewhere at $x_0 \lt 0$. On the other hand, the region over which $\rho(r,\theta)$ is most pronounced straddles the discontinuity between $\theta\sim 0$ and $\theta\sim 2\pi$ in equal measure, and in this respect $\overline{\theta}$ would seem to give an average of these two values, which is just $\pi$.
The "intuitive" result $\overline{\theta}=0$ can be obtained by simply redefining the limits of integration over $\theta$ to be from $-\pi$ to $\pi$, effectively rotating the $\theta$ discontinuity/cut angle to the $-x$-axis. That's fine for this case, but what happens when one has to deal with a normal distribution displaced to any arbitrary point in the $(r,\theta)$-plane? Or even worse, when the origin is located within a few $\sigma$ of the centroid of the pdf? Is there a generic way to handle this, or does one have to tailor the integral for each case in turn?