Some years ago, I spent considerable effort trying to prove the hypothesis below. After failing at this, I spent time trying to find a counterexample, but also with no success. I did post this as a question on mathoverflow, but it has so far received no conclusive answers. So, as far as I am aware, the following statement remains unproven either way.
Hypothesis H1 Let be such that is convex in x and right-continuous and decreasing in t. Then, for any semimartingale X, is a semimartingale.
It is well known that convex functions of semimartingales are themselves semimartingales. See, for example, the Ito-Tanaka formula. More generally, if was increasing in t rather than decreasing, then it can be shown without much difficulty that is a semimartingale. Consider decomposing as
for some process V. By convexity, the right hand derivative of with respect to x always exists, and I am denoting this by . In the case where f is twice continuously differentiable then the process V is given by Ito’s formula which, in particular, shows that it is a finite variation process. If is convex in x and increasing in t, then the terms in Ito’s formula for V are all increasing and, so, it is an increasing process. By taking limits of smooth functions, it follows that V is increasing even when the differentiability constraints are dropped, so is a semimartingale. Now, returning to the case where is decreasing in t, Ito’s formula is only able to say that V is of finite variation, and is generally not monotonic. As limits of finite variation processes need not be of finite variation themselves, this does not say anything about the case when f is not assumed to be differentiable, and does not help us to determine whether or not is a semimartingale.
Hypothesis H1 can be weakened by restricting to continuous functions of continuous martingales.
Hypothesis H2 Let be such that is convex in x and continuous and decreasing in t. Then, for any continuous martingale X, is a semimartingale.
Hypothesis H3 Let be such that is convex in x and decreasing in t. Then, where and are convex in x and increasing in t.
Suppose that X is a Markov process and is such that is integrable. Then, we can define a function by
This is equivalent to the following two conditions,
- is a martingale over .
If we are able to determine which functions f satisfy the martingale condition 2 above, then we can reconstruct the Markov transition function and the distribution of X. For twice differentiable functions, the Kolmogorov backwards equation or Feynman-Kac formula can be used. However, in general, what regularity properties can be imposed on f? It is known that, for diffusions with smoothly defined coefficients, f will be smooth. This does not hold for more general Markov processes. In the case that X is a continuous and strong Markov martingale then, if g is convex, can also be chosen to be convex in x and decreasing in t (see Hobson, Volatility misspecification, option pricing and superreplication via coupling). This is a familiar property in finance, where call and put options have a price which is convex in the underlying asset price and increasing with time to maturity. Although convex and monotonic functions need not be differentiable, the derivatives and can be interpreted in a measure theoretic sense, and extensions of the backwards equation can be applied. I did post a paper on the arXiv, Fitting Martingales to Given Marginals, using these ideas to show that continuous and strong Markov martingales are uniquely determined by their 1-dimensional marginal distributions (for diffusions with smooth coefficients, this is a well known property of local volatility models used in finance). However, the proof is rather complicated due to the fact that, without knowing hypothesis H1 to be true, it is not even known a-priori whether or not is a semimartingale. Not having a positive answer to hypothesis H1 means that a lot of extra work is required, including the papers Nondifferentiable functions of one-dimensional semimartingales and A Generalized Backward Equation For One Dimensional Processes, which were written in order to develop an alternative technique to get around this obstacle.
Before moving on to the equivalence of the different forms of the hypothesis I will mention that, on balance, I expect that the hypothesis is probably false. As I will show below, it is equivalent to certain sets of integrals being uniformly bounded (H7), and I see no reason for such a bound to exist. However, actually finding examples making these integrals large is difficult, and I still do not have any proven counterexamples to the hypothesis.
I will state various alternative forms of hypothesis H1 in this post and give explanations showing why they are equivalent. For brevity, I do not intend to give complete and rigorous proofs here. However, by filling in the details, all of the arguments can be made fully rigorous. In total I will state nine different forms of the hypothesis, H1 through to H9, and sketch how each of the following implications holds.
