Some time ago I spent a lot of effort trying to show that the semimartingale property is preserved by certain functions. Specifically, that a convex function of a semimartingale and decreasing function of time is itself a semimartingale. This was needed for a result which I was trying to prove (more details below) and eventually managed to work around this issue, but it was not easy. For twice continuously differentiable functions this is an immediate consequence of Ito's lemma, but this cannot be applied in the general case. After failing at this task, I also spent a considerable amount of time trying to construct a counterexample, also with no success. So, my question is as follows.

1) Let $f\colon\mathbb{R}^+\times\mathbb{R}\to\mathbb{R}$ be such that $f(t,x)$ is convex in $x$ and continuous and decreasing in $t$. Then, for any semimartingale $X$, is $f(t,X_t)$ necessarily a semimartingale?

Actually, it can be assumed here that $X$ is both continuous and a martingale which, with some work, would imply the general case. As it turns out, this can be phrased purely as a real-analysis question.

2) Let $f\colon\mathbb{R}^+\times\mathbb{R}\to\mathbb{R}$ be such that $f(t,x)$ is convex in $x$ and decreasing in $t$. Can we write $f=g-h$ where $g(t,x)$ and $h(t,x)$ are both convex in $x$ and increasing in $t$?

Stated like this, maybe someone with a good knowledge of convex functions would be able to answer the question one way or the other.

For $f(t,x)$ convex in $x$ and increasing in $t$ then approximation by smooth functions and applying Ito's lemma allows us to express $f(t,X_t)$ as the sum of a stochastic integral and an increasing process $$ f(t,X_t)=\int_0^t\frac{\partial f}{\partial x}(s,X_s)\,dX_s+V_t,\qquad{\rm(*)} $$ so it is a semimartingale. If, instead, $f$ is decreasing in $t$, then an affirmative answer to question 2 will reduce it to the easier case where it is increasing in $t$, also giving a positive answer to the first question. Explaining why question 1 implies 2 is a bit trickier. If 2 was false, then it would be possible to construct martingales $X$ such that the decomposition (*) holds where the variation of $V$ explodes at some positive time.

This problem arose while I was trying to prove the following: is a continuous martingale uniquely determined by its one dimensional marginals? For arbitrary continuous martingales this is false, but is known to be true for diffusions $dX_t=\sigma(t,X_t)\,dW$ for Brownian motion $W$ and smooth parameter $\sigma$. The idea is to back out $\sigma$ from the Kolmogorov forward equation. This is well-known in finance as the local volatility model. However, I was trying to show rather more than this. All continuous and strong Markov martingales are uniquely determined by the one dimensional marginals. I was able to prove this, and the relation between the marginals and joint distributions of the martingale has many nice properties (I wrote a paper on this, submitted to the arXiv, but not published as I am still working on changes asked for by the referees). The method was to reformulate the Kolmogorov backward equation in terms of the marginals. This does use Ito's lemma, requiring twice differentiability, but can be circumvented with a bit of integration by parts as long as $f(t,X_t)$ is a semimartingale for the kinds of functions mentioned above. The question above arose from trying, and failing, to prove this. Without an answer to this question, the problem becomes much much harder, as many of the techniques from stochastic calculus can no longer be applied (and, approximating by semimartingales didn't seem to help either). The work around was very involved, part of which I published as a standalone paper here and the rest forms most of a paper I submitted to the arXiv here. Adding together those papers, it comes to maybe 50 pages of maths and a lot of effort to work around the question above.

Update: I posted more details on the various equivalent forms of this question here. I also posted a possible counterexample, without proof but with some numerical evidence suggesting that it may indeed be a counterexample.

  • $\begingroup$ @ George Lowther : Hi have you considered looking at some Stone-Weirtstrass type theorem exploiting the algebra structure of semimartingale after considering the "separated" problem where $F(t,X_t)=ft).g(X_t)$ ? I think it might worth a try. Best regards. $\endgroup$
    – The Bridge
    Oct 18, 2013 at 13:18

3 Answers 3


I still have no idea what the answer to this question is. However, it is possible to attack the problem in several different ways, and there are various different (but logically equivalent) ways of stating it. I'll post some of these as an answer now, as it seems rather long to fit in the original statement of the question. Maybe one of the reformulations below will help lead to a resolution of the problem.

