Prove the following random variable is a constant almost surely

Question

Let $\phi_X$ be the characteristic function of a random variable $X$. Prove that $|\phi_X(t)| = 1, \forall t \in R$ iff there is a $b \in R$ s.t. $P(X = b) = 1$.

The if direction is easy to prove. For the only if direction, I managed the followin part:

if $|\phi_X(t)| = 1$, then $$ E^2(\cos(tX)) + E^2(\sin(tX)) = 1, $$ which gives us $$ E(\cos^2(tX)) - Var(\cos(tX)) + E(\sin^2(tX)) - Var(\sin(tX)) = 1. $$ This gives us $$ Var(\cos(tX)) = 0 $$ and $$ Var(\sin(tX)) = 0, $$ i.e., $P(\cos(tX) = E(\cos(tX))) = 1$ and $P(\sin(tX) = E(\sin(tX))) = 1$ for all $t \in R$.

I am stuck at the last line. I don't know how to go from $\cos(tX)$ and $\sin(tX)$ being constant almost surely to $X$ being constant almost surely. I am guessing the issue is that I haven't used the condition that the above is true for all $t \in R$. It's just I don't know how to utilize this condition. Any hint would be appreciated. Thank you!

Answer 1

I think I have found a proof starting from the last line I obtained after grappling for a while:

So from $P(\cos(tX) = E(\cos(tX))) = 1$, it implies that $X$ only has positive measure on countable set. Write this set as $\{a_1, a_2, ...\}$ each with probability $p_1, p_2, ...$

Now we can directly compute the norm of the characteristic funcion of $X$: \begin{align} |\phi_X(t)| &= \left(\sum_{n=1}^\infty p_n\cos(ta_n)\right)^2 + \left(\sum_{n=1}^\infty p_n\sin(ta_n)\right)^2 \\ &=\sum_{n=1}^\infty p_n^2 + 2\sum_{i<j}^\infty p_ip_j\cos(t(a_i-a_j)) = 1. \end{align}

Now notice that $$ 1 = \left(\sum_{n=1}^\infty p_n\right)^2 = \sum_{n=1}^\infty p_n^2 + 2\sum_{i<j}^\infty p_ip_j. $$

This means that $\cos(t(a_i - a_j)) = 1$ for all $i,j$ and for all $t \in R$. This is only possible when $a_i = a_j$ for all $i, j$. Therefore, $X$ only has positive measure on the set $\{b\}$ where $b = a_1 = a_2 = \cdots$, i.e., $P(X=b)=1$ for some $b \in R$.

Answer 2

Here is another argument which is quite standard: Let $\psi (t)=|\phi(t)|^{2}$. Then $\psi$ is also a characteristic function; it is the characteristic function of $Y-Z$ where $Y$ and $Z$ are i.i.d with the same distribution as $X$. The hypothesis gives $\psi (t)=1$ for all $t$. The only random variable whose characteristic function is identically $1$ is the zero random variable. This follows from the fact that characteristic functions determine the distribution uniquely. I will given below an elementary proof of this fact.

It now follows that $Y=Z$ a.s.. But $Y$ and $Z$ are independent. Hence $Y$ is independent of itself. This implies that $Y$ is a constant. But $Y$ has same distribution as $X$ so $X$ is a.s constant.

Lemma: If $Ee^{it W}=1$ for all $t$ then $W=0$ a.s.

Proof: We have $E[1-\cos (tW)]=\Re [1- Ee^{itW}]=0$. Since $1-\cos (tW) \geq 0$ this implies that $\cos (tW)=1$ as. for each $t$. Applying this to $t=1$ and $t=\sqrt 2$ we see that $W =2n\pi$ and $\sqrt 2 W=2m\pi$ a.s. for some integers $n$ and $m$. Dividing one equation by the other we get a contradiction unless $n=m=0$.