I am having difficulties to understand the form of the Levy measure of the multivariate Levy-stable motion. Let me start by defining the one dimensional motion in order to clarify my question.

The univariate $\alpha$-stable Levy motion ($L_t$) is defined as follows:

  1. $L_0 = 0$ almost surely.
  2. For $t_0 < t_1 < \dots< t_N$, the increments $(L_{t_n} −L_{t_{n−1}})$ are independent ($n = 1,\dots, N$).
  3. The difference $(L_t − L_s)$ and $L_{t−s}$ have the same distribution: $S\alpha S((t − s)^{1/α})$ for $s < t$.
  4. $L_t$ has stochastically continuous sample paths.

Here $S\alpha S(\sigma)$ denotes the symmetric $\alpha$-stable distribution with scale parameter $\sigma$.

The Levy measure of a univariate $\alpha$-stable Levy motion is given as follows:

$$\nu(dx) = \frac1{|x|^{\alpha+1}}dx.$$

I know that this measure is related to the characteristic function of $S\alpha S(1)$, which is $\exp(-|w|^\alpha)$.

My confusion arises when we define this motion in $\mathbb{R}^d$: even though I can perfectly understand that the characteristic function of a multivariate symmetric $\alpha$-stable variable (a.k.a elliptically contoured) becomes $\exp(\|w\|^\alpha)$, I am getting confused by the corresponding Levy measure that is defined as follows:

$$\nu(dx) = \frac1{\|x\|^{\alpha+d}}dx.$$

I am not able to understand why the term $d$ appears in the exponent. I would be very happy if someone can shed some light on this issue.


Roughly speaking, the $d$ in the exponent comes into play because of the change of variables formula in $\mathbb{R}^d$: $$\int_{\mathbb{R}^d} f(ry) \, dy = r^{-d} \int_{\mathbb{R}^d} f(y) \, dy. \tag{1}$$

Denote by $$\psi(\xi) := \int_{\mathbb{R}^d \backslash \{0\}} (1-\cos(y \cdot \xi)) \frac{1}{|y|^{d+\alpha}} \, dy$$ the characteristic exponent associatd with the measure $\nu(dy) = |y|^{-d-\alpha} \, dy$. Since $\psi$ is rotationally invariant, we have $$\psi(\xi) =\psi(|\xi| e_1)= \int_{\mathbb{R}^d \backslash \{0\}} (1-\cos(|\xi| y \cdot e_1)) \frac{1}{|y|^{d+\alpha}} \, dy$$ where $e_1 = (1,0,\ldots,0)^T$ denotes the first unit vector in $\mathbb{R}^d$. Now a change of variables, $z := |\xi| y$ shows by $(1)$ that

$$\psi(\xi) = |\xi|^{\alpha} \underbrace{\int_{\mathbb{R}^d \backslash \{0\}} (1-\cos(z \cdot e_1)) \frac{1}{|z|^{d+\alpha}} \, dz}_{=: C_{d,\alpha}} = C_{d,\alpha} |\xi|^{\alpha}.$$

Note that we need the exponent $d$ in the dominator in order to cancel the $|\xi|^d$-term which comes into play because of $(1)$; otherwise we would end up with a different power of $|\xi|$.

Alternatively, you can also see from the integrability condition $\int_{\mathbb{R}^d \backslash \{0\}} \min\{1,|y|^2\} \, \nu(dy) < \infty$ (which any Lévy measure $\nu$ on $\mathbb{R}^d$ has to satisfy) that we need the exponent $d$; this is due to the fact that

$$\int_{\{y \in \mathbb{R}^d; 0<|y|<1\}} |y|^{-\beta} \,d y < \infty \iff \beta<d$$


$$\int_{\{y \in \mathbb{R}^d; |y| \geq 1\}} |y|^{-\beta} \,d y < \infty \iff \beta>d.$$

Using these characterizations you can easily show that $\nu(dy)=|y|^{-d-\alpha}$ is a $d$-dimensional Lévy measure if, and only if, $\alpha \in (0,2)$.


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