This answer is already very long, and I've been trying to shorten it as much as I can. I can't see any way of giving proper proofs of all the statements below without making it a lot longer. So, I'll only give very few details of he proofs here. I will, however, list each of the equivalent formulations H1-H6 below in the logical order, so that each statement can be proven to be equivalent to the preceding one without too much work. Let's start from statement 2 of the original question, which I will refer to as Hypothesis (H1) for the purposes of this answer.

Hypothesis (H1): Let $f\colon\mathbb{R}^+\times\mathbb{R}\to\mathbb{R}$ be such that $f(t,x)$ is convex in $x$ and decreasing in $t$. Then, $f=g-h$ where $g(t,x)$ and $h(t,x)$ are both convex in $x$ and increasing in $t$.

The decomposition in (H1) exists if and only if it exists locally. Letting $I=[0,1]$ denote the unit interval, we get the following equivalent statement.

Hypothesis (H2): Let $f\colon I^2\to\mathbb{R}$ be such that $f(t,x)$ is convex and Lipschitz continuous in $x$ and decreasing in $t$. Then, $f=g-h$ where $g(t,x)$ and $h(t,x)$ are convex in $x$ and increasing in $t$.

The truth of this statement remains unchanged if it is restricted to functions ƒ which are zero on the three edges $I\times\{0\}$, $I\times\{1\}$, $\{0\}\times I$ of the unit square. I'll use $D$ to denote the set of such functions satisfying the conditions of (H2). Then, whenever the decomposition in (H2) exists, it is always possible to choose $g$, $h$ to be zero on $I\times\{0\}$, $I\times\{1\}$ and $\{1\}\times I$. From now on, whenever the decomposition $f=g-h$ in (H2) is referred to, it will be assumed that $g$, $h$ are chosen to satisfy these conditions.

We can strengthen (H2) by also placing a uniform bound on the terms $g$, $h$ in the decomposition. Here, $\lVert g\rVert$ denotes the supremum norm and $f_x$ denotes the partial derivative.

Hypothesis (H3): There is a constant $K\gt0$ such that, for all $f\in D$, the decomposition $f=g-h$ as in H2 exists and can be chosen such that $\lVert g\rVert,\lVert h\rVert\le K\lVert f_x\rVert$.

Statement (H3) is particularly convenient because of the following: to prove that (H3) holds, it is enough to look at a dense subset of functions in $D$. Taking limits of the decompositions would then extend to the result to all of $D$. So, it is enough to concentrate on, say, smooth functions or piecewise linear functions.

Next, it is useful to choose the decomposition in (H2) to minimize $\lVert g\rVert$ and $\lVert h\rVert$.

Lemma 1: Suppose that $f\in D$ and that the decomposition $f=g-h$ as in (H2) exists. Then, there is a unique maximal choice for $g$, $h$. That is, if $f=g_1-h_1$ is any other such decomposition then $g\ge g_1$ and $h\ge h_1$.

I'll refer to the decomposition in Lemma 1 as the optimal decomposition. As it not clear that any such decomposition should exist, I'll briefly sketch the argument now. The idea is to discretize time, using a partition of the unit interval $0=t_0\lt t_1\lt\cdots\lt t_r=1$. Denote the convex hull of a function $u\colon I\to\mathbb{R}$ by $v=H(u)$, which is the maximum convex function $v\colon I\to\mathbb{R}$ bounding $v$ from below, $$ \begin{align} v(x) &= \sup\left\{w(x)\colon w\le u{\rm\ is\ convex}\right\}\\ &=\inf\left\{\left((b-x)u(a)+(x-a)u(b)\right)/(b-a)\colon a\le x\le b\right\}. \end{align} $$ The optimal decomposition in discrete time can be constructed as functions $h_k\colon I\to\mathbb{R}$, starting at the final time $k=r$ and working backwards to $k=0$, $$ h_r(x) = 0,\ h_{k-1}={\rm H}\left(h_k+f(t_k,\cdot)-f(t_{k-1},\cdot)\right). $$ Interpolate this to be piecewise constant in time, defining $h(t,x)=h_k(x)$ for $t_{k-1}\lt t\le t_k$. Then, $h(t,x)$ and $g\equiv f+h$ are convex in $x$ and increasing in time $t$, restricting to times in the partition.

