Doob inequalities and Doob-Meyer decomposition

The first post I wrote on this blog was about martingales, way back in 2012 at a time when I had known what a martingale was for about a month. I now don’t have this excuse. So I’m going to write about a couple of properties of (discrete-time) martingales that came up while adjusting a proof which my thesis examiners suggested could be made much shorter as part of their corrections.

Doob’s submartingale inequality

When we prove that some sequence of processes converges to some other process, we typically want to show that this holds in some sense uniformly over a time-interval, rather than just at some fixed time. We don’t lose much at this level of vagueness by taking the limit process to be identically zero. Then, if the convergent processes are martingales or closely similar, we want to be able to bound \sup_{k\le n} |Z_k| in some sense.

Doob’s submartingale inequality allows us to do this. Recall that a submartingale has almost-surely non-negative conditional increments. You might think of it heuristically as ‘more increasing than a martingale’. If Z_n is a martingale, then |Z_n| is a submartingale. This will be useful almost immediately.

The statement is that for (Z_n) a non-negative submartingale,

\mathbb{P}\left( \sup_{k\le n} Z_k \ge \lambda\right) \le \frac{\mathbb{E}\left[Z_n\right]}{\lambda}.

The similarity of the statement to the statement of Markov’s inequality is no accident. Indeed the proof is very similar. We consider whether the event in question happens, and find lower bounds on the expectation of Z_n under both possibilities.

Formally, for ease of notation, let Z_n^* be the running maximum \sup_{k\le n}Z_k. Then, we let T:= n\wedge \inf\{k\le n, M_j\ge \lambda\} and apply the optional stopping theorem for submartingales at T, which is by construction at most n. That is

\mathbb{E}[Z_n]\ge \mathbb{E}[Z_T]=\mathbb{E}\left[Z_T\mathbf{1}_{Z_n^*<\lambda}\right] + \mathbb{E}\left[Z_T \mathbf{1}_{Z_n^*\ge \lambda}\right].

The first of these summands is positive, and the second is at least \lambda \mathbb{P}\left( Z_N^* \ge \lambda \right), from which the result follows.

We’ve already said that for any martingale Z_n, |Z_n| is a submartingale, but in fact f(Z_n) is a submartingale whenever f is convex, and \mathbb{E}|f(Z_n)|<\infty for each n. Naturally, this continues to hold when Z_n is itself a submartingale.

[Note that Z_n^* is also a submartingale, but this probably isn’t as interesting.]

A particularly relevant such function f is f(x)=x^p, for p>1. If we take Z_n a non-negative submartingale which is uniformly bounded in L^p, then by applying Holder’s inequality and this submartingale inequality, we obtain

\mathbb{E}\left( \sup_{k\le n}Z_n^p \right) \le \left(\frac{p}{p-1}\right)^p \mathbb{E}\left[ Z_n^p \right].

Since Z_n^p is a submartingale, then a limit in n on the RHS is monotone, and certainly a limit in n on the LHS is monotone, so we can extend to

\mathbb{E}\left( \sup_{k\le n}Z_\infty^p \right) \le \left(\frac{p}{1-p}\right)^p \mathbb{E}\left[ Z_\infty^p \right].

Initially, we have to define \mathbb{E}\left[ Z_\infty^p \right] through this limit, but in fact this result, Doob’s L^p inequality, shows that Z_\infty:= \lim Z_n exists almost surely as well.

Naturally, we will often apply this in the case p=2, and in the third of these three sections, we will see why it might be particularly straightforward to calculate \mathbb{E}\left[Z_\infty^2\right].

Remark: as in the case of Markov’s inequality, it’s hard to say much if the submartingale is not taken to be non-negative. Indeed, this effect can be seen even if the process is only defined for a single time step, for which the statement really is then Markov’s inequality.

Doob-Meyer decomposition

Unfortunately, most processes are not martingales. Given an discrete-time process X_n adapted to \mathcal{F}=(\mathcal{F}_n), it is a martingale if the conditional expectations of the increments are all almost surely zero. But given a general adapted process X_n which is integrable (so the increments have well-defined finite expectation), we can iteratively construct a new process M_n, where the increments are centred versions of X_n‘s increments. That is,

M_{n+1}-M_n:= X_{n+1}-X_n - \mathbb{E}\left[ X_{n+1}-X_n \,\big|\, \mathcal{F}_n\right] = X_{n+1}-\mathbb{E}\left[X_{n+1} \,\big|\, \mathcal{F}_n\right]. (*)

Then it’s immediately clear from the definition that M_n is a martingale.

