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I'm studying the book "Probability and Stochastics" by Erhan Cinlar - the probability class that I took covered until the beginning of chapter 7, and I'm now trying to do some of the exercises but having difficulties with some calculations.

The exercise 2.34 (page 337) goes as follows (the purpose is to show that $\alpha$ must be in $(0,2]$).

Fix $a >0$. Suppose that X is an $\alpha$-stable non-degenerate Levy process in $\mathbb{R}$ with characteristic exponent $\psi$.

  1. Show that $t \psi(r) = \psi(t^{1/\alpha} r)$ for $t>0, r \in \mathbb{R}$.
  2. Show that $\psi(r) = c r^\alpha$ for some complex constant $c$ for $r \in \mathbb{R}_+$.
  3. Suppose $X$ and $-X$ have the same probability law, i.e. $\psi(r) = \psi(-r)$ for all $r$. Show that $\psi(r) = c |r|^\alpha$ for all $r \in \mathbb{R}$.

As mentioned, I'm having problems with the calculations - I'm not quite sure how to work with the characteristic exponent here (although I think the exercise is supposed to be quite trivial, i.e. just some minor rearrangements should solve it).

My current work:

Recall that non-degenerate means that $X_t \neq 0$ almost surely for all $t$. Also, recall that the characteristic exponent of a Levy process is given by the Levy-Khintchine formula: $$ \psi(r) = ir \cdot b - \frac12 r \cdot v r + \int_{\mathbb{B}} \lambda(dx)(e^{ir \cdot x} - 1 - ir \cdot x) + \int_{\mathbb{B}^c}\lambda(dx) (e^{ir \cdot x} - 1),$$ for $r \in \mathbb{R}^d$, $b$ is a vector in $\mathbb{R}^d$, $v = c c^T$ (where $c$ is a $d \times d'$ matrix), $\mathbb{B}$ is the unit ball in $\mathbb{R}^d$ and $\mathbb{B}^c$ is the complement in $\mathbb{B}$ of the ball of radius $\varepsilon$.

Then $$ t \psi(r) = t \left[ ir \cdot b - \frac12 r \cdot v r + \int_{\mathbb{B}} \lambda(dx)(e^{ir \cdot x} - 1 - ir \cdot x) + \int_{\mathbb{B}^c}\lambda(dx) (e^{ir \cdot x} - 1) \right]$$ and $$\psi(t^{1/\alpha}r) = irt^{1/\alpha} \cdot b - \frac12 r t^{1/\alpha}\cdot v r + \int_{\mathbb{B}} \lambda(dx)(e^{irt^{1/\alpha} \cdot x} - 1 - irt^{1/\alpha} \cdot x) + \int_{\mathbb{B}^c}\lambda(dx) (e^{irt^{1/\alpha} \cdot x} - 1).$$ I'm not sure how to proceed next to show they are equal.

Edit: In the book, the process $X$ is said to be $\alpha$-stable if $X_t$ and $t^{1/α}X_1$ have the same distribution (or equivalently, the same characteristic exponent).

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  • $\begingroup$ Which definition of "$\alpha$-stable" are you using? $\endgroup$ – saz Dec 22 '15 at 7:21
  • $\begingroup$ Ah forgot to add that part - the process $X$ is $\alpha$-stable if $X_t$ and $t^{1/\alpha}X_1$ have the same distribution (or equivalently, the same characteristic exponent). $\endgroup$ – Olorun Dec 22 '15 at 7:22
  • $\begingroup$ I see; thanks. For a Lévy process, we have $\mathbb{E}e^{i \xi X_s} = e^{-s \psi(\xi)}$, right? So how does the characteristic function of $t^{1/\alpha} X_1$ look like? $\endgroup$ – saz Dec 22 '15 at 7:29
  • $\begingroup$ OK, I guess we have $$\mathbb{E}[e^{i \xi t^{1/\alpha}X_1}] = e^{-t^{1/\alpha}\psi(\xi)} = e^{-\psi(t^{1/\alpha}\xi)},$$ so the first result follows from the $\alpha$-stable property. $\endgroup$ – Olorun Dec 22 '15 at 8:08
  • $\begingroup$ The first "=" is not correct... but I guess you mean $e^{-t \psi(\xi)}$. $\endgroup$ – saz Dec 22 '15 at 8:52
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The Lévy-Khintchine formula states

$$\mathbb{E}e^{i \eta X_s} = e^{-s \psi(\eta)}$$

for all $s \geq 0$ and $\eta \in \mathbb{R}$. If we use this identity for $t=s$ and $\eta = \xi$, we get

$$\mathbb{E}e^{i \xi X_t} = e^{-t \psi(\xi)}. \tag{1}$$

For $s = 1$ and $\eta = t^{1/\alpha} \xi$, we find

$$\mathbb{E}e^{i \xi (t^{1/\alpha} X_1)} = e^{- \psi(t^{1/\alpha} \xi)}. \tag{2}$$

Since $X$ is $\alpha$-stable, the left-hand side of $(1)$ equals the left-hand side of $(2)$, i.e.

$$e^{-t \psi(\xi)} = e^{-\psi(t^{1/\alpha} \xi)},$$

and this implies

$$t \psi(\xi) = \psi(t^{1/\alpha} \xi).$$

Since this identity holds for any $t>0$ and $\xi \in \mathbb{R}$, we can choose $\xi = 1$ and $t = r^{\alpha}$, $r \geq 0$, to conclude

$$r^{\alpha} \psi(1) = \psi(r).$$

Setting $c := \psi(1)$ proves the second claim. The third one follows obviously from the second one.

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