Let $X=(X)_{0\le t\le 1}$ be a continuous martingale starting at $0$, then denote by $\mu$ and $\nu$ the probability laws of $\int_0^1X_tdt$ and $X_1$. Then it is easy to see that the couple $(\mu,\nu)$ is increasing in convex order, i.e.

$$\int_{R}f(x)d\mu(x)~~\le~~\int_{R}f(x)d\nu(x)$$

holds for all convex functions $f:R\to R$ of linear growth, see also "Peacocks and Associated Martingales, with Explicit Constructions" for further details. Now my question is given by probability measures $\mu$ and $\nu$ on $R$, could we give some conditions on the couple $(\mu,\nu)$ such that there exists a continuous martingale $X=(X)_{0\le t\le 1}$ satisfying

$$X_1\sim\mu~~\mbox{and}~~ \int_0^1X_tdt\sim\nu$$

Let $X=(X)_{0\le t\le 1}$ be a continuous martingale starting at $0$, then denote by $\mu$ and $\nu$ the probability laws of $\int_0^1X_t \mathrm{d}t$ and $X_1$. Then it is easy to see that the couple $(\mu,\nu)$ is increasing in convex order, i.e.

$$\int_{\mathbb{R}}f(x)\mathrm{d}\mu(x)~~\le~~\int_{\mathbb{R}}f(x)\mathrm{d}\nu(x)$$

holds for all convex functions $f:\mathbb{R}\to \mathbb{R}$ of linear growth, see also "Peacocks and Associated Martingales, with Explicit Constructions" for further details. Now my question is given by probability measures $\mu$ and $\nu$ on $\mathbb{R}$, could we give some conditions on the couple $(\mu,\nu)$ such that there exists a continuous martingale $X=(X)_{0\le t\le 1}$ satisfying

$$X_1\sim\mu~~\mbox{and}~~ \int_0^1X_t\mathrm{d}t\sim\nu ?$$