In discrete time, it is quite easily shown that a process is predictable if and only if it is natural (as seen here for example). However, in continuous time it is not nearly so clear (and certainly not clear to me). We define a process $A$, assumed to be of finite variation, as natural if and only if for every bounded RCLL martingale $M$, we have $$E(M_tA_t) = E \int_0^t M_{s-} dA_s $$ which is analogous to the discrete time definition (of coure the integral makes sense as $A_s$ is finite variation).

My question is: how does one prove that a process is natural if and only if it is predictable (i.e. measurable with respect to the sigma field generated by left continuous adapted processes)?

I have never seen the proof of this fact in continuous time, and would really like to as a pathway to understand this note on the Doob Meyer decomposition theorem. Any help would be massively appreciated, and I am sorry that I really have no idea where to start.


2 Answers 2


The books mentioned in other answers (see also this question) might be hard to get and their proofs employ some advanced theory. I think natural = predictable should be illuminated early in stochastic calculus, even before the Doob-Meyer decomposition and stochastic integration. In that light I will try to offer an original proof based on elementary methods. I have drawn heavily from the work of P. Spreij and G. Lowther. The errors are mine alone (I'm a student).


The following definition is a well known equivalent of the OP's.

$A$ is natural when it is an increasing process that satisifes \begin{equation}\tag{1} \forall t \ge 0 : \mathbb E \int_{(0,t]} M_{s} dA_s = \mathbb E \int_{(0,t]} M_{s-} dA_s \end{equation} for all bounded right continuous martingales $M$.

The integrals are Lebesgue-Stieltjes.

Predictable processes are those measurable in the predictable $\sigma$-algebra which may be defined as the one generated by the left continuous processes.

Claim. $A$ is natural if and only if $A$ is predictable.

Natural implies predictable

Let natural $A$ be decomposed as a continuous process plus a countable sum of jumps $\Delta A_T$ at stopping times $T$. Continuous processes are predictable so we need only prove that $\Delta A_T \mathbf 1_{\{t \ge T\}}$ is predictable. Notice that both $T$ and $\Delta A_T$ are strictly positive.

For convenience set $M'_s = M_{s-}$ and $B_T = \Delta A_T \mathbf 1_{\{T < \infty\}}$.

Evaluate $(1)$ for any time $t$: \begin{equation}\tag{2} \mathbb E M_T B_T \mathbf 1_{\{t \ge T\}} = \mathbb E M'_T B_T \mathbf 1_{\{t \ge T\}} \end{equation}

\begin{equation}\tag{3} \mathbb E \Delta M_T B_T \mathbf 1_{\{t \ge T\}} = 0 \end{equation}


  • $T$ is a predictable stopping time.
  • $M'_T \in \mathbb E [M_T | \mathcal F_{T-}]$ because $T$ is predictable.
  • $B_T \in \mathbb E [B_T | \mathcal F_{T-}]$ by a special choice of $M$.
  • $B_T \mathbf 1_{\{t \ge T\}}$ is predictable by $\mathcal F_{T-}$ measurability.

The key parts are really the predictability of $T$ and the $\mathcal F_{T-}$ measurability of $B_T$. The rest is boilerplate finishing with $B_T \mathbf 1_{\{t \ge T\}}$ being predictable because $\sigma(B_T \mathbf 1_{\{t \ge T\}})$ is contained in the predictable $\sigma$-algebra.

$T$ is a predictable stopping time

We appeal to a construction of $M$ related to estimation of $T - t$, in such a way that $\Delta M_T \le 0$. Finding that $\Delta M_T = 0$ using $(3)$, the announcing sequence that defines a predictable stopping time will follow.

First choose a map $f : [0,\infty) \to [0,\infty)$ that is bounded, continuous, and strictly increasing (think of sigmoids or $\tanh$ for example). The process $Y_t = E[f(T) - f(t) | \mathcal F_t]$ estimates (however poorly) the time left until stopping. What matters is that $Y$ is positive when $t < T$ and identically zero when $t=T$. Therefore $\Delta Y_T \le 0$.