Combining these implications shows that all of H1 to H9 are equivalent. In most cases, the reverse implications can also be shown without too much work, the exceptions being H3 ⇒ H1 ⇒ H2. I do not know a quick proof of the converse of these without going all the way round the circle of implications above.
As continuous martingales are special cases of semimartingales, the implication H1 ⇒ H2 is trivial. The implication H3 ⇒ H1 is also straightforward. As argued above, when f is increasing in time, equation (1) decomposes as a stochastic integral plus an increasing process. Hence, whenever decomposes as the difference of functions which are convex in x and increasing in t then (1) expresses as a stochastic integral plus a finite variation process and, if it is also right-continuous, it is a semimartingale.
In hypothesis H3, the terms g and h in the decomposition are given equal status. However, when trying to find such decompositions, I find that it is convenient to regard h as the primary function to be constructed. That is, the problem is to find a function which is convex in x such that is increasing in t. Then, the function g is defined by . Now, restricting to the unit interval , hypothesis H3 can be localized.
Hypothesis H4 Let be such that is convex and Lipschitz continuous in x and decreasing in t. Then, where and are convex in x and increasing in t.
The idea is that, if f is as in hypothesis H3, then we can apply the decompositions given by H4 on a set of overlapping rectangles covering the right half-plane, and glue these together to obtain the global decomposition. For example, suppose that we have already decomposed on a rectangle . Then, for any apply H4 to obtain the decomposition on . These can be glued together by choosing a smooth function on the reals such that is 0 for and 1 for . Setting
extends to the larger rectangle . It need not be the case that and are convex in x, but they will have second derivative bounded below and can be made convex by adding a multiple of . Continuing in this way, we can extend h to ever larger rectangles to obtain the global decomposition required by H3.
It will be useful to further restrict attention to functions which are equal to zero on the upper and lower edges of the unit rectangle. With this is mind, I make the following definitions.
- is the space of functions such that is convex in x and satisfies .
- is the space of such that is increasing in t.
- is the space of such that is decreasing in t.
In the remainder of the post, for functions of I will frequently disregard the arguments to keep the expressions reasonably short. I will also use the notation and to denote the derivatives of f with respect to its first and second argument respectively. If f is not differentiable, then these can be understood in the sense of distributions. The notation will be used for the supremum norm. In particular, is finite if and only if f is Lipschitz continuous with respect to x, with Lipschitz constant .
Hypothesis H5 Every such that decomposes as for .
For any which is convex and Lipschitz continuous in x and decreasing in t, hypothesis H5 can be used to obtain the decomposition stated in H4. By adding a constant to f, without loss of generality we can suppose that for any . We can then extend f to a larger rectangle by setting for x equal to and , and linearly interpolating in x across the intervals and . So long as , this retains convexity in x. Then, hypothesis H5 can be used to decompose and, by restricting to , we obtain the decomposition required by hypothesis H4.
Now, if we have a collection of decompositions (), as in H5, then we can set
As taking the supremum preserves both convexity in x and monotonicity in t, this gives a decomposition as in H5 with and for all . Applying this to the collection of all such decompositions gives a maximal decomposition.
Lemma 1 Suppose that is such that the decomposition in (H5) exists. Then, there exists a unique maximal choice for g, h. That is, if is any other such decomposition, then and .
Next, hypothesis H5 is equivalent to the following, apparently weaker, statement.
Hypothesis H6 There exists a constant K such that every with decomposes as for and
Clearly, hypothesis H5 follows immediately from H6 simply by dropping (2) from the conclusion. However, H6 does have one advantage — it is only necessary to prove it for a dense subset of . A function is smooth if it is continuous and all of its partial derivates exist, to all orders, on the interior of and extend continuously to . For any , it is not difficult to construct a sequence of smooth with and converging pointwise to f. So, to prove H6, it is enough to prove it just for smooth functions.
