Time derivative of signal - effect on noise distribution
Let's call $a[n]$ the samples of the angular measurement. Then, an estimate of the angular velocity (in angle/sec) is given by discrete differences, i.e., $v[n] = (a[n]-a[n-1])/t_0$, where $t_0$ is the sampling time.

Let's assume a signal pluse noise process so that $a[n] = a_0[n] + w[n]$. Inserting, we obtain $v[n] = v_0[n] + \bar{w}[n]$, where $\bar{w}[n] = (w[n]-w[n-1])/t_0$. Certainly, if $w[n]$ is zero mean, so is $\bar{w}[n]$ You can now compute the variance of your effective noise samples $\bar{w}[n]$, namely $$\mathbb{E}\{\bar{w}[n]^2\} = \frac{1}{t_0^2}\left(\mathbb{E}\{\bar{w}[n]^2\} + \mathbb{E}\{\bar{w}[n-1]^2\} -2 \mathbb{E}\{\bar{w}[n]\bar{w}[n-1]\}\right). $$ If your noise samples $w[n]$ are i.i.d. and have  constant variance $\sigma_w^2$, this leads to $\mathbb{E}\{\bar{w}[n]^2\} = \frac{2\sigma_w^2}{t_0^2}$. So there is a doubling of variance, but also some kind of rescaling. This depends of course on the unit of your velocity (whether it's angle/second or angle/sample, in the last case your variance is just $2\sigma_w^2$). NB: If they are correlated, this changes to $2\sigma_w^2(1-\rho)$, where $\rho$ is the correlation coefficient.

All this is true if the velocity is simply estimated from differences of adjacent samples. There are more elaborate schemes available, which would have an effect on the noise variance.