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I need to find the volume of the 3d space that is given by the following conditions: \begin{array}{c} 0 < x_1 < 1\\ 0 < x_2 < 1\\ 0 < x_3 < 1\\ x_1 + x_2 + x_3 < a. \end{array} I also need to solve this problem for the $n$-dimensional space. Could anybody, please, tell me if this problem is solvable analytically and how one can find the solution?

ADDED

  1. I need results only for a between 0 and 1.

  2. I know how to solve the 2D case. It is trivial. I think that I could even manage to solve the 3D case, but I need a general solution scalable to higher dimensions.

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    Continuing on my last comment... Well, since you know how to solve the 2D case (and since, according to your addendum, this is *trivial*), you could use it to decompose the 3D case as an $x_3$-integral of 2D cases. Likewise for the nD case and the (n+1)D case. (Of course, all this is rather moot since @Thomas Ahle gave$a$solution valid in any dimension, there: http://math.stackexchange.com/questions/31365).2011-08-17

2 Answers 2

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If $X_1,\ldots,X_n$ are inedependent uniform$(0,1)$ random variables, then $ {\rm P}(X_1+ \cdots +X_n \leq a) = {\rm volume}(A), $ where $ A = \{ (x_1 , \ldots ,x_n ) \in (0,1)^n :x_1 + \cdots + x_n < a\} . $ For the probability density function of the sum $X_1+ \cdots +X_n$, see this answer.

EDIT: When $0 < a \leq 1$, it holds $ {\rm volume}(A) = \frac{{a^n }}{{n!}}. $

EDIT 2: Probabilistic proof for the case $0 < a \leq 1$. Let $X_1,X_2,\ldots$ be independent uniform$(0,1)$ variables. We want to show that, for any $0 < a \leq 1$, $ {\rm P}(X_1+ \cdots +X_n \leq a) = \frac{{a^n }}{{n!}}. $ This can be easily done by induction, as follows. The case $n=1$ is trivial: ${\rm P}(X_1 \leq a) = a$. Assume that the result is true for $n$, and let $m = n+1$. By the law of total probability, $ {\rm P}(X_1+ \cdots + X_m \leq a) = \int_0^1 {{\rm P}(X_1 + \cdots + X_m \le a|X_m = u)\,du} $ $ = \int_0^a {{\rm P}(X_1 + \cdots + X_m \le a|X_m = u)\,du} = \int_0^a {{\rm P}(X_1 + \cdots + X_n \le a - u)\,du}. $ Hence, by the induction hypothesis, $ {\rm P}(X_1+ \cdots + X_m \leq a) = \int_0^a {\frac{{(a - u)^n }}{{n!}}\,du} = - \frac{{(a - u)^{n + 1} }}{{(n + 1)!}}\bigg|_0^a = \frac{{a^{n + 1} }}{{(n + 1)!}}. $ The result is thus proved.

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Here's a more geometric take. You're asking about the volume of the convex hull of the $n+1$ points ${\bf 0}, {\bf a}_1, {\bf a}_2, \ldots {\bf a}_n$, where ${\bf a}_i$ is the point with an $a$ in coordinate $i$ and $0$'s elsewhere.

The convex hull of $n+1$ affinely independent points is called an $n$-simplex. The volume of an $n$-simplex is known to equal the volume of its corresponding $n$-parallelotope, divided by $n!$.

Since in this case the corresponding $n$-parallelotope is the $n$-cube $\{ (x_1 , \ldots ,x_n ) \in (0,a)^n\}$, which obviously has volume $a^n$, the volume you want is $a^n/n!$.