1
$\begingroup$

This is a continuation of

Question abouth Skorokhod representation of random variables

Let $\mu$ and $\nu$ be two probability measures on $\mathbb R$ such that

$$\int_{\mathbb R}|x|^pd\mu(x),~ \int_{\mathbb R}|x|^pd\nu(x)<+\infty.$$

For any $\varepsilon>0$, it is easy to find some $R>0$ such that

$$\int_{\mathbb R}|x|\mathbf{1}_{\{|x|>R\}}d\mu(x),~ \int_{\mathbb R}|x|\mathbf{1}_{\{|x|>R\}}d\nu(x)<\varepsilon.\tag{1}$$

Assume further that for all $|K|\le R$ one has

$$\left|\int_{\mathbb R}(x-K)^+d\mu(x)-\int_{\mathbb R}(x-K)^+d\nu(x)\right|~<~\varepsilon,\tag{1.5}$$

$$\left|\int_{\mathbb R}(K-x)^+d\mu(x)-\int_{\mathbb R}(K-x)^+d\nu(x)\right|~<~\varepsilon.\tag{2}$$

My question is the following: Could we find some couple of random variables $X$ and $Y$ s.t. $Law(X)=\mu$, $Law(Y)=\nu$ and

$$E\left[|X-Y|\right]~\le f(\varepsilon).\tag{$*$}$$

Here $f: \mathbb R_+\to\mathbb R_+$ is some continuous function with $f(0)=0$. If $(\ast)$ can not be achieved, it is possible to get a weakened result by replacing $(\ast)$ by

$$P\left[|X-Y|>f(\varepsilon)\right]~<~f(\varepsilon)?$$

Thanks a lot for the reply!

$\endgroup$
1
  • $\begingroup$ What is $p$ and why is it needed? Can't you just assume $(1)$? $\endgroup$ Jan 13, 2016 at 3:33

1 Answer 1

1
$\begingroup$

With the added condition $(1)$, one does have $(*)$.

Assume first that $R=1$; the general case will follow by simple rescaling.

By $(1)$ (with $R=1$), without loss of generality $$\text{$|X|\le 1$ and $|Y|\le 1$}\tag{!} $$ (see details on this below, at the end of this answer), and then $(2)$ holds for all real $K$.

Let $F$ and $G$ be the distribution functions (d.f.'s) of $X$ and $Y$, respectively. If the Lévy distance $$L(F,G):=\inf\{h>0:F(x-h)-h\le G(x)\le F(x+h)+h \text{ for all real }x\} $$ between $F$ and $G$ is greater than $\delta>0$, then for some real $t$ without loss of generality $F(t-\delta)-G(t)>\delta$, whence $$\int_{t-\delta}^t[F(x)-G(x)]\,dx\ge\int_{t-\delta}^t[F(t-\delta)-G(t)]\,dx>\delta^2. $$ But $\int_{t-\delta}^t[F(x)-G(x)]\,dx=\int_{-\infty}^t[F(x)-G(x)]\,dx-\int_{-\infty}^{t-\delta}[F(x)-G(x)]\,dx. $ So, $$\Big|\int_{t-\delta}^t[F(x)-G(x)]\,dx\Big| =\Big|\int_{-\infty}^t[F(x)-G(x)]\,dx-\int_{-\infty}^{t-\delta}[F(x)-G(x)]\,dx\Big| $$ $$ =\big|\big(E(t-X)_+-E(t-Y)_+\big)-\big(E(t-\delta-X)_+-E(t-\delta-Y)_+\big)\big| \le2\epsilon $$ by $(2)$. Hence, $\delta^2\le2\epsilon$. That is, $$L(F,G)\le\sqrt{2\epsilon}. $$ That is, the supremum of the distance in the direction of the vector $(-1,1)$ between the graphs of $F$ and $G$ is $\le2\sqrt\epsilon$. (By the graph of a nondecreasing function $f$ from an open interval $I$ to $\mathbb R$ here I understand the set $\{(x,y)\colon x\in I,f(x-)\le y\le f(x+)\}$, with the corresponding vertical segments added at all points of discontinuity.)

Hence, the area between the graphs of $F$ and $G$ is $\le2\sqrt\epsilon\,\frac3{\sqrt2}=3\sqrt{2\epsilon}$, because the distance between the straight lines through points $(-1,0)$ and $(1,1)$ parallel to the vector $(-1,1)$ is $\frac3{\sqrt2}$.

Without loss of generality, re-define now $X$ and $Y$ by the formulas $X:=F^{-1}(U)$ and $Y:=G^{-1}(U)$, where $U$ is a r.v. uniformly distributed in the interval $(0,1)$ and $$F^{-1}(u):=\inf\{x\in\mathbb R\colon F(x)\ge u\} $$ for $u\in(0,1)$. Then $F$ and $G$ will indeed be the d.f.'s of $X$ and $Y$, respectively. Also, then we will have $E|X-Y|=\int_0^1|F^{-1}(u)-G^{-1}(u)|\,du$, which is the area between the graphs of $F^{-1}$ and $G^{-1}$, which latter is the same as the area between the graphs of $F$ and $G$, which we showed to be $\le3\sqrt{2\epsilon}$.

