# Distribution of uniform-normed random vector

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 $\vec{X} \in \mathbb{R}^N$.

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Choose a point uniformly on the unit sphere in $\mathbb R^N$ and inflate it until its $L^\infty$ norm is 1. –  Anthony Quas Jan 26 '12 at 16:20
What do you mean by "inflate"? –  Christian Rohlfing Jan 27 '12 at 15:39
Multiply it by a constant. (specifically, one over its $L^\infty$ norm) –  Will Sawin Jan 30 '12 at 9:17
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). –  fedja Mar 8 '12 at 16:24
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? –  fedja Mar 10 '12 at 15:54