I’ll now describe how to construct the decomposition in H5. This will always converge to the maximal decomposition whenever it exists, or diverge to if there is no such decomposition. Start by choosing a partition of the unit interval
We now construct such that is convex in x and is increasing in t. The second condition implies the inequality
Also, from the definition of , we are looking for functions satisfying . Use to denote the convex hull of a function . This is the maximum convex function bounded above by u. Then, is constructed starting at and, then, inductively for .
We can extend h in between the times of the partition however we like, so long as monotonicity in t is preserved. Choosing partitions with mesh going to zero, we do indeed get convergence to the maximal decomposition whenever it exists.
Lemma 2 Suppose that and, for each , let be a partition of the unit interval. We suppose that the partitions have mesh going to 0, and eventually include all times at which f is discontinuous. For each n, let be the function constructed as above using the n’th partition. Then, exactly one of the following holds.
- f decomposes as in (H5) and pointwise on , where is the maximal decomposition.
- f does not have a decomposition as in (H5), and for all .
Proof: For any given partition, h defined as in (3) is the maximum non-positive function such that it is convex in x and is increasing in t. This means that on whenever the partition is a refinement of . It follows that
on . Using the fact that the sequence of partitions have mesh going to zero and eventually includes each discontinuity time of f,
So, the constructions along the partitions do converge to a limit, although it could be infinite.
Now, suppose that a decomposition as in H5 does exist. Then, the constructions along partitions are bounded below, , and must converge to a finite limit. As limits of convex and monotonic functions are, respectively, convex and monotonic, the limit is convex in x and is increasing in t. As , the limit is the maximal decomposition.
Conversely, suppose that there is no such decomposition as in H5. Then, the sequence cannot converge to a finite limit everywhere. Hence for some t and y. By monotonicity in t, tends to minus infinity. By convexity, for all ,
Similarly, the same limit holds for . ⬜
The construction described above does indeed converge to a finite limit for smooth f.
Lemma 3 Suppose that is smooth. Then, the maximal decomposition exists, the derivative is bounded, and satisfies
Proof: For a given partition of the unit interval, construct h as in (2), and interpolate linearly in between the times of the partition. Setting
then is the convex hull of . The bound
is immediate, where denotes . Hence, is bounded by . So, is bounded by . Integrating over t also gives .
We can also bound above by . Taking its convex hull preserves the bound, so is bounded by . Similarly, is bounded by , so .
Next, is bounded below by and, from this, it can be shown that is bounded by . So, is bounded by .
Putting all of these together gives the following set of inequalities.
Now, if is a continuous function with convex hull v, then and, on each interval for which , we have . It follows that . Applying this to (2) gives
We can now take limits as the mesh of the partition goes to zero. The inequality shows that the construction cannot diverge and, by Lemma 2, the decomposition of hypothesis H5 exists and we have convergence to the maximal decomposition. The inequalities (5) follow for the maximal decomposition and, taking limits of (6) gives
Integrating over and applying integration by parts,
Equality (4) follows from this. ⬜
To construct the decomposition in H5, we can apply the decomposition for smooth f and then take limits to obtain the decomposition for arbitrary . The problem is that, when taking the limit, the terms g and h could diverge to minus infinity. In some cases, (4) can be used to bound h and avoid this potential divergence. However, for this to work, the following alternative version of the hypothesis is required.
Hypothesis H7 There exists a constant K such that all smooth and satisfy
If the left hand side can be replaced by a multiple of then, cancelling , this would give inequality (2) as required by hypothesis H6. To do this, we can consider applying the decomposition over a slightly larger region. For any , define on by setting
This is convex in x so long as . Applying the decomposition given by Lemma 2 to gives a such that is convex in x, is increasing in t and,
The additional factor of on the right hand side is because we are applying the decomposition over an interval of width rather than the unit interval. Using the fact that is linear over the intervals and ,
Combining with the previous inequality,
As on the edges of the rectangle , this can be integrated to give
Restricting to the unit square , this gives a convex non-positive h which is convex in x and such that is increasing in t. Setting and ,
Now, I move on to yet another form of the hypothesis. For smooth and , consider defining
for all smooth of compact support. If hypothesis H7 is true, then the inequality
would hold. Using integration by parts, can be rearranged as
This form has the advantage that it makes sense for all and without imposing any smoothness constraints. By convexity, and are well-defined bounded functions and, by monotonicity, and are well-defined measures (i.e., using Lebesgue-Stieltjes integration). I used this idea in the papers A Generalized Backward Equation For One Dimensional Processes and Fitting Martingales to Given Marginals to derive a martingale condition for where X is a continuous strong Markov martingale, and f is convex in x and decreasing in t, without imposing any differentiability constraints. For our purposes in this post, we just use it for the following form of the hypothesis.
