1
$\begingroup$

Let $\phi(t),t\in\mathbb{R}$, be the characteristic function of a random variable $X$. Show that if $\phi(t)=1$ in a neighborhood of $0$, then $X=0$ a.s.

The problem comes with the following hint: Show that $1-Re(\phi(2t))\le4(1-Re(\phi(t)))$ for $t\in \mathbb{R}$. I am stumped by this one, I am not even sure where to begin or how to prove\use use the hint, any help here would be greatly appreciated.

$\endgroup$

2 Answers 2

3
$\begingroup$

Here is another method: start by noticing that since $\cos \leq 1$, $$\Re \phi(t) = \mathbb{E}[\cos(tX)] \leq 1, \quad \forall t \in \mathbb{R}. \tag{1}$$ Now let $\def\eps{\varepsilon} (-\eps, \eps)$ be an interval on which $\phi = 1$. Then, using the hint, we have for $t \in (-\eps, \eps)$ $$0 \leq 1- \Re \phi(2t) \leq 4(1-\Re \phi(t)) = 0.$$ The first inequality follows from $(1)$ and the last equality from the assumption. This proves that $\Re \phi(2t) = 1$ for every $t \in (-\eps,\eps)$. Reiterating this process, we see that $$\Re \phi(t) = 1, \quad \forall t \in \mathbb{R}.$$ But recall that $|\phi(t)| \leq 1$. This forces $\phi(t) = 1$ for every $t \in \mathbb{R}$. Since the characteristic function determines the distribution, it follows that $X = 0$ a.s.

$\endgroup$
1
  • $\begingroup$ Beautiful, thank you! $\endgroup$ Jun 17, 2020 at 21:49
2
$\begingroup$

Let's assume that $\varphi(t) = 1$ for any $t \in [0,\delta]$. Then in particular $\varphi(\delta)=1$. We'll show that it is the case that $\mathbb P(X \in \{\frac{2k\pi}{\delta} : k \in \mathbb Z \}) = 1$ . Let $\mu_X$ be distribution of $X$.

Note that $\varphi(\delta)=1$ means: $$ 0 = 1 -\varphi(\delta) = 1 - \int_{\mathbb R} \cos(\delta x) d\mu_X(x) = \int_{\mathbb R} (1 - \cos(\delta x)) d\mu_X(x)$$ Since $1-\cos(\delta x) \ge 0$, we must have $\cos(\delta x) = 1$ , $x - d\mu_X$ almost surely, so that $x = \frac{2k\pi}{\delta}$ , $d\mu_X$ almost surely, which means $\mu_X( \{\frac{2k\pi}{\delta} : k \in \mathbb Z \})=1$.

Now note that we have only countable many points in set $\{\frac{2k\pi}{\delta} : k \in \mathbb Z \}$. For every $k \in \mathbb Z \setminus \{0\}$ we can find such $t_k \in (0,\delta)$ that $\frac{2k\pi}{\delta}$ is not equal to $\frac{2 m \pi}{t_k}$ for any $m \in \mathbb Z$ (because for every $m \in \mathbb Z$ there is at most one $s \in (0,\delta)$ such that $\frac{2m \pi}{s} = \frac{2k\pi}{\delta}$, but we have only countable many $m \in \mathbb Z$, but continuum-many $s \in (0,\delta)$, so there exists such $t_k$) which means that $\mu_X(\frac{2k\pi}{\delta}) = 0$ (for that given $k \in \mathbb Z \setminus \{0\}$, because $\varphi(t_k)=1$, so $\mu_X( \{ \frac{2m\pi}{t_k} : m \in \mathbb Z \}) = 1$, too). Since $k \in \mathbb Z \setminus \{0\}$ was arbitrary, and there are only countable many of them, we have $\mu_X( \{ \frac{2k \pi}{\delta} : k \in \mathbb Z \setminus \{0\} \} ) = 0$, so that $\mu_X(\{0\}) = 1$ what was to be proven.

EDIT: If you're interested, here's an approach with your hint. Let's prove it beforehand. $$ 1 - Re(\varphi(2t)) = \int_{\mathbb R} (1-\cos(2tx))d\mu_X(x) = 2\int_{\mathbb R} (1 - \cos^2(tx))d\mu_X(x) $$

It would be sufficient to show $1-\cos^2(s) \le 2(1- \cos(s))$ which is equivalent to $0 \le \cos^2(s) - 2\cos(s) + 1 = (\cos(s)-1)^2$, so true. Hence $$ 1- Re(\varphi(2t)) \le 4\int_{\mathbb R}(1 - \cos(tx))d\mu_X(x) = 4(1-Re \varphi(t))$$

Having lemma, it is pretty easy. Note that you have such $\delta$, that $\varphi(t) = 1$ for any $[-\delta,\delta]$. Now let's prove it is also the case for any $t \in [-2\delta,2\delta]$ using hint: Take $s \in [-2\delta,2\delta]$. We have $$ 1 - Re(\varphi(2s)) \le 4(1 - Re(\varphi(s)) = 0$$ since $s \in [-\delta,\delta]$. Moreover, $|\varphi(s)| \le 1$, so $\varphi(s) = 1$. Use it again, to prove the fact for any $[-2^k\delta,2^k\delta]$ getting $\varphi(t)=1$ for any $t \in \mathbb R$.

$\endgroup$
8
  • 1
    $\begingroup$ Oops I see we arrived at the same conclusion! :) Your first method is more powerful however as it shows that if $\phi(t) = 1$ for one $t$ then the distribution is lattice. $\endgroup$
    – Michh
    Jun 17, 2020 at 21:45
  • $\begingroup$ This is an incredible answer, thank you very much! $\endgroup$ Jun 17, 2020 at 21:48
  • 1
    $\begingroup$ Yes Michh, nice answer +1. This fact about $\varphi(t)=1$ is useful when you want to characterise all types of characteristic functions. It can be shown that either $|\varphi(t)| <1$ for every $t \in \mathbb R \setminus \{0\}$ or $|\varphi(t)|=1$ for every $t \in \mathbb R$ and then you have $X = a$ almost surely for some $a \in \mathbb R$ or finally $|\varphi(t)|=1$ for some $t \in \mathbb R_+$, but $|\varphi(s)|<1$ for $s \in (0,t)$. In the latter case it is exactly the distribution on $\{ \frac{2k\pi}{t} : k \in \mathbb Z \}$. $\endgroup$ Jun 17, 2020 at 21:49
  • $\begingroup$ @Spider Bite, I should thank you, too, because I wasn't aware of such lemma. Nice one to have in mind ^^ $\endgroup$ Jun 17, 2020 at 21:55
  • 1
    $\begingroup$ @Wave because since $\phi(\delta)=1$, then in particular it is a real number, hence imaginary part must vanish. $\endgroup$ May 9 at 14:31

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.