# A Process With Hidden Drift

Consider a stochastic process X of the form

 $\displaystyle X_t=W_t+\int_0^t\xi_sds,$ (1)

for a standard Brownian motion W and predictable process ${\xi}$, defined with respect to a filtered probability space ${(\Omega,\mathcal F,\{\mathcal F_t\}_{t\in{\mathbb R}_+},{\mathbb P})}$. For this to make sense, we must assume that ${\int_0^t\lvert\xi_s\rvert ds}$ is almost surely finite at all times, and I will suppose that ${\mathcal F_\cdot}$ is the filtration generated by W.

The question is whether the drift ${\xi}$ can be backed out from knowledge of the process X alone. As I will show with an example, this is not possible. In fact, in our example, X will itself be a standard Brownian motion, even though the drift ${\xi}$ is non-trivial (that is, ${\int\xi dt}$ is not almost surely zero). In this case X has exactly the same distribution as W, so cannot be distinguished from the driftless case with ${\xi=0}$ by looking at the distribution of X alone.

On the face of it, this seems rather counter-intuitive. By standard semimartingale decomposition, it is known that we can always decompose

 $\displaystyle X=M+A$ (2)

for a unique continuous local martingale M starting from zero, and unique continuous FV process A. By uniqueness, ${M=W}$ and ${A=\int\xi dt}$. This allows us to back out the drift ${\xi}$ and, in particular, if the drift is non-trivial then X cannot be a martingale. However, in the semimartingale decomposition, it is required that M is a martingale with respect to the original filtration ${\mathcal F_\cdot}$. If we do not know the filtration ${\mathcal F_\cdot}$, then it might not be possible to construct decomposition (2) from knowledge of X alone. As mentioned above, we will give an example where X is a standard Brownian motion which, in particular, means that it is a martingale under its natural filtration. By the semimartingale decomposition result, it is not possible for X to be an ${\mathcal F_\cdot}$-martingale. A consequence of this is that the natural filtration of X must be strictly smaller than the natural filtration of W.

The inspiration for this post was a comment by Gabe posing the following question: If we take ${\mathbb F}$ to be the filtration generated by a standard Brownian motion W in ${(\Omega,\mathcal F,{\mathbb P})}$, and we define ${\tilde W_t=W_t+\int_0^t\Theta_udu}$, can we find an ${\mathbb F}$-adapted ${\Theta}$ such that the filtration generated by ${\tilde W}$ is smaller than ${\mathbb F}$? Our example gives an affirmative answer. Continue reading “A Process With Hidden Drift”

# Do Convex and Decreasing Functions Preserve the Semimartingale Property — A Possible Counterexample

Here, I attempt to construct a counterexample to the hypotheses of the earlier post, Do convex and decreasing functions preserve the semimartingale property? There, it was asked, for any semimartingale X and function ${f\colon{\mathbb R}_+\times{\mathbb R}\rightarrow{\mathbb R}}$ such that ${f(t,x)}$ is convex in x and right-continuous and decreasing in t, is ${f(t,X_t)}$ necessarily a semimartingale? It was explained how this is equivalent to the hypothesis: for any function ${f\colon[0,1]^2\rightarrow{\mathbb R}}$ such that ${f(t,x)}$ is convex and Lipschitz continuous in x and decreasing in t, does it decompose as ${f=g-h}$ where ${g(t,x)}$ and ${h(t,x)}$ are convex in x and increasing in t. This is the form of the hypothesis which this post will be concerned with, so the example will only involve simple real analysis and no stochastic calculus. I will give some numerical calculations suggesting that the construction below is a counterexample, but do not have any proof of this. So, the hypothesis is still open.