Let $M_t \in \mathbb E[f(T) | \mathcal F_t]$ which is a bounded martingale. Given the usual condition of a right continuous filtration let $M_t$ be moreover (a.s.) right continuous. Since $M_t = Y_t + f(t)$ and $f(t)$ is continuous we have $\Delta M_T = \Delta Y_T \le 0$. Finally as $t \to \infty$ in $(3)$ it's clear that $\Delta M_T$ is zero a.e. in $\{T < \infty\}$.

The announcing sequence $T_n$ is now given by $T_n = n \wedge \inf \{t : Y_t \le 1/n\}$. The conditions $T_n < T$ and $T_n \to T$ are satisfied by continuity since (a.s.) $\Delta Y_T = \Delta M_T = 0$. Per these criteria $T$ is predictable.

$M'_T \in \mathbb E [M_T | \mathcal F_{T-}]$

We need this in the next section. It's a standard result of predictable projection theory, but here is a direct proof.

One must show that $\forall G \in \mathcal F_{T-} : \mathbb E \mathbf 1_G M'_T = \mathbb E \mathbf 1_G M_T$. First let $k,n \in \mathbb N$ such that $n > k$ and take $G \in \mathcal F_{T_k}$. Apply optional stopping then dominated convergence.

\begin{align} \mathbb E \mathbf 1_G M_T &= \mathbb E \mathbf 1_G \mathbb E \left[ \mathbb E[M_T | \mathcal F_{T_n}] \big| \mathcal F_{T_k} \right] \\ &= \mathbb E \mathbf 1_G \mathbb E [M_T | \mathcal F_{T_n}] \\ &= \mathbb E \mathbf 1_G M_{T_n}\ (\forall n > k) \\ &= \lim_{n \to \infty} \mathbb E \mathbf 1_G M_{T_n} \\ &= \mathbb E \lim_{n \to \infty} \mathbf 1_G M_{T_n} \\ &= \mathbb E \mathbf 1_G M'_T \\ \end{align}

Equality for $G \in \mathcal F_{T-}$ follows from the case $G \in \cup_{k \in \mathbb N} \mathcal F_{T_k}$ because the latter collection is a $\pi$-system and it generates $\mathcal F_{T-}$ (almost by definition). Let $\mathcal D \subseteq 2^\Omega$ comprise all sets $G$ for which the stated equality holds. Then $\mathcal D$ contains the $\pi$-system. By the linearity of the integral and by dominated convergence $\mathcal D$ is a d-system. Hence $\mathcal D = \mathcal F_{T-}$ per Dynkin's lemma.

$B_T$ is $\mathcal F_{T-}$ measurable

Let $\hat B_T = \mathbb E[B_T | \mathcal F_{T-}]$. Both $M'_T$ and $\mathbf 1_{\{t \ge T\}}$ are $\mathcal F_{T-}$ measurable, thus from $(2)$:

\begin{align} \mathbb E M_T B_T \mathbf 1_{\{t \ge T\}} &= \mathbb E M'_T B_T \mathbf 1_{\{t \ge T\}} \\ &= \mathbb E M'_T \hat B_T \mathbf 1_{\{t \ge T\}} \\ &= \mathbb E \left( \mathbb E [M_T | \mathcal F_{T-}] \hat B_T \mathbf 1_{\{t \ge T\}} \right) \\ &= \mathbb E M_T \hat B_T \mathbf 1_{\{t \ge T\}} \\ \end{align}

\begin{equation}\tag{4} \mathbb E M_T (B_T - \hat B_T)\mathbf 1_{\{t \ge T\}} = 0 \end{equation}

Finally $B_T = \hat B_T$ must hold a.e. because $M$ in $(4)$ may be chosen as follows:

  • $Z = \mathbf 1_{\{B_T > \hat B_T\}} - \mathbf 1_{\{B_T < \hat B_T\}}$
  • $M_t = \mathbb E [Z | \mathcal F_t]$
  • $M_T = Z$

Here the (bounded) martingale $M_t$ is taken (a.s.) right continuous under the usual condition of a right continuous filtration. Thus $(4)$ applies and sending $t \to \infty$ we have $B_T \ne \hat B_T$ on at most a null set within $\{T < \infty\}$. Both vanish on $\{T = \infty\}$ so they are indeed a.e. identical.

