# The Levy measure of a multivariate alpha-stable Levy motion

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

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, $$\alpha \in (0,2)$$.