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Let $X : \Omega \to \mathbb{R}$ be a random variable, and consider its characteristic function:

$$\phi_X(t) = \mathbb{E}[e^{itX}]$$

Suppose that $\phi_X(u) = 1$. How can I rigorously prove that $e^{iuX} \equiv 1$ a.s.? (More context.)

We have the obvious inequality:

$$ |\mathbb{E}[e^{itX}]| \le \mathbb{E}[|e^{itX}|] =1$$

so this is quite an intuitive result, but it's not immediately clear how to prove it. This seems especially difficult because we do not have the notion of continuity.

One idea I am considering is differentiating $\mathbb{E}[e^{itX}]$ under the integral. Since $1$ is a maximum for $|\mathbb{E}[e^{itX}]|$ by the above, the derivative at $t = u$ must be zero. But I am having some difficulty with making this work.

legionwhale
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  • (+1) what do you mean with 'we do not have the notion of continuity?' – Snoop May 18 '23 at 23:40
  • @Snoop There are similar results for real-valued integrals, where we might think about using the continuity of the function being integrated (assuming it is continuous) to assist with the proof. Here, of course, we cannot do that. – legionwhale May 18 '23 at 23:41

1 Answers1

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$E(1-\cos (tX))=\Re (1-Ee^{itX})=0$. Since $1-\cos (tX)\geq 0$, this implies $1-\cos (tX)=0$ a.s. But $\cos (tX)=1$ implies $\sin (tX)=0$ so $e^{itX}=1$ a.s.

Kavi Rama Murthy
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