$A$ is therefore predictable

It has now been shown that $B_T \mathbf 1_{\{t \ge T\}}$ is a member of a class that generates the predictable $\sigma$-algebra. Recall that by decomposition into such processes the predictability of $A$ follows.


Predictable implies natural

When $A$ is a continuous process of bounded variation the condition $(1)$ is true. This follows because the measure induced by $A$ is absolutely continuous w.r.to $dt$ and hence the countable discontinuities of the integrands fall in a set of measure zero.

So for $A$ increasing and predictable, like before we need only consider a countable sum of jumps at stopping times $T$ and prove that $\Delta A_T \mathbf 1_{\{t \ge T\}}$ is natural. Again $T$ and $\Delta A_T$ are strictly positive. It should be clear that $\mathbf 1_{\{t \ge T\}}$ by itself is also predictable.

Now it will not be a surprise that $T$ is a predictable stopping time. There is a nice theorem that relies on stochastic integrals. I will give a direct argument.

$T$ is a predictable stopping time

Consider the process $\mathbb E [T-t | \mathcal F_t]$. Pathwise, this must be positive when $t < T$ and zero when $t = T$. If it is moreover (a.s.) continuous from the left when $t = T$ the predictability of $T$ follows immediately (take decreasing hitting times as the announcing sequence). Technically the announcing sequence should also go to infinity if $T=\infty$.

These goals in mind, a weaker claim that's easy to prove is the continuity of $\mathbb E [T \wedge n - t | \mathcal F_t]$ for stopping times $(T \wedge n)_{n \in \mathbb N}$. Let $\{X^n : n \in \mathbb N\}$ and $\{Y^u : u > 0\}$ be the martingales in the next display.

\begin{equation}\tag{5} X^n_t = \mathbb E [T \wedge n | \mathcal F_t] \le \mathbb E [u \mathbf 1_{\{T \le u\}} + n \mathbf 1_{\{T > u\}} | \mathcal F_t] = Y^u_t \end{equation}

\begin{equation}\tag{6} \forall \omega \in \Omega : \forall t > 0 : X^n(t,\omega) \le \inf_{u > 0} Y^u(t,\omega) \end{equation}

Here $(6)$ merely restates the inequality $(5)$ to make a point. Fix arbitrary $\omega$ and send $t \to T(\omega) \wedge n$ from below. Assume there is a process in $\{Y^u\}$ such that $Y^u(t,\omega) - t \to 0$ continuously. Then by $(6)$ also $X^n(t,\omega) - t \to 0$ continuously.

Indeed the assumption holds for $\{Y^u : u > 0\}$. Fix $u > 0$ and send $t \to u$ from below. $Y^u_u$ must be continous from the left because $\mathbf 1_{\{T \le u\}}$ is predictable.

$$Y^u_u = \mathbb E [Y^u_u | \mathcal F_u] = \mathbb E [Y^u_u | \mathcal F_{u-}] = Y^u_{u-} \ \text{(a.s.)}$$

Moreover if $u = T(\omega) \wedge n$ then $Y^u(u,\omega) = u$ and $Y^u(t,\omega) - t = 0$ when $t=u$.

Now take $(T^n_k)_{k \in \mathbb N} = \inf \{t : X^n_t - t \le 1/k\}$ as the announcing sequence for $T \wedge n$. Clearly $n < m$ implies $X^n_t \le X^m_t$ and $T^n_k \le T^m_k < T$. Hence $(T^n_n)_{n \in \mathbb N}$ announces $T$.

$A$ is therefore natural

Finally, skipping over details elaborated in the last section, \begin{align} \forall t > 0 : \mathbb E M_T \Delta A_T \mathbf 1_{\{t \ge T\}} &= \mathbb E \left( \mathbb E [M_T \Delta A_T \mathbf 1_{\{t \ge T\}} | \mathcal F_{T-}] \right) \\ &= \mathbb E M'_T \Delta A_T \mathbf 1_{\{t \ge T\}} \\ \end{align} and condition $(1)$ has been shown.



Regarding your problem, pls refer to C. Dellacherie & P. Meyer, Probabilities and Potential B, North-Holland Publishing Company, 1982. p.126, Th. VI.2.61.


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