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.
Best Answer
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.