Show $(\mathbb{E}[Z\mid\mathcal{F}_t],\mathcal{F_t})_{t \geq0}$ is an *uniformly integrable* martingale.

Question

Suppose $Z \in \mathcal{L}^1(P)$. I want to show that $(\mathbb{E}[Z\mid\mathcal{F}_t],\mathcal{F_t})_{t \geq0}$ is an uniformly integrable martingale.

I have managed to show to it is a martingale but I am stuck trying to show is uniformly integrable as well. I have tried looking at the sufficient condition that $$ \sup_{t\geq 0} \mathbb{E}[(\mathbb{E}[Z\mid\mathcal{F}_t])^2] < \infty $$ and my idea was to use Jensens inequality but that doesn't seem to work. I would also need to somehow use that $Z \in \mathcal{L}^1(P)$ as I haven't used that yet.

Answer 1

First, it suffices to consider the case where $Z$ is non-negative. Indeed, write $Z$ as the difference of two integrable non-negative random variables and use the fact that a sum of two uniformly integrable families is uniformly integrable.

For any $t\geqslant 0$ and $R>0$, the event $\{\mathbb E\left[Z\mid\mathcal F_t\right]>R\}$ is $\mathcal F_t$ measurable hence $$ \mathbb E\left[\mathbb E\left[Z\mid\mathcal F_t\right]\mathbf{1}_{\{\mathbb E\left[Z\mid\mathcal F_t\right]>R\}}\right]=\mathbb E\left[Z\mathbf{1}_{\{\mathbb E\left[Z\mid\mathcal F_t\right]>R\}}\right]. $$ Moreover, for each positive $K$, $$ \mathbb E\left[Z\mathbf{1}_{\{\mathbb E\left[Z\mid\mathcal F_t\right]>R\}}\right]\leqslant \mathbb E\left[Z\mathbf{1}_{\{Z>K\}}\right] +K\mathbb P\left(\{\mathbb E\left[Z\mid\mathcal F_t\right]>R\}\right) $$ hence by Markov's inequality, $$ \mathbb E\left[Z\mathbf{1}_{\{\mathbb E\left[Z\mid\mathcal F_t\right]>R\}}\right]\leqslant \mathbb E\left[Z\mathbf{1}_{\{Z>K\}}\right] +\frac{K}{R}\mathbb E\left[Z\right]. $$ We thus got that for each $R,K>0$, $$ \sup_{t\geqslant 0}\mathbb E\left[\mathbb E\left[Z\mid\mathcal F_t\right]\mathbf{1}_{\{\mathbb E\left[Z\mid\mathcal F_t\right]>R\}}\right]\leqslant \mathbb E\left[Z\mathbf{1}_{\{Z>K\}}\right] +\frac{K}{R}\mathbb E\left[Z\right]. $$ Now take $K=\sqrt R$ for example.

Answer 2

Uniform integrability may be obtained from the following result.

Proposition: Let $(\Omega,\mathscr{F},\mathbb{P})$ be a probability space. Suppose that $\frak{A}$ is a collection of $\sigma$--algebras contained in $\mathscr{F}$. If $Z\in\mathcal{L}_1(\mathbb{P})$, then the family $\{\mathbb{E}[Z|\mathscr{A}]: \mathscr{A}\in\frak{A}\}$ is uniformly integrable.

Here is one way to prove this. Denote $Z_\mathscr{A}:=\mathbb{E}[Z|\mathscr{A}]$. Since $$|Z_\mathscr{A}|=|\mathbb{E}[Z\,|\mathscr{A}]\,|\leq \mathbb{E}[|Z|\,|\mathscr{A}],$$ $\|Z_\mathscr{A}\|_1=\mathbb{E}[|Z_\mathscr{A}|]\leq \|Z\|_1=\mathbb{E}[|Z|]$, and \begin{align} \int_{\{|Z_\mathscr{A}|>a\}}|Z_\mathscr{A}|\,d\mathbb{P}\leq \int_{\{\mathbb{E}[|Z|\,|\mathscr{A}]>a\}}\mathbb{E}[|Z|\,|\mathscr{A}]\,d\mathbb{P}= \int_{\{\mathbb{E}[|Z|\,|\mathscr{A}]>a\}} |Z|\,d\mathbb{P}. \end{align} Since $\mathbb{P}\big(\mathbb{E}[|Z|\,|\mathscr{A}]>a\big)\leq\frac{\mathbb{E}[|Z|]}{a}\xrightarrow{a\rightarrow\infty}0$, we conclude that $$\inf_{a>0}\sup_{\mathscr{A}\in\frak{A}} \int_{\{|Z_{\mathscr{A}}|>a\}}|Z_\mathscr{A}|\,d\mathbb{P} =0.$$ Hence $\{Z_\mathscr{A}:\mathscr{A}\in\frak{A}\}$ is uniformly integrable.