Solved – Finite $k$th moment for a random vector

If $X sim F$ where the support of $X$ is $mathbb{R}^p$. So, $X = (X_1, X_2, dots, X_p)$. Then say I assume $X$ has $k$ finite moments. When $p = 1$, I know that that means
$$int_{mathbb{R}} x^k, f(x), dx < infty, $$
where $f(x)$ is the associated density of $F$. What the mathematical equivalent of assuming $X$ has $k$ finite moments when $p > 1$?

In this link, on page 2, the authors define the $k$th moment as
$$E|X|^k = int |X|^k f(x) , dx, $$
where $| cdot|$ is the Euclidean norm.

Glen_b's answer here suggests that the $k$th moment would be
$$int x_1^kx_2^k dots x_p^k , f(x) dx. $$

Does assuming one to be finite imply the other is finite?

The answer is in the negative, but the problem can be fixed up.

To see what goes wrong, let $X$ have a Student t distribution with two degrees of freedom. Its salient properties are that $mathbb{E}(|X|)$ is finite but $mathbb{E}(|X|^2)=infty$. Consider the bivariate distribution of $(X,X)$. Let $f(x,y)dxdy$ be its distribution element (which is singular: it is supported only on the diagonal $x=y$). Along the diagonal, $||(x,y)||=|x|sqrt{2}$, whence

$$mathbb{E}left(||(X,X)||^1right) = mathbb{E}left(sqrt{2}|X|right) lt infty$$


$$iint x^1 y^1 f(x,y) dx dy = int x^2 f(x,x) dx = infty.$$

Analogous computations in $p$ dimensions should make it clear that $$intcdotsint |x_1|^k|x_2|^kcdots |x_p|^k f(x_1,ldots, x_p)dx_1cdots dx_p$$

really is a moment of order $pk$, not $k$. For more about multivariate moments, please see Let $mathbf{Y}$ be a random vector. Are $k$th moments of $mathbf{Y}$ considered?.

To find out what the relationships ought to be between the multivariate moments and the moments of the norm, we will need two inequalities. Let $x=(x_1, ldots, x_p)$ be any $p$-dimensional vector and let $k_1, k_2, ldots, k_p$ be positive numbers. Write $k=k_1+k_2+cdots k_p$ for their sum (implying $k_i/k le 1$ for all $i$). Let $q gt 0$ be any positive number (in the application, $q=2$ for the Euclidean norm, but it turns out there's nothing special about the value $2$). As is customary, write

$$||x||_q = left(sum_i |x_i|^qright)^{1/q}.$$

First, let's apply the AM-GM inequality to the non-negative numbers $|x_i|^q$ with weights $k_i$. This asserts that the weighted geometric mean cannot exceed the weighted arithmetic mean:

$$left(prod_i (|x_i|^q)^{k_i}right)^{1/k} le frac{1}{k}sum_i k_i|x_i|^q.$$

Overestimate the right hand side by replacing each $k_i/k$ by $1$ and take the $k/q$ power of both sides:

$$prod_i |x_i|^{k_i} = left(left(prod_i (|x_i|^q)^{k_i}right)^{1/k}right)^{k/q} le left(sum_i |x_i|^qright)^{k/q} = ||x||_q^k.tag{1}$$

Now let's overestimate $||x||_q$ by replacing each term $|x_i|^q$ by the largest among them, $max(|x_i|^q) = max(|x_i|)^q$:

$$||x||_q le left(sum_i max(|x_i|^q)right)^{1/q} = left(p max(|x_i|)^qright)^{1/q} = p^{1/q} max(|x_i|).$$

Taking $k^text{th}$ powers yields

$$||x||_q^k le p^{k/q} max(|x_i|^k) le p^{k/q} sum_i |x_i|^k.tag{2}$$

As a matter of notation, write

$$mu(k_1,k_2,ldots,k_p) = intcdots int |x_1|^{k_1}|x_2|^{k_2}cdots|x_p|^{k_p} f(x),dx.$$

This is the moment of order $(k_1,k_2,ldots,k_p)$ (and total order $k$). By integrating aginst $f$, inequality $(1)$ establishes

$$mu(k_1,ldots,k_p) le intcdotsint ||x||_q^k f(x),dx = mathbb{E}(||X||_q^{k})tag{3}$$

and inequality $(2)$ gives $$mathbb{E}(||X||_q^{k})le p^{k/q}left(mu(k,0,ldots,0) + mu(0,k,0,ldots,0) + cdots + mu(0,ldots,0,k)right).tag{4}$$

Its right hand side is, up to a constant multiple, the sum of the univariate $k^text{th}$ moments. Together, $(3)$ and $(4)$ show

  • Finiteness of all univariate $k^text{th}$ moments implies finiteness of $mathbb{E}(||X||_q^{k})$.

  • Finiteness of $mathbb{E}(||X||_q^{k})$ implies finiteness of all $mu(k_1,ldots,k_p)$ for which $k_1+cdots +k_p=k$.

Indeed, these two conclusions combine as a syllogism to show that finiteness of the univariate moments of order $k$ implies finiteness of all multivariate moments of total order $k$.


For all $q gt 0$, the $k^text{th}$ moment of the $L_q$ norm $mathbb{E}(||X||_q^{k})$ is finite if and only if all moments of total order $k$ are finite.

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