So, if $(!)$ holds, then $(*)$ holds with $f(\epsilon):=3\sqrt{2\epsilon}$.

Without assuming $(!)$, but still assuming $(1)$ and $(2)$, one can replace $X$ and $Y$ by their truncated versions $X_1:=1\wedge((-1)\vee X)$ and $Y_1:=1\wedge((-1)\vee Y)$, so that $|X_R|\le 1$ and $|Y_R|\le 1$. Then $(2)$ will hold for all real $K$ if $\epsilon$, $\mu$, and $\nu$ are replaced there, respectively, by $2\epsilon$, the distribution of $X_1$, and the distribution of $Y_1$. Also, $E|X-Y|\le E|X_1-Y_1|+E|X-X_1|+E|Y-Y_1|$, $E|X-X_1|=E(-1-X)_++E(X-1)_+=E(-1-X)I\{X<-1\}+E(X-1)I\{X>1\}\le E(-X)I\{X<-1\}+EXI\{X>1\}=E|X|I\{|X|>1\}\le\epsilon$ (by $(1)$, with $R=1$), and similarly $E|Y-Y_1|\le\epsilon$. So, $(*)$ will hold in general for $R=1$ with $f(\epsilon):=3\sqrt{2\times2\epsilon}+2\epsilon=6\sqrt{\epsilon}+2\epsilon.$

To rescale from $R=1$ to general $R>0$, replace in the above reasoning $X$, $Y$, $\epsilon$ by $\tilde X:=X/R$, $\tilde Y:=Y/R$, $\tilde\epsilon:=\epsilon/R$, respectively, so that the conditions $(1)$ and $(2)$ hold with $1$ in place of $R$, $\tilde\epsilon$ in place of $\epsilon$, and with the distributions of $\tilde X$ and $\tilde Y$ in place of $\mu$ and $\nu$. Then, by the above, $E|\tilde X-\tilde Y|\le6\sqrt{\tilde\epsilon}+2\tilde\epsilon$, which can be rewritten as $$E|X-Y|\le6\sqrt{R\epsilon}+2\epsilon.\tag{!!}$$

It is also seen that, instead of $(1)$, the following weaker condition will suffice: $$E(-R-X)_++E(X-R)_+\le\epsilon,\quad E(-R-Y)_++E(Y-R)_+\le\epsilon. $$

Condition $(1.5)$ is not needed.


Let us now show that the upper bound in $(!!)$ is best possible in terms of $R$ and $\epsilon$, up to a universal constant factor. Let $F$ be the d.f.\ of the uniform distribution on $(-R,R)$. For any natural $n$ and all $k=1,\dots,n$, let $G(x):=F(-R+R\,\frac{2k-1}{n})$ for $x\in[-R+R\,\frac{2k-2}{n},-R+R\,\frac{2k}{n})$, with $G=0$ on $(-\infty,-R)$ and $G=1$ on $[R,\infty)$. Then $G$ is the d.f. of a r.v. $Y$, $|E(X-t)_+-E(Y-t)_+|=|\int_t^\infty[F(x)-G(x)]\,dx|\le\frac12\,\frac{2R}{2n}\frac1n=\frac{R}{2n^2}=:\epsilon$ for all real $t$, $|E(t-X)_+-E(t-Y)_+|=|\int_{-\infty}^t[F(x)-G(x)]\,dx|\le\frac12\,\frac{2R}{2n}\frac1n=\frac{R}{2n^2}=\epsilon$ for all real $t$, $E|X|I\{|X|>R\}=E|Y|I\{|Y|>R\}=0$, whereas the least possible value here of $E|X-Y|$ is the Wasserstein distance between $F$ and $G$, which equals $$d(F,G):=\int_0^1|F^{-1}(u)-G^{-1}(u)|\,du $$ -- see e.g. (2) in [1]; hence, this distance also equals $\int_{\mathbb R}|F(x)-G(x)|\,dx=2n\epsilon=2\sqrt{\frac{R}{2\epsilon}}\,\epsilon=\sqrt{2R\epsilon}$, which shows that the upper bound in $(!!)$ is indeed best possible, up to a universal constant factor, for (say) $\epsilon\le R$.

$\endgroup$
6
  • $\begingroup$ Very nice solution! Thanks so much! (Condition (1.5) is given by my real problem and is shown that it is not needed.) $\endgroup$
    – CodeGolf
    Jan 13, 2016 at 7:15
  • $\begingroup$ I have added details to the proof and made it more self-contained. $\endgroup$ Jan 13, 2016 at 14:22
  • $\begingroup$ Thank you so much for your answer. In the estimation, $f$ is parallel to $R\sqrt{\varepsilon}$ is not quite satisfying as $R\sqrt{\varepsilon}$ does not converge to zero as $\varepsilon$ goes to zero. Could you show that the Levy-Prokhorov distance between $\mu$ and $\nu$ is always parallel to $\sqrt{\varepsilon}$? Thanks! $\endgroup$
    – CodeGolf
    Jan 13, 2016 at 15:36
  • $\begingroup$ I have corrected the account of the effect of the truncation. $\endgroup$ Jan 13, 2016 at 17:57
  • $\begingroup$ I have improved the upper bound and also showed that it cannot be further improved, up to a universal constant factor. $\endgroup$ Jan 13, 2016 at 19:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.