# Distribution of infinity-norm over the unit sphere

I need to compute probabilities of the form $P( \Vert X \Vert_\infty < r ),$ where $X$ is a random variable of dimension $n$, drawn with a uniform distribution on the unit sphere $\mathcal{S}_{n-1}$. Clearly, the distribution of $Z = \Vert X \Vert_\infty$ is supported by the interval $[\frac{1}{\sqrt{n}},1]$.

Any hint to get a simple formula (ideally a closed-form expression or a one-dimensional integral) will be greatly appreciated !

Note: Using the fact that $X \sim \frac{Y}{\Vert Y\Vert_2}$ for $Y \sim \mathcal{N}(0,1)$, at first I thought I could first compute the probabilities $P_i:=P( |X_i| < r \Vert X \Vert_2)$, which can be expressed as a quantile of the F-distribution, but then I realized that those events are not independent, so $P( \Vert X \Vert_\infty < r ) \neq \prod_{i=1}^n P_i$.

• If $r>\frac{1}{\sqrt{2}}$ we have $$P(\|X\|_\infty>r)=nP(X_1>r)=n\int_r^1 \sqrt{1-t^2}^{n-2}C(n)dt.$$ For $r<\frac{1}{\sqrt{2}}$ one could use inclusion-exclusion principle to compute probabilities. However I doubt there exist a closed form solution. Jul 6, 2015 at 12:12
• Thanks ! This a a nice trick for the case $r>\frac{1}{\sqrt{2}}$ indeed. What is the function $C(n)$ in the integral ? Jul 7, 2015 at 8:44
• $C(n)$ is the measure of the $n-2$-dimensionale unit sphere. Jul 7, 2015 at 19:41
• @user35593 More precisely, twice your number, no? Because one also has to count $P(X_1<-r)$ Dec 10, 2022 at 18:37
• @user35593 In fact, for the 1-sphere one must obtain $\frac4\pi\arccos(r)$ and I cannot obtain this from your formula? Dec 10, 2022 at 19:08

There is a trick to reduce these kinds of questions to questions about independent normals, as in my ancient preprint. For large $$n,$$ concentration of measure will presumably give you easy estimates.

• I'm not sure how to use your result, because the function $\mathbf{1}_{\{\Vert x \Vert_\infty < r\}} (x)$ is not homogeneous of degree $d$ for any $d$ ? Jul 7, 2015 at 8:46

Edit: As I write at the end. This is wrong for low dims, but approximately correct for more than a few hundreds, when I guess the coordinates become less dependent.

You can start by uniformly selecting points on the (n − 1)-sphere and then ask what is the infinity norm of such points.

Given a vector $$Z$$ with $$n$$ coordinates taken i.i.d. from the normal distribution $$Z_i \sim \mathcal{N}(0, 1)$$. The CDF of the absolute maximum of those is given by considering the CDF of $$n$$ points drawn from the half normal distribution $$F_{||Z||_{\infty}}(x)={\operatorname{erf}\left( \frac{ x }{\sqrt 2} \right) }^{n}$$. Now we need to normalize the points, so they'll be on the unit sphere. Given that the positive squared root of the sum of squared normal variables has a Chi distribution with $$n$$ degress of freedom, $$||Z||_2\sim \chi_n$$, we have:

$$P(||X||_{\infty} < r\;|\;||X||_2=1) = \\P(||Z||_{\infty}/||Z||_2 < r) =$$ [wrong transition! (Though it still works in high-dims)] $$\int_{0}^{\infty} P(||Z||_{\infty}/ ||Z||_2 < r\; |\; ||Z||_2 = x) f_{\chi_n}(x)\;dx = \\ \\\int_{0}^{\infty} F_{||Z||_{\infty}}(x \cdot r) f_{\chi_n}(x)\;dx = \\\int_{0}^{\infty} {\operatorname{erf}\left( \frac{ x \cdot r}{\sqrt 2} \right) }^{n} \cdot \dfrac{x^{n-1}e^{-x^2/2}}{2^{n/2-1}\Gamma\left(\frac{n}{2}\right)}\; dx$$

Update

Alternatively, I'm not entirely sure we can do this, but I think we can just consider the expected scale of $$r$$, instead of calculating it as above. It is the mean of $$\chi_n$$, so the above should be the same as:

$$F_{||Z||_{\infty}}(\mu(\chi_n) \cdot r)$$

Update2

These formulations seem to be correct for large enough $$n$$, but not in general:

• It cannot be true for all values of $r$ because the last integral is certainly analytic when $r>0$, while the answer definitely isn't. Dec 9, 2022 at 23:54
• The integral is defined over the normalization factor to get to the n-sphere so it is fine I don't understand the meaning of this phrase. Neither do I understand how it becomes constant $1$ for large $r$ (I mean the integral with $erf$, of course) Dec 10, 2022 at 2:32
• Yes, the entire formula: the last one in the chain. If $r\ge 1$, the whole sphere is there, so the output should be $1$ regardless of $r$. Dec 10, 2022 at 2:42
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• @მამუკაჯიბლაძე Marsaglia (1972), en.m.wikipedia.org/wiki/… Dec 10, 2022 at 15:57