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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?

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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.

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  • $\begingroup$ 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? $\endgroup$
    – blahblah
    Commented Jun 10, 2020 at 7:03
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    $\begingroup$ @НикитаВасильев See my edited answer. $\endgroup$
    – saz
    Commented Jun 10, 2020 at 20:14

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