# Characteristic functions inequality

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?

You can use the inequality $$|\mathbb{E}[Z]|\leq \mathbb{E}|Z|$$ just like before to get$$|\phi_X(t)-\phi_Y(t)| = |\mathbb{E}[e^{itX}-e^{itY}]| \leq \mathbb{E}|e^{itX}-e^{itY}| \leq \mathbb{E}[2\cdot 1_{X\neq Y}] = 2P(X\neq Y)$$
• Thank you. Could you explain why did $1_{X \neq Y}$ appear in the last inequality? – arm1223 Jun 19 at 13:01
• Because $|e^{itX}-e^{itY}|\leq 2$ for $X\neq Y$ and $|e^{itX}-e^{itY}| = 0$ for $X = Y$. – Jakobian Jun 19 at 13:13
$$E[f(X,Y)]=E[f(X,Y)|X=Y]P(X=Y)+E[f(X,Y)|X\ne Y]P(X\ne Y)$$
, given any function $$f$$ of two variables. Can you take it from here ?