2
$\begingroup$

I'm looking at this theorem:

Let $(X_t, \mathcal{F}_t)_{t \in \mathbb{R}_+}$ be a submartingale, where $\mathcal{F}$ is right-continuous and complete, and the function $\mu(t) := E[X_t]$ is right-continuous. Then there exists a submartingale $(Y_t, \mathcal{F}_t)_{t \in \mathbb{R}_+}$ satisfying $X_t = Y_t$ a.s. for all $t$ whose paths $t \mapsto Y_t(\omega)$ are rcll (right continuous with left limit) a.s.

The proof constructs $Y_t$ such that $Y_t = \underset{s\in \mathbb{Q}_+, s \searrow t}{\lim X_s}$ a.s., which shows that the paths of $Y_t$ are right-continuous. But it never proves the existence of the left limit. Is the family $(X_r, \mathcal{F}_r)_{r<t}$ a uniformly integrable submartingale? If it is, I can see that we can apply the Submartingale Convergence Theorem to show that the left limit exists. I apologize if this is an obvious question. Here's my attempt so far:

1) $E[X_r] \leq E[X_t] < \infty$ for all $r < t \implies \sup_{r<t} E[X_r] < \infty$

2) Let $\epsilon > 0$. We want to show that there exists $\delta > 0$ such that for all $A \in \mathcal{F}_t$$\mathbb{P}(A) < \delta \implies \sup_{r<t} E[X_r\mathbb{1}_A] < \epsilon$. Since $(X_t)$ is a submartingale, we have $E[X_t|X_r] \geq X_r$ a.s., and so $E[X_r\mathbb{I}_A] \leq E[X_t\mathbb{I}_A]$ for all $A \in \mathcal{F}_r$. And I'm stuck...

$\endgroup$

1 Answer 1

1
$\begingroup$

Basically your idea is correct: Since $$\mathbb{E}(X_t \mid \mathcal{F}_r) \geq X_r, \qquad r \leq t, $$ the sequence $(X_r)_{r \leq t}$ is "dominated" by the conditional expectations of $X_t$ and therefore uniformly integrable. But, in order to show uniform integrability, we have to estimate $|X_r|$ from above and this makes things a bit more complicated:

Fix $t \geq 0$ and a sequence $(t_n)_{n \in \mathbb{N}}$ with $t_n \downarrow t$. It follows from the right-continuity that

$$M := \sup_{n \in \mathbb{N}} \mathbb{E}|X_{t_n}|<\infty.$$

For given $\epsilon>0$ we can pick $N \in \mathbb{N}$ such that

$$\mathbb{E}X_{t_n} \geq \mathbb{E}X_{t_N}- \epsilon \qquad \text{for all} \, \, n \geq N \tag{1}$$

(as $\mathbb{E}X_{t_n}$ is decreasing as $n \to \infty$). The submartingale-property implies

$$\begin{align*} \int_{|X_{t_n}| > c} |X_{t_n}| \, d\mathbb{P} &= \int_{X_{t_n}<c} (-X_{t_n}) \, d\mathbb{P} + \int_{X_{t_n}>c} X_{t_n} \, d\mathbb{P} \\ &= \int_{X_{t_n} \geq -c} - \mathbb{E}X_{t_n} + \int_{X_{t_n}>c} X_{t_n} \, d\mathbb{P} \\ &\leq \int_{X_{t_n} \geq -c} X_{t_N} \, d\mathbb{P} - \mathbb{E}X_{t_n} + \int_{X_{t_n} >c} X_{t_N} \,d \mathbb{P}. \end{align*}$$

By $(1)$,

$$\begin{align*} \int_{|X_{t_n}| > c} |X_{t_n}| \, d\mathbb{P} &\leq \int_{X_{t_n} \geq -c} X_{t_N} \, d\mathbb{P}-\mathbb{E}X_{t_N} + \epsilon + \int_{X_{t_n} >c} X_{t_N} \, d\mathbb{P} \\ &\leq \int_{|X_{t_n}| \geq c} X_{t_N} \, d\mathbb{P}+ \epsilon. \end{align*}$$

Hence, by Markov's inequality

$$\begin{align*} \int_{|X_{t_n}| > c} |X_{t_n}| \, d\mathbb{P} &\leq \int_{\{|X_{t_n}| \geq c\} \cap \{|X_{t_N} > R\}} X_{t_N} \, d\mathbb{P} + \int_{\{|X_{t_n}| \geq c\} \cap \{|X_{t_N} \leq R\}} X_{t_N} \, d\mathbb{P} +\epsilon \\ &\leq \int_{|X_{t_N}| \geq R} X_{t_N} \, d\mathbb{P} + R \mathbb{P}(|X_{t_n}| \geq c) + \epsilon \\ &\leq \int_{|X_{t_N}| \geq R} X_{t_N} \, d\mathbb{P} + R \frac{M}{c} + \epsilon \\ &\stackrel{c \to \infty}{\to} \int_{|X_{t_N}| \geq R} X_{t_N} \, d\mathbb{P}+\epsilon. \end{align*}$$

Letting $R \to \infty$ and $\epsilon \to 0$ finishes the proof.


Remark: Note that the existence of the limit $$Y_t = \lim_{\mathbb{Q} \ni s \downarrow t} X_s$$ also follows directly from Doob's upcrossing estimate. Applying Doob's upcorssing estimate is much more easier than proving uniform integrability.

$\endgroup$
1
  • $\begingroup$ Wow, thank you for taking the time answer this lengthy question. It was very helpful! $\endgroup$
    – user207886
    Feb 13, 2015 at 0:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .