Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am not expecting a "realistic" answer to my question, since it is based on an impossible scenario. What I'm waiting for is a theoretic explanation/interpretation so that I can sleep at night :)

Let's take n reals from the interval (0; 1), evenly distributed. What is the probability of the sum of squares being less than 1? With a geometric approach it is relatively easy to see that the solution is $$P\left(\sum_{i=1}^nx_i^2<1\right)={\pi^{\frac n 2}\over 2^n\cdot{\frac n 2}!}=\frac {volume \;of\;hypersphere}{volume\;of\;hypercube}$$ This solution works fine for all nonnegative integer n (for odd n we compute the factorial using the $\Gamma$ function).

But what if n could be something else, namely a real from (0; 1)? In that case the resulting probability is greater than 1, (properly) indicating that something went wrong. Is there a way to make sense of this result?

share|improve this question
    
What do you mean by non integral $n$? What is \12 of a number? What is $\sqrt 2$ of a number? –  Ross Millikan Nov 9 '12 at 19:17
    
You may as well choose $2$ out of $-1$ objects in ${{-1}\choose 2}=-1$ ways... –  tomasz Nov 9 '12 at 19:36
    
@tomasz You have to be careful there since $\binom{-1}{2}$ involves (-1)! and (-3)!, which cannot be defined. You can take a limit though, which gives +1. How did you compute that -1 ways? –  Dave Nov 9 '12 at 22:51
    
Dave: For every n, n!=n(n-1)! hence (-1)!=(-1)(-2)(-3)!=2(-3)!, whatever (-3)! is. –  Did Nov 10 '12 at 12:05
1  
@Dave: ${\alpha \choose k}=\frac{\alpha^{\underline k}}{k!}$ (see e.g. Wikipedia). It's an extrapolation that, at first glance, doesn't make much combinatorial sense – just like yours. Although the generalization I mentioned actually does make sense in context of infinite series expansions, which I wouldn't be too sure about in your case. That's how I computed it, although I made a mistake, it's actually $1$, not $-1$. :) ${-1 \choose 1}=-1$, though... –  tomasz Nov 10 '12 at 12:43

2 Answers 2

The result can make sense, but I think the probability part is a red herring. You have a function which is relevant to your probability function for integer $n$: \begin{equation}f\left(n\right)=\frac{\pi^{\frac{n}{2}}}{2^{n}\cdot\frac{n}{2}!}\end{equation} where $\frac{n}{2}!$ is defined for odd $n$ based on $\frac{1}{2}!=\frac{\sqrt{\pi}}{2}$. If you choose to extend this with the $\Gamma$ function, then $f$ goes above $1$.

However, the important point to note is that this isn't a completely arbitrary choice/occurrence. If you want to extend $f$ to a nice (i.e. analytic) function on $\mathbb{R}$ (or some big interval thereof), then since $f(0)=f(1)=1$, and it can't be constant on $[0,1]$, it's forced to go above $1$ somewhere.

As an aside, if you want an aesthetic reason to use the Gamma function in particular, the Bohr–Mollerup theorem says that the Gamma function (on the positive reals) is the only function $g$ with $g(1)=1$ (corresponding to $0!=1$), obeys the functional identity $g(x+1)=xg(x)$ (which applies even for half-integer factorials), and is "log-convex" (so that $\log g(x)$ is convex).

share|improve this answer

The half-diagonal of a hypercube of sidelength $2$ is $\sqrt n$.

If your non-integral "dimension" is smaller than 1, then clearly the half-diagonal is smaller than the radius 1 and so the "sphere" is not contained in the "cube" anymore, so it is quite consistent that now you would reverse your quotient to calculate the probability that a point in one set is included in the other.

share|improve this answer
    
Isn't that simply $\sqrt n$? BTW +1, you got a valid point there :) –  Dave Dec 4 '12 at 20:01
    
Yes, thanks, corrected it. –  Phira Dec 4 '12 at 23:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.