As we know, a RCLL submartingale on [0,T], $Y$, in class D can be decomposed as: $$Y_t=Y_0+M_t+A_t,\ a.s.,$$ where $M$ is a martingale and $A$ is an increasing previsible process.

In my question, I have some specific assumptions:

(i) $Y$ is continuous in t, and is bounded;

(ii) the filtration is generated by Brownian motion, so both $M$ and $A$ are continuous.

My question is whether we have any result about the integrability of $A^2$ (of course, if it holds, it must result from assumption (i)), so that $M$ is a square integrable martingale ? Thanks !


1 Answer 1


Since $(Y_t)_{t \geq 0}$ is a bounded submartingale, there exists $Y_{\infty} \in L^1$ such that $Y_t \to Y_{\infty}$ in $L^1$ as $t \to \infty$. Moreover, by the monotone convergence theorem,

$$\mathbb{E}\left( \sup_{t \geq 0} A_t \right) = \sup_{t \geq 0} \mathbb{E}(A_t),$$

i.e. $A_t \to A_{\infty} := \sup_{t \geq 0} A_t$ in $L^1$. The decomposition

$$Y_t = Y_0 + M_t+A_t$$

yields $M_t \to M_{\infty}$ in $L^1$ for some random variable $M_{\infty}$ and $(M_t)_{t \in (0,\infty]}$ is a martingale; in particular uniformly integrable. Now the following theorem applies:

Theorem Let $Y_t = M_t+A_t$ where $(M_t)_{t \geq 0}$ is a uniformly integrable martingale and $(A_t)_t$ an increasing predictable process with $A_0=0$ a.s. If $(Y_t)_t$ is bounded, then $\mathbb{E}(A_{\infty}^p)<\infty$ for any $p>0$.

Proof: Fix $C>0$ such that $|Y_t| \leq C$ for all $t \geq 0$. For $\lambda>0$ we define a stopping time $$T:= \inf\{t \geq 0; A_t \geq \lambda\}.$$ Then

$$\begin{align*} \mathbb{P}(A_{\infty} \geq \lambda+4C) &= \mathbb{P}(A_{\infty} \geq \lambda+C, T<\infty) \leq \mathbb{P}(A_{\infty}-A_T \geq 4C, T<\infty) \\ &\leq \frac{1}{4C} \mathbb{E}((A_{\infty}-A_T) \cdot 1_{\{T<\infty\}}). \tag{1}\end{align*}$$

Now let $$T_n := \inf\{t \geq 0; A_t \geq \lambda-1/n\}.$$

Then $T_n \uparrow T$ and $\mathcal{F}_{T_n} \subseteq \mathcal{F}_T$. For $j \geq k$, we have $\{T_k <N\} \in \mathcal{F}_{T_k} \subseteq \mathcal{F}_{t_j}$ and therefore, by the tower property,

$$\begin{align*} \mathbb{E}((A_{\infty}-A_{T_j}) \cdot 1_{\{T_k<N\}}) &= \mathbb{E}\bigg[ \mathbb{E}(A_{\infty}-A_T \mid \mathcal{F}_{T_k}) \cdot 1_{\{T_k<N\}} \bigg] \\ &= \mathbb{E}\bigg[ \mathbb{E}(Y_{\infty}-Y_T \mid \mathcal{F}_{T_k}) \cdot 1_{\{T_j<N\}} \bigg] \\ &\leq 2C \mathbb{P}(T_k<N). \end{align*}$$

Here we used in the second step that $-M_t = A_t-Y_t$ is a uniformly integrable martingale. Applying Fatou's lemma yields for $j \to \infty$

$$ \mathbb{E}((A_{\infty}-A_T) \cdot 1_{\{T_k<N\}}) \leq 2C \mathbb{P}(T_k<N).$$

Applying again Fatou's lemma (for $k \to \infty$) we see that

$$\mathbb{E}((A_{\infty}-A_T) \cdot 1_{\{T<N\}}) \leq 2 C \mathbb{P}(T \leq N).$$

Finally, by the monotone convergence theorem,

$$\mathbb{E}((A_{\infty}-A_T) \cdot 1_{\{T<\infty\}}) \leq 2 C \mathbb{P}(T < \infty). \tag{2}$$

Combining $(1)$ and $(2)$ we obtain

$$\mathbb{P}(A_{\infty} \geq \lambda+4C) \leq \frac{1}{2} \mathbb{P}(T<\infty) = \frac{1}{2} \mathbb{P}(A_{\infty} \geq \lambda).$$

In particular, for $\lambda = 4kC$, $k \in \mathbb{N}_0$,

$$\mathbb{P}(A_{\infty} \geq (k+1) 4C) \leq \frac{1}{2} \mathbb{P}(A_{\infty} \geq 4kC).$$

Iterating this inequality yields

$$\mathbb{P}(A_{\infty} \geq 4kC) \leq \frac{1}{2^{k-1}}.$$

Now the claim follows from the fact that

$$\mathbb{E}(A_\infty^p) = p \int_{(0,\infty)} x^{p-1} \mathbb{P}(A_{\infty} \geq x) \, dx.$$

Remark The proof is adapted from Proposition 16.32 in Stochastic Processes by Richard F. Bass.

(... and I guess that there is an easier (or more elegant) way to prove the claim; this one seems rather like overkill to me.)


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