Let $(X_k)_{k=0,1,..., n}$ be a discrete martingale defined on some probability space $(\Omega,\mathcal{F},\mathbb{P})$. I would like to know whether there exists a (continuous) martingale $(\tilde{X}_t)_{0\le t\le n}$ defined on some probability space such that $(X_0, X_1,\dots, X_n)$ and $(\tilde{X}_0, \tilde{X}_1,\dots, \tilde{X}_n)$ have the same law? Thanks a lot for the reply!

1$\begingroup$ I changed your title to one that's more descriptive. Feel free to edit if you don't like it. $\endgroup$– Nate EldredgeMar 6, 2014 at 13:58

$\begingroup$ It is quite more reasonable for this title, thx $\endgroup$– CodeGolfMar 6, 2014 at 15:04

2$\begingroup$ Isn't this equivalent to: given a random variable $Z$ with $\mathbb E[Z]=0$, does there exist a continuous martingale $X=(X_t:t \geq 0)$ such that $X_1$ has law $Z$? $\endgroup$– BatiMar 6, 2014 at 15:07
2 Answers
Martingales on the Brownian filtration are continuous, so we construct one with the same law on discrete times as $X$. We construct this process in a similar way to how one might simulate $X$, given its conditional distributions and a sequence of normals.
For simplicity, assume $X_0$ is constant. Let $(W_t)_{t\ge 0}$ be a Brownian motion on $(C[0,\infty), \mathcal{W}, \mathbb{Q})$, with natural filtration $(\mathcal{W}_t)_{t\ge 0}$. For $i=1,\cdots, n$, let $$N_i = W_i  W_{i1}.$$
Claim: For $k=0,\cdots,n$, there is a $\sigma(N_1,\cdots,N_k)$measurable random variable $Y_k$ such that $(X_0,\cdots, X_n)$ has the same law as $(Y_0,\cdots, Y_n)$.
$Y$ has the same law as $X$, so it is a martingale (on its natural filtration). Also it is (we may check) a discretetime martingale on the Brownian filtration, $$Y_k = \mathbb{E}[Y_n\mathcal{W}_k],$$ so we may extend the definition of $Y_k$ to all $k$ in $[0,n]$. $Y$, being a martingale on the Brownian filtration, is continuous.
Proof of claim: Without loss of generality, assume $X_k$ is the $k$th coordinate map on a probability space $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n), \mathbb{P})$. Let $\phi$ be the CDF of a standard normal random variable. For $i=0,\cdots,N1$, let $\mathcal{G}_i =\sigma(X_1,\cdots, X_{i})$, and let $Q_i:\Omega\times \mathcal{F}\to[0,1]$ be the regular conditional probability of $\mathbb{P}$ with respect to $\mathcal{G}_{i1}$. Let $$F_{X_iX_0,\cdots, X_i}(x x_0, \cdots, x_{i1}) = Q((x_0,\cdots, x_i,0,\cdots,0), \{X_i\le x_i\}).$$
For $G \in\mathcal{G}$, $\omega\mapsto Q(\omega,G)$ is $\mathcal{G}$measurable, so $Q$ only depends on the first $i$ components of $\omega$. Denote $F_{X_iX_0,\cdots, X_{i1}}$ by $F_i$. Let $$G_i(y x_0, \cdots, x_{i1}) = \inf\{x:F_i(x x_0,\cdots, x_{i1})\ge y\}$$ Let $Y_0$ equal the constant $X_0$ and $$Y_i = G_i(\phi(N_i) Y_0,\cdots, Y_{i1})$$
$X_0$ has the same law as $Y_0$. Suppose $(X_0, \cdots,X_{i1})$ has the same law as $(Y_0, \cdots,Y_{i1})$. Fix $x_0, \cdots, x_i$. $$ \begin{align}\{Y_i\le x_i\} &= \{G_i(\phi(N_i), Y_0,\cdots, Y_{i1}) \le x_i\}\\ &= \{F_i(x_i Y_0,\cdots, Y_{i1})\ge \phi(N_i)\}\\ &= \{Q((Y_0,\cdots,Y_{i1}, x_i, 0,\cdots, 0), \{X_i\le x_i\})\ge \phi(N_i)\} \end{align} $$ Similarly, $$ \begin{align}\mathbb{Q}[Y_0\le x_0, \cdots, Y_i\le x_i] &= \mathbb{Q}[Y_0\le x_0,\cdots,Y_{i1}\le x_{i1}, Q((Y_0,\cdots,Y_{i1}), \{X_i\le x_i\})\ge \phi(N_i)]\\ &= \mathbb{E}[Q((Y_0,\cdots, Y_{i1}),\{X_i\le x_i\})\mathbf{1}_{Y_0\le x_0}\cdots\mathbf{1}_{Y_{i1}\le x_{i1}}]\\ &= \mathbb{E}[Q((X_0,\cdots, X_{i1}),\{X_i\le x_i\})\mathbf{1}_{X_0\le x_0}\cdots\mathbf{1}_{X_{i1}\le x_{i1}}]\\ &= \mathbb{P}[X_0\le x_0, \cdots, X_i\le x_i] \end{align} $$