There’s a temptation to tie oneself up in knots with the dependence. We might have that increments of the original process X_n depend on the current value of the process. And is it necessarily clear that we can recover the current value of the original process from the current value of M_n? Well, this is why we demand that everything be adapted, rather than just Markov. It’s not the case that M_n should be Markov, but it clearly is adapted.

Now we look at the middle expression in (*), and in particular the term we are subtracting, namely the conditional expectation. If we define, in the standard terminology, A_0=0 and

A_{n+1}-A_n:= \mathbb{E}\left[ X_{n+1}-X_n \,\big|\, \mathcal{F}_n\right],

then we have decomposed the original process X_n as the sum of a martingale M_n, and this new process A_n. In particular, note that the increment A_{n+1}-A_n given above is adapted to \mathcal{F}_n, which is a stronger condition than being adapted to \mathcal{F}_{n+1} as we would expect a priori. This property of the process (A_n) is called predictability (or possibly previsibility).

This decomposition X_n=X_0+M_n+A_n as just defined is called the Doob-Meyer decomposition, and there is a unique such decomposition where M_n is a martingale, and A_n is predictable. The proof of uniqueness is very straightforward. We look at the equalities given above as definitions of M_n,A_n, but then work in the opposite direction to show that they must hold if the decomposition holds.

I feel a final heuristic is worthwhile, using the term drift, more normally encountered in the continuous-time setting to describe infinitissimal expected increments. The increments of A_n represent the drift of X_n, and the increments of M_n are what remains from X_n after subtracting the drift. In general, the process to be subtracted to turn a non-martingale into a martingale is called a compensator, and the existence or otherwise of such processes is important but challenging for some classes of continuous-time processes.

In particular, note that when X_n is itself a martingale, then A_n\equiv 0. However, probably the most useful case is when X_n is a submartingale, as then the drift is always non-negative, and so A_n is almost surely increasing. The converse holds too.

This is relevant because this Doob-Meyer decomposition is obviously only a useful tool for treating X_n if we can handle the two processes M_n,A_n easily. We have tools to bound the martingale term, but this previsible term might in general be tricky, and so the case where X_n is a submartingale is good, as increasing processes are much easier than general processes, since bounding the whole process might involve only bounding the final term in many contexts.

Predictable quadratic variation

A particularly relevant example is the square of a martingale, that is X_n=M_n^2, where M_n is a martingale. By the convexity condition discussed earlier, X_n is a submartingale (provided it is integrable, ie M_n is square-integrable), and so the process A_n in its Doob-Meyer decomposition is increasing. This is often called the (predictable) quadratic variation of (X_n).

This predictable quadratic variation is sometimes denoted \langle X_n\rangle. This differs from the (regular) quadratic variation which is defined as the sum of the squares of the increments, that is [X_n]:= \sum_{k=0}^{n-1} (X_{k+1}-X_k)^2. Note that this is adapted, but obviously not previsible. The distinction between these two processes is more important in continuous time. There, they are almost surely equal for a continuous local martingale, but not for eg a Poisson process. (For a Poisson process, the PQV is deterministic, indeed linear, while the (R)QV is almost surely equal to the Poisson process itself.) In the discrete time setting, the regular quadratic variation is not relevant very often, while the predictable quadratic variation is useful, precisely because of this decomposition.

Whenever we have random variables which we then centre, there is a standard trick to apply when treating their variance. That is

A_{n+1}-A_n= \mathbb{E}\left[ M^2_{n+1}-M^2_n \,\big|\, \mathcal{F}_n\right]
= \mathbb{E}\left[ M^2_{n+1}\,\big|\, \mathcal{F}_n\right] - 2M_n^2 +M_n^2
= \mathbb{E}\left[ M^2_{n+1}\,\big|\, \mathcal{F}_n\right] - 2M_n \mathbb{E}\left[ M_{n+1}\,\big|\, \mathcal{F}_n\right] + M_n^2
= \mathbb{E}\left[ \left(M_{n+1}-M_n\right)^2\,\big|\, \mathcal{F}_n\right].

