# Show that if $\phi_{X}(t)=1$ in a neighborhood of $0$, then $X=0$ a.s.

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.

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.

• Beautiful, thank you! Jun 17, 2020 at 21:49

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 $$t \in \mathbb R$$.

• 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. Jun 17, 2020 at 21:45
• This is an incredible answer, thank you very much! Jun 17, 2020 at 21:48
• 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 $\{ \frac{2k\pi}{t} : k \in \mathbb Z \}$. Jun 17, 2020 at 21:49
• @Spider Bite, I should thank you, too, because I wasn't aware of such lemma. Nice one to have in mind ^^ Jun 17, 2020 at 21:55
• @Wave because since $\phi(\delta)=1$, then in particular it is a real number, hence imaginary part must vanish. May 9 at 14:31