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. $\endgroup$ – Ton 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 $\tilde \lambda =\lambda t^{1/\alpha}$. $\endgroup$ – Ton May 15 '18 at 23:06
  • $\begingroup$ Many thanks @Ton. $\endgroup$ – Squird37 May 15 '18 at 23:12

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