Vrftjkry

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: $Salpha S((t − s)^{1/α})$ for $s < t$.


4. 
   $L_t$ has stochastically continuous sample paths.

Here $Salpha S(sigma)$ 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 $Salpha S(1)$, 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

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

and

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

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