One consequence is seen by taking an ‘overall’ expectation. Because M_n^2-A_n is a martingale,

\mathbb{E}\left[M_n^2\right] = \mathbb{E}\left[A_n\right] = \mathbb{E}\left[M_0^2\right] + \sum_{k=0}^{n-1} \mathbb{E}\left[A_{k+1}-A_k\right]
= \mathbb{E}\left[ M_0^2\right] + \sum_{k=0}^{n-1}\mathbb{E}\left[ \left(M_{k+1}-M_k\right)^2 \right]. (**)

This additive (Pythagorean) property of the square of a martingale is useful in applications where there is reasonably good control on each increment separately.

We can also see this final property without the Doob-Meyer decomposition. For a martingale it is not the case that the increments on disjoint intervals are independent. However, following Williams 12.1 [1], disjoint intervals are orthogonal, in the sense that

\mathbb{E}\left[(M_t-M_s)(M_v-M_u)\right]=0,

whenever s\le t\le u\le v. Then, when we square the expression M_n=M_0+\sum M_{k+1}-M_k, and take expectations, all the cross terms vanish, leaving precisely (*).

References

[1] Williams – Probability with Martingales

I also followed the notes I made in 2011/12 while attending Perla Sousi’s course on Advanced Probability, and Arnab Sen’s subsequent course on Stochastic Calculus, though I can’t find any evidence online for the latter now.

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Brownian Motion is not finite variation

There is a natural definition of ‘pathwise’ stochastic integrals of a certain type of ‘simple’ process with respect to cadlag non-decreasing processes. It can be a shown that a function is of finite variation iff it can be expressed as the difference of two such functions. Hence, these finite variation processes can be used as variable of integration via an obvious linear extension. One direction of this result is obvious; the other is fiddly. To proceed, we show that the total valuation process is cadlag (and, obviously, increasing), and then check that a'=\frac12(v+a),a''=\frac12(v-a) are processes satisfying the conditions of the result.

Our overall aim is to define integrals with respect to Brownian Motion since that is (in a sense to be made precise through the Dubins-Schwarz theorem later) the canonical non-trivial stochastic process with non-zero quadratic variation. The result we demonstrate shows that it is not possible to define the integral with respect to BM through pathwise finite variation integrals.

Theorem: M\in\mathcal{M}_{c,loc},M_0=0 a.s. is of finite variation. Then M is indistinguishable from 0.

We will show this for M a bounded martingale with bounded variation. Why does this suffice? In general, set S_n:=\inf\{t,V_t\leq n\}, noting that V is continuous adapted non-decreasing. If M^{S_n}\equiv 0\,\forall n, then we are done, as the S_ns are increasing. But this is a bounded martingale with bounded variation.

To prove this, we make use of the orthogonality relation which is a key trick for this sort of result: If M is a martingale, with M_s,M_t\in L^2, for s<t, then just by multiplying out:

\mathbb{E}[(M_t-M_s)^2|\mathcal{F}_s]=\mathbb{E}[M_t^2-M_s^2|\mathcal{F}_s] a.s.

Now, for this particular result, we decompose

\mathbb{E}[M_t^2]=\mathbb{E}\left[\sum_{k=0}^{2^n-1}(M_{(k+1)2^{-n}t}^2-M_{k2^{-n}t}^2)\right]=\mathbb{E}[\sum (M_{(k+1)2^{-n}t}-M_{k2^{-n}t})^2]

and then we bound this last term as

\leq \mathbb{E}\left[\sup_k [M_+-M_-]\sum_k |M_+-M_-|\right]

Now, as n\uparrow\infty, we have \sum_k |M_+-M_-|\uparrow V_t\leq N by the boundedness assumption. Furthermore, M is almost surely continuous on [0,t] and so it is in fact uniformly continuous, which allows us to conclude that

\sup_k |M_+-M_-|\downarrow 0

By bounded convergence, this limit applies equally under the expectation, and so conclude that \mathbb{E}M_t^2=0 for each time t, and so for each time t the martingale is almost surely equal to 0. In the usual, can lift this to rational points by countability, then to all points by continuity.