# (Theoretic) probability greater than 1

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?

-
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
@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

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 $n$: $$f\left(n\right)=\frac{\pi^{\frac{n}{2}}}{2^{n}\cdot\frac{n}{2}!}$$ 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).

-

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.

-
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