TITLE: Can every discrete martingale be embedded in a continuous martingale?
QUESTION [10 upvotes]: 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!

REPLY [5 votes]: 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 
    $$N_i = W_i - W_{i-1}.$$ 
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 discrete-time 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 $i=0,\cdots,N-1$, 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 $\mathcal{G}_{i-1}$. Let 
$$F_{X_i|X_0,\cdots, X_i}(x| x_0, \cdots, x_{i-1}) = Q((x_0,\cdots, x_i,0,\cdots,0), \{X_i\le x_i\}).$$ 
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 $F_{X_i|X_0,\cdots, X_{i-1}}$ by $F_i$. Let 
$$G_i(y| x_0, \cdots, x_{i-1}) = \inf\{x:F_i(x| x_0,\cdots, x_{i-1})\ge y\}$$
Let $Y_0$ equal the constant $X_0$ and
$$Y_i = G_i(\phi(N_i)| Y_0,\cdots, Y_{i-1})$$
$X_0$ has the same law as $Y_0$. Suppose $(X_0, \cdots,X_{i-1})$ has the same law as $(Y_0, \cdots,Y_{i-1})$. Fix $x_0, \cdots, x_i$.
$$
\begin{align}\{Y_i\le x_i\} &= \{G_i(\phi(N_i), Y_0,\cdots, Y_{i-1}) \le x_i\}\\
&= \{F_i(x_i| Y_0,\cdots, Y_{i-1})\ge \phi(N_i)\}\\
&= \{Q((Y_0,\cdots,Y_{i-1}, x_i, 0,\cdots, 0), \{X_i\le x_i\})\ge \phi(N_i)\}
 \end{align}
$$
 Similarly, 
$$
 \begin{align}\mathbb{Q}[Y_0\le x_0, \cdots, Y_i\le x_i] &= \mathbb{Q}[Y_0\le x_0,\cdots,Y_{i-1}\le x_{i-1}, Q((Y_0,\cdots,Y_{i-1}), \{X_i\le x_i\})\ge \phi(N_i)]\\
 &= \mathbb{E}[Q((Y_0,\cdots, Y_{i-1}),\{X_i\le x_i\})\mathbf{1}_{Y_0\le x_0}\cdots\mathbf{1}_{Y_{i-1}\le x_{i-1}}]\\
 &=  \mathbb{E}[Q((X_0,\cdots, X_{i-1}),\{X_i\le x_i\})\mathbf{1}_{X_0\le x_0}\cdots\mathbf{1}_{X_{i-1}\le x_{i-1}}]\\
 &= \mathbb{P}[X_0\le x_0, \cdots, X_i\le x_i]
 \end{align}
$$