$\begingroup$ @CodeGolf You're welcome. Did a particular question motivate this problem? $\endgroup$– BenMar 7, 2014 at 20:42

1$\begingroup$ @CodeGolf If you are satisfied with an answer to a question you wrote, you can click on the little check mark underneath the up and downvotes, to "accept" it.... $\endgroup$– UwFApr 13, 2014 at 13:53
Let $(W_t)_{t\geq 0}$ be a Brownian motion. By the Skorokhod embedding theorem there exist a stopping time $T$ such that $W_T \,{\buildrel d \over =}\ X_1X_0$, see e.g. "The Skorokhod embedding problem and its offspring", Obłój, (link) (Actually, we may choose $T$ in such a way, that $W^T = (W_{T \wedge t})$ is a uniformly integrable martingale). Denote $\phi(t) = \frac{t}{1t}$ and define for $t \in [0;1)$
$$ \tilde{X} _t = W^T_{\phi (t)} = W _{\phi (t)}I_{\{\phi (t) \leq T \}}+ W _{T}I_{\{\phi (t) > T \}}, $$
and let $\tilde{X} _1  \tilde{X}_0 = W _{T}$. Then $(\tilde{X} _t)_{0\leq t\leq1}$ is a continuous uniformly integrable martingale with respect to the filtration $(\tilde{\mathscr{F}} _t)_{0\leq t\leq1}$, where $\tilde{\mathscr{F}} _t = \mathscr{F} _{ \phi (t)} \vee \sigma (X_0)$, $\mathscr{F} _{ t}$ is the completion of $\mathscr{F} _{ t+}^W$.
Taking n independent Brownian motions, we may extend $\tilde{X}$ to $[0;n]$, and $\tilde{X}$ will even be uniformly integrable.
EDIT: Apparently, this construction is not correct. We also need to ensure that $(X_0,X_1X_0) \,{\buildrel d \over =}\, (\tilde{X}_0,W _{T})$, which is possible. But then $\tilde{X} _t = \tilde{X} _0+ W^T_{\phi (t)}$, need not be a martingale, because $\tilde{X} _0$ and $W$ are not independent, $\sigma (\tilde{X}_0)$ may contain some information about the "future".
Let $Q(x,A)$ be a transition probability kernel such that $Q(X_0,A)$ is a version of $ P\{X_1  X_0 \in A  X_0 \} $, and let $B$ be a Brownian motion, independent of $\tilde{X} _0$. For each $x\in R$, define a stopping time $T_x$ in such a way, that $B_{T_x}$ has distribution $Q(x,\cdot)$. Then, if we could provide measurability of $T_x$ by $x$, or alternatively if we could show, that $T_{X_0}$ is a random variable, then the process
$$\tilde{X} _t =\tilde{X} _0 + B _{\phi (t)}I_{\{\phi (t) \leq T_{X_0} \}}+ B _{T_{X_0}}I_{\{\phi (t) > T_{X_0} \}}, t\in [0;1), \tilde{X} _1 = B _{T_{X_0}} $$
would have desired properties. So, this should do for some particular cases, e.g. when $P\{ X_0 \in C \} =1$ for some countable set $C$. But I don't see how to extend this approach to the general case.