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.

Brownian Bridge Construction

The example of a process with hidden drift will be based around the decomposition of a Brownian motion into Brownian bridges or, conversely, the composition of a standard Brownian motion from Brownian bridges. So, I first quickly go over this standard material.

Let us start with a Brownian motion {X} defined over an interval {[t_0,t_1]}. We consider the linear interpolant across the interval,

\displaystyle  \hat X_t=\frac{t_1-t}{t_1-t_0}X_{t_0}+\frac{t-t_0}{t_1-t_0}X_{t_1}

and the deviation of the process from this interpolation,

\displaystyle  B_t=X_t-\hat X_t.

It can be seen that B is independent of X at the interval endpoints. It is straightforward to compute the covariances of {\hat X} and {B_t}, which are zero. As the increments of X are jointly normal, independence follows. In fact, B is a Brownian bridge. The process

\displaystyle  W_t=B_t+\int_{t_0}^t\frac{B_s}{t_1-s}ds,

is a Brownian motion over {[t_0,t_1]} which, again, can be shown by computing covariances. Differentiating, the Brownian bridge B is given from the Brownian motion W via the SDE

\displaystyle  dB_t=dW_t-\frac{B_t}{t_1-t}dt.

This can be solved to express B in terms of W,

\displaystyle  B_t=(t_1-t)\int_{t_0}^t\frac{dW_s}{t_1-s} (3)

Now, suppose we are given the following, for times {t_0 < t_1}, defined on the same probability space.

  • A random variable {X_{t_0}}.
  • A standard normal random variable {U}.
  • A Brownian motion {W} over the interval {[t_0,t_1]}.

We furthermore suppose that {X_{t_0}}, U and W are independent. The idea is to construct a new Brownian motion X on the interval {[t_0,t_1]}. Taking {X_{t_0}} as specified, we need {X_{t_1}-X_{t_0}} to be normal with variance {t_1-t_0}. This is done by setting

\displaystyle  X_{t_1}=X_{t_0}+(t_1-t_0)^{1/2}U.

We interpolate X across the interval by inverting the Brownian bridge decomposition above. Constructing a Brownian bridge by (3) and adding this to the linear interpolation of X gives

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle X_t &\displaystyle = \frac{t_1-t}{t_1-t_0}X_{t_0}+\frac{t-t_0}{t_1-t_0}X_{t_1}+(t_1-t)\int_{t_0}^t\frac{dW_s}{t_1-s}\smallskip\\ &\displaystyle =X_{t_0}+\frac{t-t_0}{\sqrt{t_1-t_0}}U+(t_1-t)\int_{t_0}^t\frac{dW_s}{t_1-s}. \end{array} (4)

for all times {t_0\le t < t_1}. This is our desired Brownian motion. Expressing things slightly differently, we have

\displaystyle  dX_t=dW_t+\xi_tdt (5)

where {\xi} is the process

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \xi_t &\displaystyle =\frac{X_{t_1}-X_t}{t_1-t}\smallskip\\ &\displaystyle =\frac{U}{\sqrt{t_1-t_0}}-\int_{t_0}^t\frac{dW_s}{t_1-s}. \end{array} (6)

I will also note that, as standard Brownian motion has independent normal increments, for any fixed time {t\in[t_0,t_1]} the random variable

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle U_t &\displaystyle = (t_1-t)^{-1/2}(X_{t_1}-X_t)\smallskip\\ &\displaystyle =\sqrt{\frac{t_1-t}{t_1-t_0}}U-\sqrt{t_1-t}\int_{t_0}^t\frac{dW_s}{t_1-s} \end{array} (7)

will be standard normal and independent of {X_s} over {s\le t}.

Solution to the Hidden Drift Problem

We assume are given a standard Brownian motion W. Start by choosing a strictly increasing sequence of positive times {\{t_n\}_{n\in{\mathbb Z}}\in{\mathbb R}} such that {t_n\rightarrow\infty} as {n\rightarrow\infty} and {t_n\rightarrow0} as {n\rightarrow-\infty}. For example, we can simply take {t_n=2^n}.

