For a Brownian motion $B_t$ in $\mathbb R^d$, the transition density of $B_t$ is the normal distribution $$P_x[B_t\in dy]=(2\pi t)^{-d/2}e^{-\frac{|x-y|^2}{2t}}dy$$ and obviously the density is decreasing with $|x-y|$.

My question is does it also hold for rotational $\alpha$-stable process? ($\alpha\in(0,2)$). $X_t$ is called a rotational $\alpha$-stable process, if it is a Levy process and the characteristic function of $P_0[X_t\in dy]$ has the form $e^{-t |z|^\alpha}$.

Then by the inverse Fourier transformation, the transition density has the form $$(2\pi)^{-d}\int_{\mathbb R^d}e^{-i(x-y,z)}e^{-t|z|^\alpha} \, dz$$

I tried but can't figure out from this form whether it is decreasing with $|x-y|$. Any help please.

  • $\begingroup$ Yes, it's decreasing in $|x-y|$. Do you know about subordination? $\endgroup$ – saz May 5 '16 at 5:59
  • $\begingroup$ @saz Yes, I know subordination. But how to get the conclusion? $\endgroup$ – Danielsen May 5 '16 at 12:03

Any rotationally invariant $\alpha$-stable Lévy process can be written as a subordinate process. More precisely, if $(B_t)_{t \geq 0}$ is a ($d$-dimensional) Brownian motion and $(S_t)_{t \geq 0}$ an $\frac{\alpha}{2}$-stable subordinator such that $(B_t)_{t \geq 0}$ and $(S_t)_{t \geq 0}$ are independent, then

$$X_t := B_{S_t}, \qquad t \geq 0,$$

is a rotationally invariant $\alpha$-stable process. Because of the independence of $(B_t)_{t \geq 0}$ and $(S_t)_{t \geq 0}$, we have

$$\mathbb{P}(X_t \in B) = \int_{(0,\infty)} \mathbb{P}(B_s \in B) \, \mathbb{P}(S_t \in ds). \tag{1}$$

Plugging in the transition density of $(B_t)_{t \geq 0}$, we find

$$\mathbb{P}(X_t \in B) = \int_{(0,\infty)} \int_B (2\pi s)^{-d/2} \exp \left(- \frac{|x|^2}{2s} \right) \, dx \, \mathbb{P}(S_t \in ds).$$

By Tonelli's theorem, we may interchange the integrals and therefore

$$\mathbb{P}(X_t \in B) = \int_B \underbrace{ \int_{(0,\infty)} (2\pi s)^{-d/2} \exp \left(- \frac{|x|^2}{2s} \right) \, \mathbb{P}(S_t \in ds)}_{=: p_t(x)} \, dx$$

which shows that $X_t$ is absolutely continuous with respect to Lebesgue measure with density

$$p_t(x) = \int_{(0,\infty)} (2\pi s)^{-d/2} \exp \left(- \frac{|x|^2}{2s} \right) \, \mathbb{P}(S_t \in ds).$$

It follows directly from this representation and the monotonicity of $\exp$ that $p_t$ is decreasing in $|x|$.

Remark: In $(1)$ we have used that $\mathbb{P}(S_t = 0)=0$ for all $t \geq 0$. To prove this, note that

$$\{S_t = 0\} \subseteq \{X_t = 0\}$$

and therefore $\mathbb{P}(S_t=0) \leq \mathbb{P}(X_t=0) = 0$ since $X_t$ is absolutely continuous with respect to the Lebesgue measure (just note that the characteristic function of $X_t$ equals $e^{-t |\xi|^{\alpha}}$, and this function is obviously Lebesgue-integrable.)

  • $\begingroup$ Didn't know the relation between subordinator and rotational stable process before. Thank you very much! $\endgroup$ – Danielsen May 5 '16 at 16:39

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