Lemma 2: Suppose that $f\in D$ and let $0=t_{n,0}\lt t_{n,1}\lt\cdots\lt t_{n,r_n}=1$ be a sequence of partitions of the unit interval. For each $n$ let $h^n(t,x)$ be the function corresponding to the partition and piecewise constant in $t$, as constructed above. We also suppose that the partitions have mesh going to zero and eventually include all times at which $f$ is discontinuous. Then one of the following holds.

  • $f$ decomposes as in (H2) and $h^n(t,x)\to h(t,x)$ pointwise on $I^2$, where $f=g-h$ is the optimal decomposition.
  • $f$ does not have a decomposition as in (H2) and $h^n(0,x)\to-\infty$ for all $0\lt x\lt1$.

The idea is that, if $h^n$ has any limit point $h$, then $h(t,x)$ and $g=f+h$ will be convex in $x$ and increasing in $t$, giving the decomposition required by (H2). Furthermore, by construction, if $f=g^\prime-h^\prime$ is any other such decomposition, then $h^n\ge h^\prime$ at times in the partition, so $h\ge h^\prime$. This shows that the decomposition is optimal and, as the optimal decomposition is unique, all limit limit points of $h^n$ are the same, so $h^n\to h$. The only alternative is that $h^n$ has no limit points, in which case the second statement of the Lemma holds. Using this construction of the optimal decomposition, (H3) can be shown to be equivalent to the following.

Hypothesis (H4): There is a constant $K\gt0$ such that, for all smooth functions $f,g\colon I^2\to\mathbb{R}$ with $\lVert f\rVert$, $\lVert g\rVert$, $\lVert f_x\rVert$, $\lVert g_x\rVert$ bounded by 1 and $f(t,x)$, $g(t,x)$ convex in $x$ and respectively decreasing and increasing in $t$, then, $$\int_0^1\int_0^1 f_{xx}g_t\,dxdt \le K.$$

As this statement is quite different from the preceding ones, I'd better give some explanation now. The idea is to use integration by parts, $$ \begin{align} \int_0^1\int_0^1(f_{xx}g_t+g_{xx}f_t)\,dxdt &= \left[\int_0^1(f_xg_t+g_xf_t)\,dt\right]_{x=0}^1-\left[\int_0^1f_xg_x\,dx\right]_{t=0}^1\\ &\le 6(\Vert f_x\Vert \Vert g\Vert + \Vert g_x\Vert \Vert f\Vert). \end{align} $$ If $f$ and $g$ are increasing in time then the terms on the left hand side are both positive, so we get bounds for the integrals of $f_{xx}g_t$ and $g_{xx}f_t$ individually. Hypothesis (H3) extends this to the case where $f$ is decreasing in time, implying (H4).

Conversely, suppose that (H4) holds. Letting $f=g-h$ be the decomposition computed along a partition as described above, we can use the fact that the convex hull $v=H(u)$ of a function $u$ satisfies $v_{xx}(u-v)=0$ to get the equality $(h_{k-1})_{xx}(h_k-h_{k-1}+f(t_k,\cdot)-f(t_{k-1},\cdot))=0$. This leads to the following inequalities,

$$ \begin{align} \frac12\Vert h\Vert^2&\le\frac12\int_0^1 h_x(0,x)^2\,dx\le\sum_{k=1}^r \int_0^1(h_{k-1})_{xx}(h_{k-1}-h_k)\,dx\\ &=-\sum_{k=1}^r\int_0^1 (h_{k-1})_{xx}(f_k-f_{k-1})\,dx. \end{align} $$ Hypothesis (H4) can be used to bound the final term, showing that $h$ cannot diverge as the mesh of the partition goes to zero, so we get convergence to an optimal decomposition satisfying a bound as in (H3).

The hypothesis can now be formulated as a statement about martingales.

