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Let $X$ be a random variable with characteristic function $\phi(\ )$ satisfying $|\phi(t)|=1$ for all $|t|\leq 1/T$ with some $T>0$. Show that $X$ is degenerate, i.e., there is $c$ such that $P(X=c)=1$.

My try :

$|\phi(t)|^2=1 \implies (\mathbb{E}(\cos tX))^2+(\mathbb{E}(\sin tX))^2=1=\mathbb{E}(\cos^2 tX+\sin^2 tX)=\mathbb{E}(\cos^2 tX)+\mathbb{E}(\sin^2 tX)$ so we can say that $\sin tX=\mathbb{E}(\sin tX), \cos tX=\mathbb{E}(\cos tX)$, that is $\phi(t)=\rm{e}^{\rm{i}tX}$ for $|t|\leq 1/T$. But I cannot go anywhere from here, can someone help me? Thanks.

Edit : I found out this fact. Let $\psi(t)=|\phi(t)|^2$ which is a characteristic function and its real, and $\psi(t)=1, |t|\leq 1/T$. Now employ the inequality $\Re(1-\psi(t))\leq 4\Re(1-\psi(t/2))$ now apply this $n$ times we get $(1-\psi(t))\leq 4^n\left(1-\psi\left(\dfrac{t}{2^n}\right)\right)$ now for any $t$ we get rhs goes to $0$. But since $\psi$ is real it is also $\le 1$ so $\psi(t)=1$ for all $t$. Now we have $|\phi(t)|=1$ for all $t$. I know this is pretty pointless, but I don't understand any of the proofs given below, if someone would clearly explain why the sets mentioned have only one element in common, I would be grateful.

Did
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shadow10
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3 Answers3

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Let $Y$ be an independent copy of $X$, i.e. $Y$ has the same distribution as $X$ and $Y$ is independent of $X.$ Define $Z=X-Y.$ The characteristic function of $Z$ is given by $$E(e^{itZ})=E(e^{itX})E(e^{-itY})=\phi(t)\phi(-t)=|\phi(t)|^2.$$ Therefore, for $t\in (-1/T,1/T),$ the characteristic function of $Z$ matches with the characteristic function of a random variable $Z'$ degenerate at $0.$ This implies that $Z$ has the same distribution as $Z'$, which means that distribution of $Z$ is degenerate at $0.$ In other words, $X = Y$ with probability 1.

If $F(x) = \Pr(X \leq x)$ be the CDF of $X,$ then using the independence of $X, Y$ and the fact that $Y=X$ with probability 1, we get $$F(x) = \Pr(Y = X \leq x) = \Pr(X \leq x)\Pr(Y \leq x) = F(x)^2$$ which implies that for every $x,$ $F(x)$ is either 0 or 1. This clearly says that $X$ must have a degenerate distribution.

Aditya Ghosh
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  • From $\hat(Z)(t)=1$, we can only get $Z\overset{d}{\to}0$, i.e. $X-Y\overset{d}{\to}0$. Furthermore, $X\overset{d}{\to}Y$. How can you say $X=Y$ with probability 1. Your "In other words" is not true? Because "with probability 1" means almost surely which is stronger than weak convergence(convergence in distribution). If I'm wrong, please correct me. – suineg Nov 28 '24 at 17:18
  • My comment "$Z\overset{d}{\to}0$" is incorrect, because $Z$ is not a sequence. The following is my real question. From $\hat(Z)(t)=1$, we can only get that $Z$ degenerates at 0. How can you say $X=Y$ with probability 1? "with probability 1" means almost surely, which is a strong claim. – suineg Nov 28 '24 at 17:33
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I found an answer to this problem here: http://www.les-mathematiques.net/phorum/read.php?12,488850,489042

As French language may be a problem, here's a translation of this solution:

Let $t\in\left[-\frac{1}{T},\frac{1}{T}\right]$. If $|\phi(t)|=1$, then there exists $\theta=\theta(t)$ such that $\phi(t)=e^{i\theta}$. So:

$$E[1-e^{i(tX-\theta)}]=0$$

This leads to:

$$E[\operatorname{Re}(1-e^{i(tX-\theta)})]=0$$

As $\operatorname{Re}(1-e^{i(tX-\theta)})\geq 0$, when then have:

$$e^{i(tX-\theta)}=1\text{ a.s}$$

Thus:

$$X\in\frac{\theta}{t}+\frac{2\pi\mathbb{Z}}{t}$$

Now we can do the same thing with $t'$ such that $|t'|\leq\frac{1}{T}$ and $t$ and $t'$ are rationally independent.

$$X\in\left(\frac{\theta}{t}+\frac{2\pi\mathbb{Z}}{t}\right)\cap\left(\frac{\theta}{t'}+\frac{2\pi\mathbb{Z}}{t'}\right)$$ which is a singleton since $t$ and $t'$ are rationally independent.

Augustin
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    there is a problem with your proof, how can you say the last set is a singleton, because $\theta$ is a function of $t$, so we don't really know anything about it do we?@Augustin – shadow10 Sep 02 '15 at 11:13
  • Sorry for reviving this, but how can you conclude that $e^{i(tX-\theta)}=1$ a.s., when you didn't look at the imaginary part of it? We know that $E[Im(1-e^{i(tX-\theta)})]=E[-\sin(tX-\theta)]=0$, but that doesn't tell us much about it. – DDomjosa Nov 11 '24 at 16:56
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First fix $t$. From $$(\mathbb{E}(\cos tX))^2+(\mathbb{E}(\sin tX))^2 = \mathbb{E}(\cos^2 tX+\sin^2 tX)$$ you get $$\textrm{var}{(\cos{tX})} + \textrm{var}{(\sin{tX})} = 0$$

Then, almost surely $$\cos{tX} = c(t) \qquad\text{and}\qquad \sin{tX} = s(t)$$ for some constants $c(t)$ and $s(t)$ satisfying $c(t)^2+s(t)^2=1$.

Then you could say $X = \frac{1}{t}\arg(c(t)+is(t)) = 1$ almost surely.

I must admit though that the last step seems a bit weird to me. But I will post this anyway. I hope someone points a possible mistake I made.

Calculon
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  • Thanks for the response. But how is the $\text {atan2}$ function defined? And how is that so obvious from your calculation? Could you please explain? @Calculon – shadow10 Aug 26 '15 at 15:46
  • $atan2(y,x)$ is the argument of the complex number $x+iy$. I also used the fact that $\arg{(R\cos{\theta}+iR\sin{\theta})} = \theta$ for any $R\neq 0$. – Calculon Aug 26 '15 at 15:52