projection of a non-zero mean Gaussian vector into a Ball

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Let $d$ denote the dimension, $\mathbf{B}_d$ denote the ball of radius one in $\mathbb{R}^d$. For $x\in \mathbb{R}^d$ let $\Pi_{\mathbf{B}_d}(x) = \frac{x}{\max\{1,\|x\|_2\}}$.

Consider a fixed vector $x \in \mathbb{R}^d$ with $\|x\|_2=1$.

I am interested in understanding the distribution of $\Pi_{\mathbf{B}_d}(x+\xi)$ where $\xi \sim \mathcal{N}(0,\sigma^2 I_d)$ is a isotropic Gaussian vector with zero mean.

(Finding the exact distribution might be challenging, however, I am interested in any ideas how to play with this random variable.)

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Since the vector $ \bf \xi$ is isotropic, you can reduce the problem by fixing the direction of $\bf x$ to be ,e.g., along the $(1,0, \ldots,0)$, and since the magnitude of $\bf x$ is one, you can take it to be just $(1,0, \ldots,0)$.

Then consider that $\bf \xi$ projects onto $\bf x$ with a component of magnitude $\xi _{//}$, and orthogonally to that with a component of magnitude $\xi _ \bot$.

The absolute (unsigned) value of these will be random variables, independent, respectively distributed as $\chi _1$ and $\chi _{d-1}$, with the same sigma as each component of $\xi$.

For the parallel component you shall consider the signed value which, being symmetric, will be distributed as a bilateral $\chi _1$, i.e. normal.

You end up with a noncentral normal summed to a central $\chi _{d-1}$, orthogonal to each other and independent. To the magnitude of the resulting vector you shall then apply the projection inside the ball.