Discrete-time sampling of filtered white noise
Hmm. For standard deviation, I see https://dsp.stackexchange.com/a/8632/829 which states for uniform power spectral density $N_0/2$, the standard deviation is

$$\sigma^2  = \int_{-\infty}^\infty \frac{N_0}{2}|H(f)|^2\,\mathrm df$$

which in my case would yield

$$\begin{align}
\sigma^2  &amp;= \int_{-\infty}^\infty \frac{N_0}{2}|H(f)|^2\,df \\
          &amp;= \int_{-\infty}^\infty \frac{N_0}{2}|H(\omega/2\pi)|^2\,d(\omega/2\pi) \\
          &amp;= \frac{1}{2\pi}\int_{-\infty}^\infty \frac{N_0}{2}|H(\omega)|^2\,d\omega \\
          &amp;= \frac{1}{2\pi}\int_{-\infty}^\infty \frac{N_0}{2}\frac{1}{(\omega\tau)^2+1}\,d\omega \\
          &amp;= \frac{1}{2\pi\tau}\int_{-\infty}^\infty \frac{N_0}{2}\frac{1}{(\omega\tau)^2+1}\,d(\omega\tau) \\
          &amp;= \frac{1}{2\pi\tau}\int_{-\infty}^\infty \frac{N_0}{2}\frac{1}{u^2+1}\, du \\
          &amp;= \frac{N_0}{4\pi\tau}\left[\tan^{-1} u\right]_{-\infty}^{\infty} \\
          &amp;= \frac{N_0}{4\pi\tau}\left[\pi/2 - (-\pi/2)\right] \\
          &amp;= \frac{N_0}{4\tau},
\end{align} 
$$

and with unit WGN $N_0/2 = 1$ so $\sigma = \sqrt{1/2\tau}$.

Not sure how to compute the PSD of a sampled random process, however.