# 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 ${\mathcal G_{t_n}\not\subset\mathcal F_{t_n}}$.