I am presented with the definition of a joint moment generating function:
The joint moment generating function (joint MGF) of a random vector $\mathbf{X} = (X_1, \dots, X_k)$ is the function $M$ defined by
$$M(\mathbf{t}) = E(e^{\mathbf{t}' \mathbf{X}}) = E(e^{t_1 X_1 + \dots + t_kX_k}),$$
for $\mathbf{t} = (t_1, \dots, t_k) \in \mathbb{R}^k$. We require this expectation to be finite in a box containing the origin in $\mathbb{R}^k$; otherwise we say the joint MGF does not exist.
Isn't a "box" (a cube?) a geometric object in $\mathbb{R}^3$? I'm confused how it make sense to talk about a "box" for $\mathbb{R}^k$? Thank you.
A box $B$ in $\mathbb R^k$ (also called a $k$-dimensional box) is the product of $k$ real intervals $J_1,\ldots,J_k$ of finite length: $$B = J_1 \times \ldots \times J_k. $$ A box is open / closed if all intervals are open / closed. The definition of Nch is just a special case of a closed box $B$ which is centered at the origin: $B = [-a_1,a_1] \times \ldots \times [-a_k,a_k]$. A cube is a box such that all $J_i$ have the same length.
The phrase "... finite in a box containing the origin in $\mathbb R^k$ ..." is not precise enough. I believe the box is tacitly assumed to be open. If we allow closed boxes, then we should additionally require that the origin is contained in the interior of the box.
In fact, the following are equivalent:
The expectation is finite in an open box containing the origin.
The expectation is finite in a closed box containing the origin in its interior.
The expectation is finite in an open cube centered at the origin.
The expectation is finite in a closed cube centered at the origin.
The expectation is finite in an open ball centered at the origin.
The expectation is finite in a closed ball centered at the origin.
The proof is an easy exercise. Note that
is not equivalent to the other conditions. As an example take $[0,1] \times \ldots \times [0,1]$. This contains the origin, but does not contain any neighborhood of $0$.