I'm trying to prove that, for a $\alpha-$stable subordinator, with $\alpha\in(0,1)$ the next equality holds:$X_{t}\overset{d}{=}t^{1/\alpha}X_{1}.$

The definition of $\alpha-$stable subordinator is the next:

A stable subordinator with index $\alpha\in(0,1)$ is a subordinator with zero drift and Lévy measure $\frac{c}{x^{1+\alpha}}1_{\{(0,\infty)\}}(x)dx,$ where $c$ is a positive constant.

I've computed the Laplace transform of such subordinator,which is $E(e^{-\lambda X_{t}})=e^{-tc\Gamma(1-\alpha)\lambda^{\alpha}}.$ So, I'd like to prove that this Laplace transform is equal to the Laplace transform of $t^{1/\alpha}X_{1}$ but I don't get such equality.Prove the behind finishes the proof because of the uniqueness of Laplace transforms and distribution functions.

By other hand, I was thinking that,a subordinator is an increasing Lévy process,so It is Infinitely Divisible; then it satisfies $X_{t}\overset{d}{=}tX_{1},$ but I don't find how this equality in distribution helps to get $X_{t}\overset{d}{=}t^{1/\alpha}X_{1}.$

Any kind of help is thanked in advanced.

  • $\begingroup$ Infinite divisibility does not imply $X_t=tX_1$. You can see that it does not hold for Brownian motion. $\endgroup$ – Rgkpdx May 15 '18 at 22:55

$$E(e^{-\lambda X_{t}})=e^{-tc\Gamma(1-\alpha)\lambda^{\alpha}}=e^{-c\Gamma(1-\alpha)(\lambda t^{1/\alpha})^{\alpha}}=E(e^{-\lambda t^{1/\alpha} X_{1}}).$$

  • $\begingroup$ Thanks to answer @Ton. I don't get the last equality. I had thought the same but I don't know why. Could you explain it? $\endgroup$ – Squird37 May 15 '18 at 22:58
  • $\begingroup$ Compute $E(e^{-\tilde \lambda X_1})=e^{-c\Gamma(1-\alpha)\tilde \lambda^\alpha}$, and now pick $\tilde \lambda =\lambda t^{1/\alpha}$. $\endgroup$ – Rgkpdx May 15 '18 at 23:06
  • $\begingroup$ Many thanks @Ton. $\endgroup$ – Squird37 May 15 '18 at 23:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.