1
$\begingroup$

What is the pdf of $\vec{Y} = \frac{\vec{X} }{\lVert \vec{X} \rVert_\infty}$ with $\vec{X}$ a random vector following a multivariate standard normal distribution (zero-mean $\vec{\mu} = 0$ and covariance-matrix $\Sigma = I$ ($I$ denotes identity matrix))?

$\lVert \vec{X} \rVert_\infty$ denotes the uniform-norm with $\lVert \vec{X} \rVert_\infty = \max(\lvert X_1 \rvert,...\lvert X_n \rvert,...\lvert X_N \rvert)$ with $X_n$ the $n$th entry of vector $\vec{X} \in \mathbb{R}^N$.

$\endgroup$
9
  • $\begingroup$ Choose a point uniformly on the unit sphere in $\mathbb R^N$ and inflate it until its $L^\infty$ norm is 1. $\endgroup$ Jan 26, 2012 at 16:20
  • $\begingroup$ What do you mean by "inflate"? $\endgroup$
    – rohrspecht
    Jan 27, 2012 at 15:39
  • $\begingroup$ Multiply it by a constant. (specifically, one over its $L^\infty$ norm) $\endgroup$
    – Will Sawin
    Jan 30, 2012 at 9:17
  • 2
    $\begingroup$ Just think of how a small piece on the cube boundary is projected to the unit sphere (the Jacobian is just the cosine of the angle between the radius-vector and the coordinate direction determining the current face of the cube divided by the length of the vector to the power $n-1$. Normalize it properly and here is your pdf). $\endgroup$
    – fedja
    Mar 8, 2012 at 16:24
  • 2
    $\begingroup$ Draw a 2D picture. You need a distribution on the boundary of the square that, after projecting to the unit circle, becomes a uniform distribution on the circle. Assume that your pdf is $f$. Note that if you have a small interval on the side of the square around a point $x$, then the image on the circle is a small arc, so $f(x)$ must be approximately the ratio of the length of that arc divided by $2\pi$ to the length of the interval. That should remind you of the general change of variable formula in the integral where the Jacobian is what you need. Is that much clear? $\endgroup$
    – fedja
    Mar 10, 2012 at 15:54

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.