Hypothesis H8 There exists a constant K such that
for all , and smooth of compact support.
As the terms and have opposite signs, putting a bound on their sum does not imply any bound on the individual terms. Instead, consider choosing a partition , let be the mid-points, and define and by
Now, we have over the intervals and over the intervals . So,
Applying hypothesis H8 to the functions and gives
Taking the limit as the mesh of the partition goes to zero gives the inequality
and hypothesis H7 follows immediately from this.
I now move on to the final form of the hypothesis, which re-introduces the stochastic calculus element.
Hypothesis H9 There exists a constant such that, for every and continuous martingale with , has mean variation
In particular, this implies that Y is a quasimartingale. Suppose that hypothesis holds. Given any smooth with on the interior of and , define the function by
This is convex in x and increasing in t. Now, consider the stochastic differential equation
for a Brownian motion W and
We only consider solving (8) up until the first time at which X hits 0 or 1, after which X is constant. If the initial distribution of X is chosen so that
for , then this holds for all . This is a well known result, used in financial option pricing by the local volatility model, where represents the price of a call option of strike price x and maturity t. For any bounded measurable , the following identities hold,
Now, for smooth , (1) expresses as a martingale plus a finite variation term
Now, using the identities above, for bounded measurable ,
The left hand side is bounded by the mean variation of Y whenever is bounded by 1 so, if hypothesis H9 holds, it is bounded by . Scaling g and gives,
Approximating arbitrary f and g by smooth functions implies hypothesis H8.
It only remains to show how hypothesis H2 implies H9, which I will do now. This is the most difficult of the implications shown in this post, and I will instead show the contrapositive. That is, supposing that H9 is false, we show that H2 is also false. The idea is to construct a continuous martingale X and such that the process V in decomposition (1) has infinite variation. This will imply that is not a semimartingale.
Start by choosing a sequence of continuous martingales with and smooth such that the mean variation of is greater than . Passing to a larger probability space if necessary, we suppose that we have a doubly indexed sequence of independent continuous martingales, where has the same distribution as . We then decompose
The variation, , of over then has expectation equal to the mean variation of . By the weak law of large numbers,
with probability at least 1/2, for large enough . We also suppose that . Setting for and otherwise. Then,
and is an integer. Rearranging the non-zero terms gives a singly indexed sequence with
where is an integer, are smooth with , is an independent sequence of continuous martingales with decomposition
and has variation over .
Let be the rectangle . By extrapolating and applying a change of variables, we can suppose that are continuous martingales with and , and is convex in x and decreasing in t, such that on the boundary of and at .
Now define the sequence of times . Define the martingale M by
for . This is a random walk, interpolated by the continuous martingales . The times are increasing to 1 and,
This is -bounded so, by martingale convergence, the limit exists in and with probability 1, giving a continuous -bounded martingale .
The fact that is an integer implies that the support of is contained in the set or for each .
Define by for and . We interpolate between these times by setting
for and , some . This is convex in x and decreasing in t. It can be seen that
Then, if we decompose
the process V is continuous with variation
This is finite, but tends to infinity almost surely as n goes to infinity.
We can now conclude that is not a semimartingale as, otherwise, it would decompose uniquely as
for a continuous local martingale N and continuous FV process A with . Comparing with (9) over each interval for gives and, hence, A has almost surely infinite variation on , contradicting the fact that it is an FV process. This then contradicts hypothesis H2.