Rate of change of direction of momentum correlation in Brownian motion

Imagine a very small thin shelled sphere with radius R filled with an ideal gas of temperature T and density N molecules per unit volume that can move independently (Brownian motion), each identical molecule having a mass of M. At any one time the probability of the combined momentum of the particles having an upward component is 1/2. My question is twofold: 1) What would be the average frequency of this occurrence (the spectrum probably resembling white noise) and 2) What is the average magnitude of this upward momentum and/or force exerted on the shell - which could be thought of as the center of gravity randomly wobbling up an down (as well as in all other directions). The solution is probably quite simple and previously answered but please enlighten me. Jens

Since you have finite number of particles, they are not performing a Brownian motion. Between two shocks, the particles experience no force and have therefore a straight trajectory along a distance $\ell$ call the mean free path. For a gas of hard spheres of radius $a$ with $N$ spheres per unit volume, the mean free path is $$\ell=\frac{1}{4\pi\sqrt{2}N a^2}.$$ If the spheres have a mass $M$, the average speed is $\bar v=\sqrt{\frac{8k_{\rm B}T}{\pi M}}$. The average collision rate is therefore $$r_{\rm collision}=\frac{\bar v}{\ell}=16\sqrt{\pi}\,Na^2\sqrt{\frac{k_{\rm B}T}{M}}.$$ For each collision, the change of direction has a probability $1/2$ to occur, thus $$r_{\rm direction\;change}=\frac12r_{\rm collision}=8\sqrt\pi\,Na^2\sqrt{\frac{k_{\rm B}T}M}.$$ If you want to find the result if the sphere follow a pure Brownian motion, you should take the limit when $\ell\to0$ (which is equivalent to $N\to\infty$), and you find $r_{\rm direction\;change}\to\infty$.
The force exerted on the shell sphere on a element of surface is $\delta\vec F=-P\vec{\mathrm dS}$ where $\vec{\mathrm dS}$ is the element of surface normal vector oriented toward the sphere's interior. Whatever the shape of the volume, as the pressure $P$ is uniform inside, we have $\vec F=\vec0$. The fluctuations of $\vec F$ are known to be Gaussian, their relative fluctuations scale as $\frac{1}{\sqrt{VN}}$ ($V=\frac43\pi R^3$ is the volume of the sphere). Since the pressure is $P=Nk_{\rm B}T$, we find that the fluctuations of pressure are of the order of $$\sigma_P=\sqrt{\frac NV}k_{\rm B}T.$$ The fluctuation of displacement of the sphere are thus of order $$\sigma_x\approx\frac{\sigma_P\times S}{M}\frac{1}{(r_{\rm collision})^2}= \frac{\sqrt3}{128\sqrt\pi}\frac{\sqrt{R}}{N^{3/2}a^4}.$$ $S=4\pi R^2$ is the area of the sphere. In the Brownian limit, $P\to\infty$ (can you understand why ?) and the fluctuations $\sigma_P$ and $\sigma_x$ are vanishingly small.
Note that it is remarkable that the fluctuations of the positions are independent of $M$ and $T$.