Hypothesis (H5): There is a constant $K\gt0$ such that, for all $f\in D$ and martingales $0\le X_t\le1$, $f(t,X_t)$ decomposes as $M+V$ where $M$ is a martingale and $V$ has variation $$ \mathbb{E}\left[\int_0^1\,\vert dV\vert\right]\le K\Vert f_x\Vert. $$

The idea is that $g(t, x)\equiv\mathbb{E}[(X_t-x)_+]$ is convex in $x$ and increasing in $t$. In the case where $f$, $g$ are smooth and $X$ is a continuous martingale, Ito's formula can be used to split $f(t,X_t)$ into a martingale term plus the sum of the increasing process $\frac12\int f_{xx}(t,X_t)\,d[X]_t$ and the decreasing process $\int f_t(t,X_t)\,dt$, which have expectations $\iint f_{xx}g_t\,dxdt$ and $\iint f_tg_{xx}\,dxdt$ respectively.

Finally, by adding a randomly occurring jump term to any semimartingale, and with a bit more work, it is possible to reduce this to the martingale case. This gives the statement asked in the original question.

Hypothesis (H6): Let $\colon\mathbb{R}^2\to\mathbb{R}$ be such that $f(t,x)$ is convex in $x$ and continuous and decreasing in $t$. Then, for any semimartingale $X$, $f(t,X_t)$ is a semimartingale.


What follows is too long to fit in a comment.

Some thoughts on the second problem. Let us consider the problem for $t$ taking values in the compact interval $[0,T]$. The general case perhaps can be approximated by this case. If the $t$ variable were discrete and finite the problem would be: we have a sequence of functions $f_1 \ge f_2 \ge f_3 \cdots \ge f_n$ (let us make the simplifying assumption that $f_i$ are all positive; otherwise, replace $f_i$ with $f_i +C$ where $C$ is $ C = -\min_{i,x} f_i(x)$), can one find convex and increasing $\{h_i\}$ and $\{g_i\}$ such that $f_i = h_i - g_i$? The answer to this question is obviously yes. For example, $h_i = \sum_{j\le i} f_i$ and $g_i = \sum_{j < i} f_i$ is one possible solution. The problem with this solution when $t$ is continuous is that the $h_i$ and $g_i$ would explode as the discretization of $t$ is refined. Thus, one needs to choose $h$ and $g$ in a way that they increase slowly. How slowly can this increase be? We can start with $h_1 = f_1$ and $g_1 = 0$. $h_2$ and $g_2$ will be of the following form: $$ h_2 = h_1 + S_2 = f_1 + S_2, $$ and $$ g_2 = g_1 + R_2 = R_2 $$ such that: 1) $S_2, R_2 \ge 0$, 2) $h_2 = f_1 + S_2$, $g_2 = R_2$ are convex and 3) $h_2 - g_2 = f_1 + S_2 - R_2 = f_2$. The last of these is equivalent to $R_2 = f_1 - f_2 + S_2$. Thus, what we are looking for is a function $S_2$ satisfying the above conditions. There will be many such $S_2$, the goal is to choose $S_2$ in a minimal way so that $h_i$ and $g_i$ grow slowly.The best would be: $S_2 = 0$, which is indeed a solution when $f_1 - f_i$ is convex. If $f_0 - f_t$ is convex for all $t \in [0,T]$ then this solution directly generalizes to the original continuous time problem (i.e., if $f_0-f_t$ is convex then $h(t,x) = f(0,x)$ and $g(t,x) = f(0,x) - f(t,x)$ is a solution).

Now, let us consider the case when $f(t,x)$ is such that $\frac{\partial^2}{\partial x^2}f(t,x)$ is continuous in $(t,x)$ and $x$ also takes values in a compact set $K$. The following type of argument quickly comes to mind. Define $$\tau \doteq \{t: \exists g, h:[0,t]\times K\rightarrow {\mathbb R}, f = h-g, h,g \text{ convex in } x \text{ and increasing in } t \}.$$ $0$ is clearly a member of this set. One can perhaps argue that $\sup \tau$ must be $T$ as follows. If $t_0 = \sup \tau < T$ then one can slightly modify $h(t_0,x)$ and $g(t_0,x)$ to obtain functions $h(t_0+\delta,x)$ and $g(t_0+\delta,x)$ whose difference will be $f(t_0+\delta,x)$ [the slight modification is made possible by the fact that the second derivative of $f$ with respect to $x$ barely changes between $t_0$ and $t_0 + \delta$].

