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everyone.

Given a probabilistic space $(\Omega, \mathcal{F}_t, \mathbb{P})$ and a martingale $(M_t)_{t\leq 1}$ on it. Suppose

$$M_1\stackrel{\mathbb{P}}{\sim}\mu$$

where $\mu$ is a probability distribution on $\mathbb{R}$. Now given another probability distribution $\nu$ such that $d(\mu,\nu)<\epsilon$ where $d$ is Levy metric. We want to construct another martingale $N$ such that $M_1$ is "close" to $N_1$.

Consider $\tilde{\Omega}=\Omega\times \mathcal{C}[0,1]$ and $\tilde{\omega}=(\omega,w)\in\tilde{\Omega}$. Let $\mathcal{G}_t$ be the natural filtration generated by $w\in\mathcal{C}[0,1]$ and set $\mathcal{H}_t=\sigma(\mathcal{F}_t,\mathcal{G}_t)$. Let $\mathbb{P}_0$ be Wiener measure and set $\tilde{\mathbb{P}}=\mathbb{P}\times\mathbb{P}_0$. Since $\mathcal{F}$ and $\mathcal{G}$ are independent, thus $\tilde{M}_t(\tilde{\omega}):=M_t(\omega)$ is a $\mathcal{H}$ martingale and $B_t(\tilde{\omega}):=w_t$ is a Brownian motion.

By a general result cited in A.V. Skorokhod, On a representation of random variables, there exists a measurable function $f$ such that

$$N:=f(M_1, B_1)$$

and

$$N\stackrel{\tilde{\mathbb{P}}}{\sim}\nu,\ \ \tilde{\mathbb{P}}(|N-M_1|\geq\epsilon)\leq\epsilon$$

then set $\tilde{N}_t=E^{\tilde{\mathbb{P}}}[N|\mathcal{H}_t]$, then we get the need martingale. Now my question is if we have two marginals $\mu_1$ and $\mu_2$

$$M_1\stackrel{\mathbb{P}}{\sim}\mu_1,\ \ M_2\stackrel{\mathbb{P}}{\sim}\mu_2$$

$$d(\mu_1,\nu_1)<\epsilon,\ \ d(\mu_2,\nu_2)<\epsilon$$

and $\nu_1$, $\nu_2$ satisfy the condition called "increasing convex order": for any convex function $f$, we have $\nu_1(f)\leq\nu_2(f)$, which ensures the existence of martingales having respectively marginal distributions $\nu_1$ and $\nu_2$.

Can we get another martingale $N$ such that(defined maybe on another space)

$$N_i\stackrel{\mathbb{P}}{\sim}\nu_i,\ \ i=1,2$$

and

$$\mathbb{P}(|M_2-N_2|\geq\epsilon)\leq\epsilon$$

Thanks a lot for your help!

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1 Answer 1

Not without more conditions. Let $B_t$ be a Brownian motion and $M_t = c B_t$ for small $c$ to be chosen later. Then $\mu_1 = N(0, c^2)$ and $\mu_2 = N(0, 2c^2)$. Choose $c$ so small that $d(\mu_1, \delta_0) < \epsilon$ and $d(\mu_2, \delta_0) < \epsilon/2$, where $\delta_0$ is the Dirac measure at 0. Then choose a sufficiently small nonzero $a$ such that $d(\delta_0, \delta_a) < \epsilon/2$. Now $\nu_1 = \delta_0$, $\nu_2 = \delta_a$ satisfy the hypotheses, but there is no martingale $N$ with $N_1 \sim \nu_1$, $N_2 \sim \nu_2$.

Being a martingale is not an open condition.

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Thanks a lot for your answer. But if the two marginal distributions are continuous, could we have some conditions under which such construction is possible? Or more precisely, we are given the joint distributions $\mu(dx_1, dx_2)$ and $\nu(dx_1, dx_2)$ which are both continous –  CodeGolf Dec 2 '13 at 21:03
1  
At a bare minimum, they'd better have the same mean. –  Nate Eldredge Dec 2 '13 at 21:04
    
, can we construct such martingale? Thanks a lot! –  CodeGolf Dec 2 '13 at 21:04
    
Sure $\nu_1$ and $\nu_2$ satisfy a condtion called "increasing in convex order", which means for any convex function $f$, we have $\nu_1(f)\leq\nu_2(f)$ –  CodeGolf Dec 2 '13 at 21:05
    
Under this condition, there exists a martingale with respectively marginal distributions as $\nu_1$ and $\nu_2$ –  CodeGolf Dec 2 '13 at 21:06

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