In this post, I asked about the calculation for the joint distribution of two iid $\chi^2$ random variables $X,Y$ and their sum $S=X+Y$, when there are constraints to the three variables. In particular, I needed to pay attention to the bounds of integration.
I want to consider the case with three $\chi^2$ random variables $X,Y,Z$ and their sum, $S=X+Y+Z$ and the probability: $$ P(X+Y+Z>s,X\leq c,Y\leq c, Z\leq c) $$
As before, we can use independence to arrive at the integral: $$ P(X+Y+Z>s,X\leq c,Y\leq c, Z\leq c)=\int_{\mathcal{X}\times\mathcal{Y}\times\mathcal{Z}} f_X(x)f_Y(y)f_Z(z)~dx~dy~dz $$ where $f(\cdot)$ is the $\chi^2$ pdf and $\mathcal{X}\times\mathcal{Y}\times\mathcal{Z}$ is the integration region.
When $s<0$ or $s>3c$, the probabilities are 1 and 0, respectively. Therefore, we consider the following relationships between $s$ and $c$:
Case 1: $0\leq s\leq c$
Like in the post, we can consider the bounds formed by the lower right triangular region in each direction and subtract it from 1. Therefore, we have $$ P(X+Y+Z>s,X\leq c,Y\leq c, Z\leq c)=1-\int_{0}^s\int_{s-z}^c\int_{s-(y+z)}^c \dots $$
Case 3: $2c < s \leq 3c$
We can consider the bounds formed by the upper right triangular projection in each direction. Therefore, we have $$ P(X+Y+Z>s,X\leq c,Y\leq c, Z\leq c)=\int_{s-2c}^c\int_{s-z}^c\int_{s-(y+z)}^c \dots $$
Case 2: $c< s\leq 2c$
Question Here: The plane $x+y+z=s$ intersects the cube formed by $[0,c]^3$ so that the frustrum we want has a hexagonal base with coordinates \begin{gather} (s-c,0,c)\\ (c,0,s-c)\\ (c,s-c,0)\\ (s-c,c,0)\\ (0,c,s-c)\\ (0,s-c,c) \end{gather}
This is where I'm stuck, as I'm having trouble deriving the bounds of integration for this case.
EDIT
I've thought a little bit more about Case 2 and have the following additional work:
Case 2: $c< s\leq 2c$
First, the hexagonal slice defines the bounds for $z$: $s-(x+y)\leq z\leq c$. For the $x$ and $y$ bounds, consider the projection of the hexagonal slice onto the $xy$-plane. On one hand you have the area defined by $y\leq s-x$ and on the other hand, you have $y\geq s-c-x$ because the lines $x+y=s$ and $x+y=s-c$ create three disjoint regions. Integrating over the desired region requires two pieces so that: $$ P(X+Y+Z>s,X\leq c,Y\leq c, Z\leq c)=\int_0^{s-c}\int_{s-c-x}^c\int_{s-(x+y)}^c \dots + \int_{s-c}^c\int_0^{s-x}\int_{s-(x+y)}^c \dots $$