Uniform integrability of martingale projection Suppose $X$ is a martingale relative to a filtration $(F_t)$ and that $(G_t)$ is a sub-filtration of $F$. Define a new martingale $Y$ by $Y_t := E(X_t | G_t) $.
Question: does "$Y$ is uniformly integrable" imply "$X$ is uniformly integrable"?
Many thanks.

As saz pointed out, the answer is no. If $\mathcal{G}_t := \left\{\emptyset,\Omega\right\}$, then $Y_t=\mathbb E\left[X_t\right]=\mathbb E\left[X_0\right]$ whatever the martingale $X$ is. In particular, $X$ may not be uniformly integrable.
