projection of a non-zero mean Gaussian vector into a BallLet d denote the dimension, B d denote the ball of radius one in R d . For x ∈ R d let Π B d ( x ) = x max { 1 , ‖ x ‖ 2 } .Consider a fixed vector x ∈ R d with ‖ x ‖ 2 = 1.I am interested in understanding the distribution of Π B d ( x + ξ ) where ξ ∼ N ( 0 , σ 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.)

Question

projection of a non-zero mean Gaussian vector into a BallLet d denote the dimension,             B        d   denote the ball of radius one in             R        d  . For   x  ∈            R        d   let       Π                            B                d              (  x  )  =      x          max      {      1      ,      ‖      x              ‖        2            }      .Consider a fixed vector   x  ∈            R        d   with   ‖  x      ‖    2    =  1.I am interested in understanding the distribution of       Π                            B                d              (  x  +  ξ  ) where   ξ  ∼      N    (  0  ,      σ    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.)

Jaxson White · Accepted Answer

Step 1Since the vector   ξ is isotropic, you can reduce the problem by fixing the direction of x to be ,e.g., along the (1,0,…,0), and since the magnitude of x is one, you can take it to be just (1,0,…,0).Then consider that   ξ projects onto x with a component of magnitude       ξ                  /                    /            , and orthogonally to that with a component of magnitude       ξ    ⊥  .The absolute (unsigned) value of these will be random variables, independent, respectively distributed as       χ    1   and       χ          d      −      1      , with the same sigma as each component of   ξ.Step 2For the parallel component you shall consider the signed value which, being symmetric, will be distributed as a bilateral       χ    1  , i.e. normal.You end up with a noncentral normal summed to a central       χ          d      −      1      , orthogonal to each other and independent. To the magnitude of the resulting vector you shall then apply the projection inside the ball.

Let d denote the dimension, B_d denote the ball of radius one in R^d. For x in R^d let \Pi_{\mathbf{B}_d}(x) = \frac{x}{\max\{1,\||x\||_2\}}. Consider a fixed vector x in R^d with ∥x∥_2=1.

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