# Characteristic function of stable distribution $\neq 0 \forall t \in \mathbb{R}$

Let $$\varphi$$ be a characteristic function of stable distribution $$X$$. Show that $$\forall t \in \mathbb{R}$$ we have $$\varphi_X(t) \neq 0$$.

I tried playing with the characteristic function to get some contradictions, but with no effect.

First, since all $$X_i$$ are i.i.d.: $$\varphi_{X_1+X_2}(t)=\varphi_{X_1}(t)\varphi_{X_2}(t) = (\varphi_{X_1}(t))^2$$ on the other hand, $$X$$ is stable so for some $$c$$ and $$\gamma$$: $$\varphi_{X_1+X_2}(t)=\varphi_{cX_1 + \gamma}(t)= e^{it\gamma}\varphi_X(ct)$$ If I assume that for some $$t$$ our $$\varphi_{X_1}(t)=0$$ can I get a contradiction here? At first I assumed that $$0 = (\varphi_{X_1}(t))^2 = e^{it\gamma}\varphi_X(ct) \implies \forall c \varphi_{X_1}(ct)=0 \implies \forall t\varphi_{X_1}(t)=0$$ but from characteristic function's properties: $$\varphi_{X_1}(0)=1$$ which would give us the contradiction. But this does not seem right - I think my implications are wrong.

How does one prove this?

The problem with your reasoning is that $$0 = (\varphi_{X_1}(t))^2 = e^{it \gamma} \varphi_X(ct)$$ holds only for some $$t \in \mathbb{R}$$ and some $$c>0$$; therefore you cannot deduce immediately that $$\varphi_{X_1}(ct)=0$$ for all $$c$$.

There is, however, a way to fix this issue. Suppose that $$\varphi_{X}(t)=0$$ for some $$t \in \mathbb{R}$$. Since $$(\varphi_{X}(t))^2 = e^{it \gamma} \varphi_X(ct), \qquad t \in \mathbb{R},\tag{1}$$ for some $$\gamma \in \mathbb{R}$$ and $$c>0$$ (as you proved in your question), it follows that $$\varphi_X(ct)=0$$. Hence, $$\tilde{t} := ct$$ is another root of $$\varphi$$. Assume, for the moment, that $$c \in (0,1)$$. Using $$(1)$$ with $$t$$ replaced by $$ct$$, it follows that

$$0 = (\varphi_X(ct))^2 = e^{i(ct)\gamma} \varphi_X(c(ct)),$$

i.e. $$\varphi_X(c^2 t)=0$$. Proceeding by iteration, we find that $$\varphi_X(c^n t)=0$$ for any $$n \in \mathbb{N}$$. By the continuity of $$\varphi_X$$, this implies $$\varphi_X(0) = \lim_{n \to \infty} \varphi_X(c^n t)=0,$$ in contradiction to $$\varphi_X(0)=1$$.

If $$c>1$$, then we note that $$(1)$$ implies

$$(\varphi_X(t/c))^2 = e^{it\gamma/c} \varphi_X(t),$$

and so $$\varphi_X(t)=0$$ implies $$\varphi_X(t/c)=0$$. Now we can proceed as above, to find that $$\varphi_X(t/c^n)=0$$ for all $$n \in \mathbb{N}$$, which gives again a contradiction.

Finally, the case $$c=1$$ can happen only if $$X$$ is constant. Indeed, if $$c=1$$, then by $$(1)$$,

$$(\varphi_X(t))^2 = e^{it \gamma} \varphi_X(t).$$

As $$\varphi_X(0)=1$$ and $$\varphi_X$$ is continuous, there is some $$r>0$$ such that $$\varphi_X(t) \neq 0$$ for all $$|t| \leq r$$. Thus,

$$\varphi_X(t) = e^{it \gamma}, \qquad |t| \leq r,$$

i.e. $$Y:=X-\gamma$$ satisfies

$$\mathbb{E}e^{i Yt}=1, \qquad |t| \leq r.$$

The only characteristic function which equals $$1$$ in a neighbourhood of zero is the function which is constant one, and so $$Y=0$$ almost surely, i.e. $$X=\gamma$$ a.s. In particular, $$\varphi_X(t)=e^{it \gamma}$$ does not have any roots.

• Thank you for the fix. I don't see the last part though why $c = 1$ can happen only with constant $X$ and makes the assertion obvious? Commented Jun 10, 2020 at 7:03
• @НикитаВасильев See my edited answer.
– saz
Commented Jun 10, 2020 at 20:14