  • $\begingroup$ Thanks for looking at this. I agree that discretizing it is a good first step (it also helps to restrict to a compact interval for x). However, it works out a bit better if you solve for g,h by starting at the last time and working backwards. Doing this, you can show that it converges to the continuous time decomposition if and only if there is a continuous time decomposition. I'm going to post an answer myself, when I have time, describing a few different ways to reformulate this problem. $\endgroup$ Oct 30, 2010 at 21:38
  • $\begingroup$ Also, if f is either twice differentiable in x or once in t, then it will have the required decomposition. The problem appears with nondifferentiable functions. You can always approximate by smooth functions but, then, it is possible that the decompositions of these smooth approximations diverge when you take the limit. $\endgroup$ Oct 30, 2010 at 21:42
  • $\begingroup$ You are welcome; your questions above and your related results are very interesting, thanks for sharing. What makes the problem difficult seems to be that there are many things that influence how $g$ and $h$ are to evolve in time and there seems to be many choices. The optimization problem in the discrete formulation (the choice of $S_2$) referred to in my answer seems to be nontrivial (I think it can be formulated as a control problem). $\endgroup$
    – has2
    Oct 31, 2010 at 7:45
  • $\begingroup$ Another thing that seem relevant is the following: the region where $x \rightarrow f(x,0) - f(x,t)$ is not convex must be small, because otherwise $f(x,0)$ could not dominate $f(x,t)$. $\endgroup$
    – has2
    Oct 31, 2010 at 7:50


Maybe I haven't fully understood your question but I think there is a counter-example (in a simpler context) given in Protter's book on Stochastic Integration in the form of a theorem. It's theorem 71, chapter IV, section 7 (Local Times).

It's claimed there that for any $\alpha \in(0,1/2)$ and any continuous local martingale $X_t$, then $Y_t=|X_t|^\alpha$ is not a semi-martingale unless it is a null process.

The proof is based on local time and Tanaka-Meyer's theorem.

Hope it helps, (but if not I would appreciate that you explain where I went wrong).

Best regards.

  • $\begingroup$ The function $f(x)=\vert x\vert^\alpha$ is not convex for $\alpha < 1$. Convex functions do preserve the semimartingale property (see the Ito-Tanaka formula for example). It is only when you have convex in space and decreasing in time that it gets difficult. $\endgroup$ Oct 24, 2010 at 15:56
  • $\begingroup$ Actually, I think what I just called the Ito-Tanaka formula is the same thing as Tanaka-Meyer's theorem that you just mentioned. It is also called the Ito-Tanaka-Meyer formula. $\endgroup$ Oct 24, 2010 at 15:59
  • $\begingroup$ Sorry to insit but $f$ is concave though (right ?),so taking $-f$ should do the trick. I must miss something about the convex function definition here. Regards $\endgroup$
    – The Bridge
    Oct 24, 2010 at 16:19
  • $\begingroup$ No. f is concave individually on the intervals $[0,\infty)$ and $(-\infty,0]$. It is neither convex nor concave on any open interval about zero. And that is the point of the result you refer to. In any case, a local martingale restricted to being either nonnegative everywhere or nonpositive everywhere must remain at 0 once it hits zero. It can only escape 0 if it can only take both positive and negative values, so the behaviour of f in a neighbourhood of 0 is what is important here. $\endgroup$ Oct 24, 2010 at 16:29
  • $\begingroup$ One further point: I don't have Protter's book to hand right now but, if you look at the proof of the result you just quoted, I expect that it is writing the drift term of f(X) as $\int L^x_t f^{\prime\prime}(x)\,dx$ where $L^x_t$ is the local time at x. This explodes if $L^0>0$ in your example but, if f was either convex or concave, $f^{\prime\prime}$ would be locally integrable and you get that f(X) is indeed a semimartingale. $\endgroup$ Oct 24, 2010 at 16:39

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