Although the construction given here will be self-contained, it is worth noting that it is connected to the example of a martingale which moves along a deterministic path. If ${\{M_t\}_{t\in[0,1]}}$ is the martingale constructed there, then

$\displaystyle C(t,x)={\mathbb E}[(M_t-x)_+]$

defines a function from ${[0,1]\times[-1,1]}$ to ${{\mathbb R}}$ which is convex in x and increasing in t. The question is then whether C can be expressed as the difference of functions which are convex in x and decreasing in t. The example constructed in this post will be the same as C with the time direction reversed, and with a linear function of x added so that it is zero at ${x=\pm1}$. Continue reading “Do Convex and Decreasing Functions Preserve the Semimartingale Property — A Possible Counterexample”

# A Martingale Which Moves Along a Deterministic Path

In this post I will construct a continuous and non-constant martingale M which only varies on the path of a deterministic function ${f\colon{\mathbb R}_+\rightarrow{\mathbb R}}$. That is, ${M_t=f(t)}$ at all times outside of the set of nontrivial intervals on which M is constant. Expressed in terms of the stochastic integral, ${dM_t=0}$ on the set ${\{t\colon M_t\not=f(t)\}}$ and,

 $\displaystyle M_t = \int_0^t 1_{\{M_s=f(s)\}}\,dM_s.$ (1)

In the example given here, f will be right-continuous. Examples with continuous f do exist, although the constructions I know of are considerably more complicated. At first sight, these properties appear to contradict what we know about continuous martingales. They vary unpredictably, behaving completely unlike any deterministic function. It is certainly the case that we cannot have ${M_t=f(t)}$ across any interval on which M is not constant.

By a stochastic time-change, any Brownian motion B can be transformed to have the same distribution as M. This means that there exists an increasing and right-continuous process A adapted to the same filtration as B and such that ${B_t=M_{A_t}}$ where M is a martingale as above. From this, we can infer that

$\displaystyle B_t=f(A_t),$

expressing Brownian motion as a function of an increasing process. Continue reading “A Martingale Which Moves Along a Deterministic Path”

# Failure of the Martingale Property For Stochastic Integration

If X is a cadlag martingale and ${\xi}$ is a uniformly bounded predictable process, then is the integral

 $\displaystyle Y=\int\xi\,dX$ (1)

a martingale? If ${\xi}$ is elementary this is one of most basic properties of martingales. If X is a square integrable martingale, then so is Y. More generally, if X is an ${L^p}$-integrable martingale, any ${p > 1}$, then so is Y. Furthermore, integrability of the maximum ${\sup_{s\le t}\lvert X_s\rvert}$ is enough to guarantee that Y is a martingale. Also, it is a fundamental result of stochastic integration that Y is at least a local martingale and, for this to be true, it is only necessary for X to be a local martingale and ${\xi}$ to be locally bounded. In the general situation for cadlag martingales X and bounded predictable ${\xi}$, it need not be the case that Y is a martingale. In this post I will construct an example showing that Y can fail to be a martingale. Continue reading “Failure of the Martingale Property For Stochastic Integration”

# Martingales with Non-Integrable Maximum

It is a consequence of Doob’s maximal inequality that any ${L^p}$-integrable martingale has a maximum, up to a finite time, which is also ${L^p}$-integrable for any ${p > 1}$. Using ${X^*_t\equiv\sup_{s\le t}\lvert X_s\rvert}$ to denote the running absolute maximum of a cadlag martingale X, then ${X^*}$ is ${L^p}$-integrable whenever ${X}$ is. It is natural to ask whether this also holds for ${p=1}$. As martingales are integrable by definition, this is just asking whether cadlag martingales necessarily have an integrable maximum. Integrability of the maximum process does have some important consequences in the theory of martingales. By the Burkholder-Davis-Gundy inequality, it is equivalent to the square-root of the quadratic variation, ${[X]^{1/2}}$, being integrable. Stochastic integration over bounded integrands preserves the martingale property, so long as the martingale has integrable maximal process. The continuous and purely discontinuous parts of a martingale X are themselves local martingales, but are not guaranteed to be proper martingales unless X has integrable maximum process.

