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We want to show if $X, Y$ are random variables defined on a common probability space, with characteristic functions $f, g$ respectively, then the following inequality is valid:

$$\sup |f(x)-g(x)| \le 2P(X\neq Y).$$

This is from an old qualifying exam and I cannot solve it. I tried to analyze the difference in probability measures, but to no avail. Any help would be greatly appreciated.

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Got something from an answer below? – Did Aug 11 '12 at 9:21

We have for two real numbers $a$ and $b$ that $$|e^{ia}-e^{ib}|=|e^{ia/2}-e^{i(b-a/2)}|=|e^{\frac{a-b}2i}-e^{\frac{b-a}2i}|=2\left|\sin\left(\frac{a-b}2\right)\right|.$$ This gives \begin{align} |f(x)-g(x)|&\leq \int_{\Omega}|e^{iX(\omega)x}-e^{iY(\omega)x}|d\mathbb P\\ &\leq 2\int_{\Omega}\left|\sin\left(\frac{X(\omega)-Y(\omega)}2x\right)\right|d\Bbb P(\omega)\\ &=2\int_{\{X\neq Y\}}\left|\sin\left(\frac{X(\omega)-Y(\omega)}2x\right)\right|d\Bbb P(\omega)\\ &\leq 2\mathbb P(X\neq Y). \end{align} The constant $2$ can't be improved. This can be seen taking Dirac measures at distinct points.

Note that the bound can be taken quite straightforwardly considering the first line, integrating over $\{X\neq Y\}$ and bounding the integrand by $2$. But the equality in the first displayed equation helps us to see why $2$ is optimal.

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Very cool, thanks very much. – James Jul 28 '12 at 20:39

Almost content-free, since once the inequality in 1. is stated the rest is routine:

  1. For every real numbers $a$ and $b$, $\color{red}{|\mathrm e^{\mathrm ia}-\mathrm e^{\mathrm ib}|\leqslant2\cdot[a\ne b]}$.
  2. Apply 1. to $a=xX$ and $b=xY$, for some real $x$. This yields $|\mathrm e^{\mathrm ixX}-\mathrm e^{\mathrm ixY}|\leqslant2\cdot[X\ne Y]$ almost surely.
  3. Integrate 2. This yields $|\mathrm E(\mathrm e^{\mathrm ixX})-\mathrm E(\mathrm e^{\mathrm ixY})|\leqslant2\cdot\mathrm P(X\ne Y)$.
  4. The inequality in 3. holds for every $x$ uniformly hence $$ \sup\limits_x|\mathrm E(\mathrm e^{\mathrm ixX})-\mathrm E(\mathrm e^{\mathrm ixY})|\leqslant2\cdot\mathrm P(X\ne Y).$$
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