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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.

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Dom Fomello
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  • It just mean a neighborhood / open-ball in $\mathbb{R}^k$? – BGM Feb 23 '20 at 06:14
  • Box in $\mathbb R^k$ containing the origin is the set of $t_1,\ldots, t_k$ such that $$|t_1|\leq a_1, ; |t_2|\leq a_2, ;\ldots,; |t_k|\leq a_k$$ – NCh Feb 23 '20 at 06:32
  • @BGM Can you define that more rigorously? – Dom Fomello Feb 23 '20 at 07:00
  • @NCh Can you please elaborate on this? – Dom Fomello Feb 23 '20 at 07:01
  • Have I not answered your question above? – NCh Feb 23 '20 at 07:33
  • @NCh Well, your answer and user BGM's are different, and since I'm the one asking the question, I have no way of knowing who is correct (or if they're both correct). An elaborated explanation for why what you're saying is the correct answer would help me understand this. – Dom Fomello Feb 23 '20 at 07:50
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    Box is a rectangle. No need to explain smth since either you imagine box as a circle, every rectangle containes a circle and is contained in a circle. So these are equaivalent poits of view. – NCh Feb 23 '20 at 08:06

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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:

  1. The expectation is finite in an open box containing the origin.

  2. The expectation is finite in a closed box containing the origin in its interior.

  3. The expectation is finite in an open cube centered at the origin.

  4. The expectation is finite in a closed cube centered at the origin.

  5. The expectation is finite in an open ball centered at the origin.

  6. The expectation is finite in a closed ball centered at the origin.

The proof is an easy exercise. Note that

  1. The expectation is finite in a closed box containing the origin.

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 $0$.

Paul Frost
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