I'm interested in how the expected cosine proximity between a vector $x\in \mathbb{R}^N$ and a perturbed version of it, $\tilde{x}=x+\varepsilon$, where $x\sim \mathcal{N}(0,\sigma I_N)$ and $\varepsilon\sim \mathcal{N}(0,\tilde{\sigma}I_N)$, scales a a function of $N$.
(EDIT: Note that I should have written $\sigma^2, \tilde{\sigma}^2$, but as the answer below uses the above notation, I will keep this as is. So, for the following, $\sigma$ denotes the variance.)
That is, what is
$$ \mathbb{E}_{x,\varepsilon}[\cos(\theta)] = \mathbb{E}_{x,\varepsilon}\left[\left\langle \frac{x}{\| x\|} \cdot \frac{\tilde{x}}{\| \tilde{x}\|} \right \rangle \right] $$
as a function of $N$?
Without loss of generality, we can assume that $\sigma = \| x \| = 1$. We cannot assume that $\tilde{\sigma} \ll \sigma$, but - if it's helpful - we could assume that $\tilde{\sigma} \leq \sigma$.
I have tried writing out the integrals, but it's a mess, and I have the feeling that there should exist a much more elegant geometric solution, which eludes me at the moment.
If there is no immediate closed-form solution to the general problem, the situation where $N\gg 1$ is the most relevant for my specific application.
EDIT: Perhaps this could be useful: Expected value of dot product between a random unit vector in $\mathbb{R}^N$ and another given unit vector
Another interesting aspect is how this expected value changes as a function of of the ratio between the two standard deviations. I have done some simulations on this:
For some ratio, the expected value always converges to some number for $N\rightarrow \infty$. In the plot below, the converged value (for $N=100$, which seemed to be more than enough) is plotted as a function of the ratio.
We clearly see that, as the size of the perturbation (captured by $\tilde{\sigma}$) increases, $x$ and $\tilde{x}$ become independent, and therefore orthogonal, which is expected.
Any insights would be appreciated!
The expectation is undefined, as $\tilde x$ could be equal to $-x$. However, that event has measure zero so you can condition on it not being the case, or define the cosine to be zero in that case.
Let us assume $\|x\|=1$, as you did. No matter what vectors you sample, rotating them will not change the cosine and your distributions are invariant under orthogonal transformations, so you can assume $x=e_1=(1,0,0,\dots)$. Thus, using $\epsilon$ to denote the rotated vector in a slight abuse of notation, $$ \cos(\theta) = \frac{1+\epsilon_1}{\|\tilde x\|}. $$ So $$ \mathbb E[\cos(\theta)] = \mathbb E\left[\frac{1+\epsilon_1}{\|\tilde x\|}\right] $$
$$ = \frac{\mathbb E\left[1+\epsilon_1\right]}{\mathbb E\left[\|\tilde x\|\right]} - \mathbb E\left[(1+\epsilon_1)\left(\frac{1}{\|\tilde x\|}-\frac{1}{\mathbb E[\|\tilde x\|]}\right)\right] $$
$$ = \Theta\left(\frac{1}{(\sigma+\tilde\sigma)\sqrt{N}}\right) - \mathbb E\left[(1+\epsilon_1)\left(\frac{1}{\|\tilde x\|}-\frac{1}{\mathbb E[\|\tilde x\|]}\right)\right]. $$ I think it is clear that the r.h.s. of the difference vanishes with $N$.
For the second equality see
For the expected value of the norm see this.