Setting {t_n^\prime=2t_n-t_{n-1}}, we have an infinite sequence of overlapping intervals {[t_{n-1},t^\prime_n]} containing {t_n} in their interiors. The idea is to use the Brownian Bridge construction described above to construct X across each of these intervals. To do this, for each n, we need to construct a standard normal random variable {U_n} which is independent of {\{X_s\colon s\le t_n\}}. Given that we have already applied the above construction across the interval {[t_{n-1},t_n^\prime]}, starting with a standard normal {U_{n-1}}, we can construct the new standard normal with the required independence property by (7),

\displaystyle  U_n=\sqrt{\frac{t^\prime_n-t_n}{t^\prime_n-t_{n-1}}}U_{n-1}-\sqrt{t^\prime_n-t_n}\int_{t_{n-1}}^{t_n}\frac{dW_s}{t^\prime_n-s} (8)

Since {(t^\prime_n-t_n)/(t^\prime_n-t_{n-1})=2^{-1}}, we can solve this recursive formula as

\displaystyle  U_n=2^{(m-n)/2}U_m-\sum_{k=m+1}^n2^{(k-n)/2}\sqrt{t_k-t_{k-1}}\int_{t_{k-1}}^{t_k}\frac{dW_s}{t^\prime_k-s}

for any {m < n}. Noting that {2^{(m-n)/2}\rightarrow0} as {m\rightarrow-\infty}, we can take the limit to obtain

\displaystyle  U_n=-\sum_{k=-\infty}^n2^{(k-n)/2}\sqrt{t_k-t_{k-1}}\int_{t_{k-1}}^{t_k}\frac{dW_s}{t^\prime_k-s}. (9)

I take (9) as the definition of {U_n}, which we note are a sequence of standard normal random variables satisfying (8), and that {U_n} is {\mathcal F_{t_n}}-measurable. Next, we define X across the interval {[t_{n-1},t_n]\subseteq[t_{n-1},t^\prime_n]} so that it obeys (4) above,

\displaystyle  X_t=X_{t_{n-1}}+\frac{t-t_{n-1}}{\sqrt{t^\prime_n-t_{n-1}}}U_{n-1}+(t^\prime_n-t_{n-1})\int_{t_{n-1}}^t\frac{dW_s}{t^\prime_n-s}.

From the construction above, this ensures that {U_n} is independent of the increments of {X_s} over {s\le t}, and that X is a standard Brownian motion. From (5) and (6), we obtain

\displaystyle  X_t=W_t-W_s+X_s+\int_s^t\xi_udu (10)

for all times {0 < s < t}, with the drift process

\displaystyle  \xi_t =\frac{U_{n-1}}{\sqrt{t^\prime_n-t_{n-1}}}-\int_{t_{n-1}}^t\frac{dW_s}{t^\prime_n-s} (11)

over {t_{n-1}\le t < t_n}. In order to take the limit {s\rightarrow0} in (10), we need to show that {\int\lvert\xi\rvert dt} is almost surely finite. From (11), {\xi_t} is standard normal with variance {(t_n^\prime-t)^{-1}} and, hence, {{\mathbb E}[\lvert\xi_t\rvert]=\sqrt{2/\pi(t^\prime_n-t)}} giving,

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle {\mathbb E}\left[\int_0^{t_n}\lvert\xi_t\rvert dt\right] &\displaystyle =\int_0^{t_n}{\mathbb E}[\lvert\xi_t\rvert]dt =\sqrt{\frac2\pi}\sum_{m=-\infty}^n\int_{t_{m-1}}^{t_m}(t_m^\prime-t)^{-1/2}dt\smallskip\\ &\displaystyle =\sqrt{\frac8\pi}\sum_{m=-\infty}^n\left(\sqrt{t^\prime_m-t_{m-1}}-\sqrt{t^\prime_m-t_m}\right)\smallskip\\ &\displaystyle =(\sqrt2-1)\sqrt{\frac8\pi}\sum_{m=-\infty}^n\sqrt{t_m-t_{m-1}} \end{array}

So long as {\sum_{n < 0}\sqrt{t_n-t_{n-1}}} is finite, this guarantees that {\int_0^t\lvert\xi_s\rvert ds} is almost surely finite. This is true for the simple case with {t_n=2^n} and, hence, we can take the limit {s\rightarrow0} in (10) to obtain

\displaystyle  X_t=W_t+X_0+\int_0^t\xi_udu.

Choose {X_0=0} giving (1) as required.

I finally note that the original motivation for this example was to find a process X satisfying (1) whose natural filtration is strictly smaller than that of W. Letting {\mathcal G_t} be the filtration generated by X, we see that {U_n} is {\mathcal F_{t_n}}-measurable but is independent of {\mathcal G_{t_n}}, directly demonstrating that {\mathcal G_{t_n}\not\subset\mathcal F_{t_n}}.

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