The aim of this post is to show, by means of some examples, that a cadlag martingale need not have an integrable maximum. Continue reading “Martingales with Non-Integrable Maximum”

# The Optimality of Doob’s Maximal Inequality

One of the most fundamental and useful results in the theory of martingales is Doob’s maximal inequality. Use ${X^*_t\equiv\sup_{s\le t}\lvert X_s\rvert}$ to denote the running (absolute) maximum of a process X. Then, Doob’s ${L^p}$ maximal inequality states that, for any cadlag martingale or nonnegative submartingale X and real ${p > 1}$,

 $\displaystyle \lVert X^*_t\rVert_p\le c_p \lVert X_t\rVert_p$ (1)

with ${c_p=p/(p-1)}$. Here, ${\lVert\cdot\rVert_p}$ denotes the standard Lp-norm, ${\lVert U\rVert_p\equiv{\mathbb E}[U^p]^{1/p}}$.

An obvious question to ask is whether it is possible to do any better. That is, can the constant ${c_p}$ in (1) be replaced by a smaller number. This is especially pertinent in the case of small p, since ${c_p}$ diverges to infinity as p approaches 1. The purpose of this post is to show, by means of an example, that the answer is no. The constant ${c_p}$ in Doob’s inequality is optimal. We will construct an example as follows.

Example 1 For any ${p > 1}$ and constant ${1 \le c < c_p}$ there exists a strictly positive cadlag ${L^p}$-integrable martingale ${\{X_t\}_{t\in[0,1]}}$ with ${X^*_1=cX_1}$.

For X as in the example, we have ${\lVert X^*_1\rVert_p=c\lVert X_1\rVert_p}$. So, supposing that (1) holds with any other constant ${\tilde c_p}$ in place of ${c_p}$, we must have ${\tilde c_p\ge c}$. By choosing ${c}$ as close to ${c_p}$ as we like, this means that ${\tilde c_p\ge c_p}$ and ${c_p}$ is indeed optimal in (1). Continue reading “The Optimality of Doob’s Maximal Inequality”

# Failure of the Martingale Property

In this post, I give an example of a class of processes which can be expressed as integrals with respect to Brownian motion, but are not themselves martingales. As stochastic integration preserves the local martingale property, such processes are guaranteed to be at least local martingales. However, this is not enough to conclude that they are proper martingales. Whereas constructing examples of local martingales which are not martingales is a relatively straightforward exercise, such examples are often slightly contrived and the martingale property fails for obvious reasons (e.g., double-loss betting strategies). The aim here is to show that the martingale property can fail for very simple stochastic differential equations which are likely to be met in practice, and it is not always obvious when this situation arises.

Consider the following stochastic differential equation

 $\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle dX = aX^c\,dB +b X dt,\smallskip\\ &\displaystyle X_0=x, \end{array}$ (1)

for a nonnegative process X. Here, B is a Brownian motion and a,b,c,x are positive constants. This a common SDE appearing, for example, in the constant elasticity of variance model for option pricing. Now consider the following question: what is the expected value of X at time t?

The obvious answer seems to be that ${{\mathbb E}[X_t]=xe^{bt}}$, based on the idea that X has growth rate b on average. A more detailed argument is to write out (1) in integral form

 $\displaystyle X_t=x+\int_0^t\,aX^c\,dB+ \int_0^t bX_s\,ds.$ (2)

The next step is to note that the first integral is with respect to Brownian motion, so has zero expectation. Therefore,

$\displaystyle {\mathbb E}[X_t]=x+\int_0^tb{\mathbb E}[X_s]\,ds.$

This can be differentiated to obtain the ordinary differential equation ${d{\mathbb E}[X_t]/dt=b{\mathbb E}[X_t]}$, which has the unique solution ${{\mathbb E}[X_t]={\mathbb E}[X_0]e^{bt}}$.

