# 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?

Contents

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\$\$

whereas

\$\$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\$.

Thus,

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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