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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?

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You can use the inequality $|\mathbb{E}[Z]|\leq \mathbb{E}|Z|$ just like before to get$$|\phi_X(t)-\phi_Y(t)| = |\mathbb{E}[e^{itX}-e^{itY}]| \leq \mathbb{E}|e^{itX}-e^{itY}| \leq \mathbb{E}[2\cdot 1_{X\neq Y}] = 2P(X\neq Y)$$

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  • $\begingroup$ Thank you. Could you explain why did $1_{X \neq Y}$ appear in the last inequality? $\endgroup$ – arm1223 Jun 19 at 13:01
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    $\begingroup$ Because $|e^{itX}-e^{itY}|\leq 2$ for $X\neq Y$ and $|e^{itX}-e^{itY}| = 0$ for $X = Y$. $\endgroup$ – Jakobian Jun 19 at 13:13
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Law of total expectation formalizes things fine and I think should be cited:

$E[f(X,Y)]=E[f(X,Y)|X=Y]P(X=Y)+E[f(X,Y)|X\ne Y]P(X\ne Y)$

, given any function $f$ of two variables. Can you take it from here ?

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  • $\begingroup$ Thank you, it's also a nice way to show it! $\endgroup$ – arm1223 Jun 19 at 13:27

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