I was looking at the following question (link). To summarise, the aim is to find, for standard uniform distributions $U_i$,
$$P(U_1+\ldots+U_n \leq 1)$$
for $0\leq t\leq1$. A nice proof by induction is shown in that question. However, if we consider the 2-dimensional case, we have
$$P(U_1 + U_2 \leq t) = t^2 / 2$$ which is the area of an isosceles triangle with base and height of length $t$. For the 3D case we have a pyramid. Can this geometric argument be extended to $n$-dimensions?
Yes. The set of all values for the $U$s is $[0, 1] \times [0, 1] \times \ldots \times [0, 1]$, an $n$-dimensional cube.
The "face" defined by $$ U_1+\ldots+U_n = 1 $$ is a pyramid-like shape in $n$-dimensions known as an $(n-1)$-simplex, which is the thing that mathematicians generally agree is the generalization of a triangle to higher dimensions.
Let's call this specific $(n-1)$-simplex by the name $K$.
Every point in the set defined by $$ U_1+\ldots+U_n \leq 1 $$
is actually in the "cone" on $K$ from the origin, i.e., it's pyramid-like, with the origin as the apex. (BTW, the "pyramid" in the 3D case isn't a traditional Egyptian pyramid -- it's actually a tetrahedron, but I think that was what you intended).
That "cone on the set $K$" is defined as the set of all linear combinations of the form $$ (1-t)\cdot (0, \ldots, 0) + t \cdot (v_1, \ldots, v_n), $$ where $t$ ranges from $0$ to $1$ and $(v_1, \ldots, v_n) $ is a point of $K$. If you have some nonzero point $$ (u_1, \ldots, u_n) $$ in the set defined by $$ U_1+\ldots+U_n \leq 1 $$ you can let $$ t = u_1+\ldots+u_n $$ and $ v_i = \frac{u_i}{t} $, which gives the correspondence between this cone and the set you're interested in.
If you actually draw out all this stuff in the cases $n = 2, 3$, you'll pretty quickly see how it generalizes.
