Let $X\in \mathbb{R}^n$ and $Y\in \mathbb{R}^n$ be random variables with characteristic functions $\phi_X(t)$ and $\phi_Y(t)$, respectively.

Suppose that \begin{align} \sup_{t \in \mathbb{R}^n} \frac{|\phi_X(t)-\phi_Y(t)|}{\|t\|} \le \epsilon. \quad (*) \end{align}

Question: Can we say something about how close the distributions of $X$ and $Y$ are in some metric over probability spaces? In other words, suppose that $(*)$ small what other distance would be small too.

For example, for $n=1$. There exists the following inequality \begin{align} L^2(P_X,P_Y) \le 2 \sup_{t \ge 0} \frac{|\phi_X(t)-\phi_Y(t)|}{t} \end{align} where $L$ is the L\'evy distance.

However, I was not able to locate similar inequality for $n>1$. One distance that I have taken a look at extensively is the Levy-Prokhorov metric. However, whenever I find anything it always depends on the derivatives of characteristic functions.


There is a good reason why you cannot get anything for the standard Levy-Prokhorov distance in high dimensions. Let's consider the uniform distribution on the sphere of radius $R$ in $\mathbb R^3$ and the mixture (with weight $\frac 12$ for each) of the uniform distributions on the spheres of radii $R-r$ and $R+r$ where $R>2r$. Then the difference of the characteristic functions is $|F(R|t|)-\frac 12[F((R-r)|t|)+F((R+r)|t|)]|\le r^2|t|^2\max_{[(R-r)|t|,(R+r)|t|]}|F''|$ where $F(u)=\frac{\sin u}{u}$. However, $|F''(u)|\le \frac Cu$ for $u>0$, so to have your condition, it suffices to ensure that $$ \frac {Cr^2|t|^2}{R|t|/2}\le\varepsilon |t|, $$ i.e. $2Cr^2/R\le\varepsilon$, which still allows $r$ to grow without bound as $R\to+\infty$ for any fixed $\varepsilon>0$.

So, you'll have to either settle for some cruder distance that gives you less control at infinity, or find a way to get that control by some alternative means from other assumptions you may have in your problem.

  • $\begingroup$ Ok. Thanks. This is a lot more clear to me now. Are you aware of an alternative distance that would work? $\endgroup$ – Boby Jan 9 at 13:23
  • $\begingroup$ @Boby I would look at it from the other end: what is the weakest distance you would be happy with? $\endgroup$ – fedja Jan 9 at 18:00
  • $\begingroup$ Something that matrizes the weak convergence. Or is that too much to ask for? $\endgroup$ – Boby Jan 9 at 18:01
  • $\begingroup$ @Boby If I understand it right, the weak convergence of probability measures (over the class of continuous bounded functions) is metrized by anything that ensures that the characteristic functions converge pointwise, so that is rather "too little" to ask for. $\endgroup$ – fedja Jan 9 at 18:11
  • $\begingroup$ I don't know many probabilistic distance in high dimensions. Do you have any suggestion on which distance is small if $ \sup_{t \in \mathbb{R}^n} \frac{| \phi_X(t) - \phi_Y(t)|}{\|t\|}$ is small? $\endgroup$ – Boby Jan 14 at 13:31

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