# Error for the convergence by distribution

A sequence of random variables $X_n$ converges in distribution to $X$, if there is pointwise convergence of its characteristic functions, i.e. $\lim_{n\rightarrow\infty}\phi_{X_n}(\lambda) = \phi_X(\lambda)$ for all $\lambda \in \mathbb R$.

Imagine we do not only have pointwise, but uniform convergence of the characteristic functions ($\lim_{n\rightarrow\infty}\|\phi_{X_n}-\phi_X\|_\infty = 0$). Can we estimate the error in the distributions by knowing $\|\phi_{X_n}-\phi_X\|_\infty$, i.e. do we have $$|P(X_n \le z ) - P(X < z)| \le \alpha\left(\|\phi_{X_n}-\phi_X\|_\infty\right)$$ for a function $\alpha(\cdot)$ and any $z \in \mathbb R$? What is the form of $\alpha(\cdot)$?

## migrated from math.stackexchange.comFeb 9 '15 at 6:35

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Assuming that you want $\alpha(0+)=0$, there is no such inequality. In fact, by Proposition 11.7.6 in Dudley's Real Analysis and Probability, the optimal $\alpha$ in your question, if taken literally, is identical to $1$ for strictly positive arguments.