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Let $\mu$ and $\nu$ be two probability measures on $\mathbb R^d$ with finite first movements, i.e.

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

$\mu$ and $\nu$ are said to be increasing in convex order, if the following inequality holds for any convex function $f$ with linear growth:

$$\int_{\mathbb R^d}f(x)~\mu(dx)\quad \le \quad \int_{\mathbb R^d}f(x)~\nu(dx).$$

Then Strassen's Theorem states that $\mu$ and $\nu$ are said to be increasing in convex order, iff there exists a martingale $(M,N)$ s.t. $M\sim\mu$ and $N\sim\nu$.

My question is the following: let $d=1$ and $\varepsilon\ge 0$. What are the conditions on $\mu$ and $\nu$ to ensure the existence of a stochastic process $(M,N)$ s.t. $M\sim\mu$, $N\sim\nu$ and

$$ M~-~\varepsilon\quad \le \quad E[N|M]\quad \le \quad M~+~\varepsilon.\quad\quad\quad\quad\quad\quad (\ast)$$

Thanks a lot for the reply!

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I encourage you to consult Strassen's paper (http://www.jstor.org/stable/2238148). The result you quote as "Strassen's theorem" is Theorem 8 therein and follows from the much more general Theorem 7, as does the extension you seek. I'll sketch an argument:

By Theorem 7, a coupling as you describe exists if and only if $\int \varphi\,d\nu \le \sup \int \varphi(y)\gamma(dx,dy)$ for every continuous $\varphi$ of linear growth, where the supremum is over all probability measures $\gamma$ on $\mathbb{R}^2$ of the form $\gamma(dx,dy) = \mu(dx)K_x(dy)$, where the kernel $K$ satisfies $\left|\int y\,K_x(dy) - x\right| \le \epsilon$ for $\mu$-a.e. $x$. Using a standard measurable selection argument, this is further equivalent to the statement that $\int \varphi\,d\nu \le \int h_{\varphi}\,d\mu$ for every continuous $\varphi$ of linear growth, where $h_{\varphi}(x) := \sup_{\eta \in M(x)}\int\varphi\,d\eta$ and where $M(x) := \sup\left\{\eta : \left|\int y\,\eta(dy) - x\right| \le \epsilon\right\}$.

To simplify this further, we find a better expression for $h_{\varphi}$. Note first that $h_{\varphi}(x) \ge \int\varphi\,d\delta_z \ge \varphi(z)$ for $|z-x| \le \epsilon$, so $h_{\varphi}(x) \ge \sup_{|z-x| \le \epsilon}\varphi(z)$. On the other hand, letting $\varphi^c$ denote the concave envelope, for each $\eta \in M(x)$ Jensen's inequality yields

$\int\varphi\,d\eta \le \int\varphi^c\,d\eta \le \varphi^c\left(\int y\,\eta(dy)\right) \le \sup_{|z-x| \le \epsilon}\varphi^c(z)$.

Thus, $h_{\varphi}(x) \le \sup_{|z-x| \le \epsilon}\varphi^c(z)$. It is a simple exercise to show (1) that $h_{\varphi}$ is concave and (2) that $x \mapsto \sup_{|z-x| \le \epsilon}\varphi^c(z)$ is concave and thus equal to the concave envelope of $x \mapsto \sup_{|z-x| \le \epsilon}\varphi(z)$. Hence, $h_{\varphi}(x) = \sup_{|z-x| \le \epsilon}\varphi^c(z)$.

Finally, we find that the your coupling exists if and only if

$\int\varphi(x)\,\nu(dx) \le \int \sup_{|z-x| \le \epsilon}\varphi(z)\,\mu(dx)$

for every concave function $\varphi$ of linear growth.

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  • $\begingroup$ Thank you very much for the reply. I've read the paper by Strassen actually and I was confused with the Hanh-Banach theorem. I've one question related to your argument: in the second paragraph, why the existence of $\gamma$ is equivalent to $\int \varphi d\nu\le \int h_{\varphi} d\mu$? $\endgroup$ – MB2009 Jul 26 '17 at 19:47
  • $\begingroup$ Actually this is the only part where I don't know how to adapt the arguments of Strassen to my case. Could you please explain a bit more? Thank you very much! $\endgroup$ – MB2009 Jul 26 '17 at 19:50
  • $\begingroup$ You can pass the supremum inside the integral, $\sup_{K : K_x \in M(x) \ \forall x}\int\mu(dx)\int K_x(dy)\varphi(y) = \int\mu(dx)\sup_{\eta \in M(x)}\int \eta(dy)\varphi(y) = \int h_\varphi\,d\mu$. The first supremum is over all measurable maps (kernels) from $\mathbb{R} \ni x \mapsto K_x \in \mathcal{P}(\mathbb{R})$ satisfying $K_x \in M(x)$ for every $x$. The supremum inside the integral in the second expression is just over $M(x)$, for $x$ fixed. See, for instance, Proposition 7.50 of the book of Bertsekas & Shreve. $\endgroup$ – Dan Jul 27 '17 at 11:47
  • $\begingroup$ Thank you so much for the prompt reply. I think I didn't formulate my question correctly. Let me reformulate: Why there exists $\pi$ s.t. $\pi$ has marginals $\mu$ and $\nu$ and satisfies the above condition, if and only if, one has $\int \varphi d\nu\le \sup \int \varphi(y)\gamma(dx,dy)$ among $\gamma(dx,dy)=\mu(dx)K_x(dy)$ satisfying $|\int y K_x(dy)-x|\le \epsilon$? $\endgroup$ – MB2009 Jul 27 '17 at 11:56
  • $\begingroup$ Basically, if one has a $\pi$ s.t. $\pi$ has marginals $\mu$ and $\nu$ and satisfies the above condition, then $\int \varphi d\nu\le \sup \int \varphi(y)\gamma(dx,dy)$ among $\gamma(dx,dy)=\mu(dx)K_x(dy)$ satisfying $|\int y K_x(dy)-x|\le \epsilon$. My question is how to show the inverse direction? $\endgroup$ – MB2009 Jul 27 '17 at 11:59

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