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If a point is chosen at random in an $N$-dimensional unit sphere, what is the probability of falling inside the sphere of radius $0.99999999$? What if $N=3$, $N=10^{23}$, or $N = \infty $?

Okay, that is the question I have to address. I don't know how to bring the 2 concepts of Probability and $N$-dimensional hypersphere together to arrive at a solution. I found similar work 1, and 2. I still don't understand the underlying concept that would allow me to find $P(x,y)$.

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related material 1, related material 2 –  Paresh471 May 10 '13 at 4:00
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One should not use lower-case $n$ and capital $N$ interchangeably. –  Michael Hardy May 10 '13 at 5:24
    
This may help you: en.wikipedia.org/wiki/Volume_of_an_n-ball –  FiveLemon May 10 '13 at 5:28
    
I will try to never make said error again. –  Paresh471 May 10 '13 at 15:33
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1 Answer 1

up vote 1 down vote accepted

Assuming a uniform distribution, the probability would be the ratio of the volumes.

In a $n$-dimensional space, the volume of a $n$-sphere of radius $r$ is a constant times $r^n$. Since we are dividing two volumes, the constant goes away, so the probability is $r^n$, where $r$ is your $0.999999$.

Various estimates can be made depending on how close $r$ is to $1$ and how large $n$ is.

If $r = 1-c$ where $c$ is small, this becomes $(1-c)^n \approx 1-n c$ for $n$ small compared with $1/c$.

Since $(1-c)^n = e^{n \ln(1-c)}$, if $n c^2$ is small, $n \ln(1-c) \approx -nc$, so $(1-c)^n\approx e^{-nc} $

If $r = 1-c/n$, this becomes $(1-c/n)^n \to e^{-c}$ as $n \to \infty$.

And so it goes.

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