Basic (continuous) martingale properties

Question

I just learned about martingales in continuous time and solved some basic exercises. But unfortunately there are some seemingly easy and surely basic things I still have problems with.

1) Let $X=(X_t)_{t\geq0}$ be a martingale that is bounded in $L^{1}$. Then $\mathbb{E}[X_n^{+}|\mathcal{F}_t]\rightarrow M_t$, where $(M_t)_{t\geq0}$ is again a martingale.

I know that $X_t\rightarrow X_{\infty}$ as $t\rightarrow\infty$ a.s., $X_{\infty}$ being an integrable r.v. But I don't know anything about convergence in $L^1$ or (apriori) that $X_t^{+}$ is convergent?

To show that $M$ is a martingale, I need to exchange $\lim$ and $\mathbb{E}$. I don't see how to use monotone or dominated convergence in this case.

2) In 1) a general question came up. If $X_t\rightarrow X_{\infty}$ in $L^1$ or a.s. does one can conclude that $\mathbb{E}[X_t|\mathcal{F}_s]\rightarrow\mathbb{E}[X_{\infty}|\mathcal{F}_s]$ or $\mathbb{E}[X_t|\mathcal{F}_{\infty}]\rightarrow\mathbb{E}[X_{\infty}|\mathcal{F}_{\infty}]$? What if I have convergence in $L_1$ and a.s?

3) If $X$ is a right-continuous martingale, then for all $t\geq0$, we have that $X_{t+\varepsilon}\rightarrow X_t$ in $L^1$. I know it is true a.s. (by assumption), but I just don't see how to conclude the $L^1$ convergence...

EDIT: 4) The statement is, that a positive, right-continuous supermartingale $(X_t)_{t\geq0}$ is uniformly integrable if $\mathbb{E}[X_{\infty}]=\mathbb{E}[X_0]$

I already managed to show that it is bounded in $L^1$, that's all unfortunately. The key should be the positivity, but I just don't get it :/

Thanks a lot for your patience and your help!

Answer 1

1. First of all, we note that $(X_n^+)_{n \in \mathbb{N}}$ is a sub-martingale. Let $$Y_n := \mathbb{E}(X_n^+ \mid \mathcal{F}_t)$$ For $n >m \geq t$ we have by the tower property, $$\mathbb{E}(Y_n \mid \mathcal{F}_m) = \mathbb{E}(X_n^+ \mid \mathcal{F}_t) \geq \mathbb{E}(X_m^+ \mid \mathcal{F}_t) = Y_m$$ i.e. $(Y_n)_{n \geq t}$ is a $L^1$-bounded sub-martingale and therefore, by the martingale convergence theorem, $\mathbb{E}(X_n^+ \mid \mathcal{F}_t) \to M_t$ as $n \to \infty$ for some random variable $M_t$. Note that $M_t \geq 0$. By Fatous' lemma, $$\mathbb{E}(M_t) \leq \liminf_{n \to \infty} \mathbb{E}(\mathbb{E}(X_n^+ \mid \mathcal{F}_t)) \leq \sup_{n \in \mathbb{N}} \mathbb{E}(|X_n|)<\infty$$ i.e. $M_t \in L^1$. From $$\mathbb{E}(X_n^+ \mid \mathcal{F}_t) \geq \mathbb{E}(X_m^+ \mid \mathcal{F}_t)$$ we see that $(Y_n)_{n \geq t}$ is increasing, i.e. $M_t = \sup_n Y_n$. Consequently, we find by the monotone convergence theorem $$\mathbb{E}(M_t \mid \mathcal{F}_s) = \lim_{n \to \infty} \mathbb{E}(\mathbb{E}(X_n^+ \mid \mathcal{F}_t) \mid \mathcal{F}_s) = \lim_{n \to \infty} \mathbb{E}(X_n^+ \mid \mathcal{F}_s) = M_s$$

2. If $X$ is a martingale, then $X_t \to X_{\infty}$ in $L^1$ implies $\mathbb{E}(X_t \mid \mathcal{F}_s) \to \mathbb{E}(X_{\infty} \mid \mathcal{F}_s)$ a.s. for any $s \leq t$. This follows readily from the definition: Fix $F \in \mathcal{F}_s$, then $$\int_F \mathbb{E}(X_t \mid \mathcal{F}_s) \, d\mathbb{P} = \int_F X_t \, d\mathbb{P} \to \int_F X_{\infty} \, d\mathbb{P} = \int \mathbb{E}(X_{\infty} \mid \mathcal{F}_s) \, d\mathbb{P}$$ Concerning $\mathbb{E}(X_t \mid \mathcal{F}_{\infty}) \to \mathbb{E}(X_{\infty} \mid \mathcal{F}_{\infty})$: Note that this is equivalent to $X_t \to X_{\infty}$ since $X_t$ and $X_{\infty}$ are $\mathcal{F}_{\infty}$-measurable.

3. Fix $t \geq 0$. Since $\mathbb{E}(M_{t+1} \mid \mathcal{F}_s) = M_s$ for any $t \leq s \leq t+1$, it follows that $(M_{t+\varepsilon})_{\varepsilon>0}$ is uniformly integrable. Hence, by the (backwards) martingale convergence theorem, $M_{t+\varepsilon} \to Y$ a.s. and in $L^1$ as $\varepsilon \to 0$. Because of the right-continuity, we find $Y = M_t$.

4. Since $X$ is a positive super-martingale, we have $X_t \to X_{\infty}$ a.s. By Fatous Lemma, the additional assumption and the super-martingale property, $$\mathbb{E}(X_0) = \mathbb{E}(X_{\infty}) \leq \liminf_{t \to \infty} \mathbb{E}(X_t) \leq \mathbb{E}(X_0)$$ i.e. all expressions in the last line are equal. Since the expectations are decreasing, this shows $\mathbb{E}(X_t) \to \mathbb{E}(X_{\infty})$. By Scheffé's theorem, we find $X_t \to X_{\infty}$ in $L^1$. This implies that $(X_t)_t$ is uniformly integrable.