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


Thanks a lot for the reply!

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

1 Answer 1


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) $\epsilon\le R$.

  • $\begingroup$ 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

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