In fact this argument is false. For ${c\le1}$ there is no problem, and ${{\mathbb E}[X_t]=xe^{bt}}$ as expected. However, for all ${c>1}$ the conclusion is wrong, and the strict inequality ${{\mathbb E}[X_t] holds.

The point where the argument above falls apart is the statement that the first integral in (2) has zero expectation. This would indeed follow if it was known that it is a martingale, as is often assumed to be true for stochastic integrals with respect to Brownian motion. However, stochastic integration preserves the local martingale property and not, in general, the martingale property itself. If ${c>1}$ then we have exactly this situation, where only the local martingale property holds. The first integral in (2) is not a proper martingale, and has strictly negative expectation at all positive times. The reason that the martingale property fails here for ${c>1}$ is that the coefficient ${aX^c}$ of dB grows too fast in X.

In this post, I will mainly be concerned with the special case of (1) with a=1 and zero drift.

 $\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle dX=X^c\,dB,\smallskip\\ &\displaystyle X_0=x. \end{array}$ (3)

The general form (1) can be reduced to this special case, as I describe below. SDEs (1) and (3) do have unique solutions, as I will prove later. Then, as X is a nonnegative local martingale, if it ever hits zero then it must remain there (0 is an absorbing boundary).

The solution X to (3) has the following properties, which will be proven later in this post.

• If ${c\le1}$ then X is a martingale and, for ${c<1}$, it eventually hits zero with probability one.
• If ${c>1}$ then X is a strictly positive local martingale but not a martingale. In fact, the following inequality holds
 $\displaystyle {\mathbb E}[X_t\mid\mathcal{F}_s] (4)

(almost surely) for times ${s. Furthermore, for any positive constant ${p<2c-1}$, ${{\mathbb E}[X_t^p]}$ is bounded over ${t\ge0}$ and tends to zero as ${t\rightarrow\infty}$.

# Failure of Pathwise Integration for FV Processes

The motivation for developing a theory of stochastic integration is that many important processes — such as standard Brownian motion — have sample paths which are extraordinarily badly behaved. With probability one, the path of a Brownian motion is nowhere differentiable and has infinite variation over all nonempty time intervals. This rules out the application of the techniques of ordinary calculus. In particular, the Stieltjes integral can be applied with respect to integrators of finite variation, but fails to give a well-defined integral with respect to Brownian motion. The Ito stochastic integral was developed to overcome this difficulty, at the cost both of restricting the integrand to be an adapted process, and the loss of pathwise convergence in the dominated convergence theorem (convergence in probability holds intead).

However, as I demonstrate in this post, the stochastic integral represents a strict generalization of the pathwise Lebesgue-Stieltjes integral even for processes of finite variation. That is, if V has finite variation, then there can still be predictable integrands ${\xi}$ such that the integral ${\int\xi\,dV}$ is undefined as a Lebesgue-Stieltjes integral on the sample paths, but is well-defined in the Ito sense. Continue reading “Failure of Pathwise Integration for FV Processes”

# Stochastic Calculus Examples and Counterexamples

I have been posting my stochastic calculus notes on this blog for some time, and they have now reached a reasonable level of sophistication. The basics of stochastic integration with respect to local martingales and general semimartingales have been introduced from a rigorous mathematical standpoint, and important results such as Ito’s lemma, the Ito isometry, preservation of the local martingale property, and existence of solutions to stochastic differential equations have been covered.

I will now start to also post examples demonstrating results from stochastic calculus, as well as counterexamples showing how the methods can break down when the required conditions are not quite met. As well as knowing precise mathematical statements and understanding how to prove them, I generally feel that it can be just as important to understand the limits of the results and how they can break down. Knowing good counterexamples can help with this. In stochastic calculus, especially, many statements have quite subtle conditions which, if dropped, invalidate the whole result. In particular, measurability and integrability conditions are often required in subtle ways. Knowing some counterexamples can help to understand these issues. Continue reading “Stochastic Calculus Examples and Counterexamples”