How to show that for any random variables $X,Y$ with characteristic functions $\phi_X, \phi_Y$ we have: $$\sup_{\xi \in \mathbb{R}} |\phi_X(\xi) – \phi_Y(\xi)| \leq 2P(X \neq Y)?$$
My attempt: First, I was considering an easier case, when $X,Y$ have densities $f_X, f_Y$. In that case, we have:
\begin{align}
|\phi_X(\xi) – \phi_Y(\xi)| = \big| \int_\mathbb{R} e^{i\xi t} (f_X(t)-f_Y(t)) \, dt \big| \leq \int_\mathbb{R} |f_X(t)-f_Y(t)| \, dt \leq\\ \leq \int_\mathbb{R} f_X(t) \, dt + \int_{\mathbb{R}} f_Y(t) \, dt = 2.
\end{align}
But since $X,Y$ have densities $Z = X-Y$ does as well, so $P(Z=z)=0$ for any $z \in \mathbb{R}$, so $P(Z \neq 0) = 1$, so $2 P(X \neq Y) = 2.$
Is that correct? If so, how to generalise it for any random variables? Or maybe is there a completely different solution?
Best Answer
You can use the inequality $|\mathbb{E}[Z]|\leq \mathbb{E}|Z|$ just like before to get$$|\phi_X(t)-\phi_Y(t)| = |\mathbb{E}[e^{itX}-e^{itY}]| \leq \mathbb{E}|e^{itX}-e^{itY}| \leq \mathbb{E}[2\cdot 1_{X\neq Y}] = 2P(X\neq Y)$$