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Let $(X_t)_{t\in[0,T]}$ be a Gauss-Markov process, that is, a Gaussian process satisfying the Markov property. On page 230 of the book "Statistical Orbit Determination" by Bob Schutz, Byron Tapley, George H. Born the authors claim that all Gauss-Markov processes obey Langevin equations: $$ dX_t=-\beta X_t dt + \sigma dW_t$$ for some $\beta \geq 0, \sigma >0$.

That would mean that all Gaussian-Markov processes are either stationary (and in particular are Ornstein-Ulenbeck processes) or scaled Brownian motion (when $\beta =0$). Is this true? If so, how do we prove that all Gaussian-Markov processes are solutions to Langevin equations?

I believe it is not true. There must be a Gauss-Markov process that is not stationary (and not scaled Brownian motion). But I cannot produce an example.

EDIT: The Brownian bridge is Gauss-Markov and does not obey a Langevin SDE. So clearly, there is something wrong with the claim. If somebody can shed some light on the type of SDE that Gauss-Markov processes satisfy (if this type of SDE exists). To define a Gauss-Markov process, we only need its covariance. If we suppose that the process is never $0$ is probability one on $(0,T)$, then $Cov(X_s,X_t)=f(\min(s,t)) g(\max(s,t))$ where $\frac f g$ is positive nondecreasing. These functions $f$ and $g$ should play a role in the potential SDE.

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2 Answers 2

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What the book you are quoting is almost true. It has in fact become very popular in applied mathematics to identify solutions to the Langevin SDE with Gauss-Markov processes.

The exact story goes as follows

According to an 80 year old theorem by J.L. Doob [1], [2] every stationary Gauss-Markov process $Y_t$ with continuous sample paths has covariance function $$\tag{1} {\rm Cov}\big[Y_t,Y_s\big]=\frac{\sigma^2}{2\beta}e^{-\beta|t-s|}\,. $$ Doob writes $\sigma_0^2=\frac{\sigma^2}{2\beta}$ and calls this $Y_t$ an O.U. process.

Ornstein and Uhlenbeck themselves have instead studied the meanreverting Langevin type SDE $$\tag{2} dX_t=\beta\,(\mu-X_t)\,dt+\sigma\,dW_t $$ which gained a lot of popularity for example in quantitative finance. Its explicit solution is $$ X_t=X_0\,e^{-\beta t}+\mu(1-e^{-\beta t})+e^{-\beta t}\int_0^te^{\beta s}\,dW_s\,. $$ When $X_0=0$ and $\mu=0$ there is a close relationship between $X$ and $Y$: Namely, Doob shows in [1] also that there exists a standard Brownian motion $B_t$ such that $$ B_{\sigma^2_0t}=\sqrt{t}\,Y_{\textstyle\frac{\log t}{2\beta}}\, $$ holds. Changing the time variable to $s=\frac{\log t}{2\beta}$ this can be written as $$\tag{3} Y_s=e^{-\beta s}B_{T_s}\,,\quad\text{ where }\quad T_s:=\textstyle\sigma^2_0\, e^{2\beta\, s}\,. $$ Lemma. The process $Y_t$ has constant variance $\sigma^2_0$ while $$\tag{4} X_t:=e^{-\beta t}B_{(T_t\color{red}{-\,\sigma^2_0})} $$ solves the special case of the SDE (2) with $X_0=0$ and $\mu=0\,.$

Proof. The time changed Brownian motion in (4) can be written as $$\tag{5} B_{(T_t\color{red}{-\,\sigma^2_0})}=\color{red}{\sigma}\int_0^te^{\beta s}\,dW_s $$ with yet another standard Brownian motion $W$. To see this note that $$ T_t-\,\sigma_0^2=\sigma^2_0\, e^{2\beta\, t}-\sigma_0^2=\sigma^2\frac{e^{2\beta t}-1}{2\beta}=\sigma^2\int_0^te^{2\beta s}\,ds\,. $$ Therefore, \begin{align} W_t=\int_0^t\frac{e^{-\beta\,u}}{\sigma}\,dB_{(T_u-\,\sigma^2_0)}= \int_0^t\frac{e^{-\beta\,u}}{\sigma}\,dB_{\textstyle(\sigma^2\int_0^ue^{2\beta s}\,ds)} \end{align} which is seen to be a continuous martingale with quadratic variation $t$ and therefore a Brownian motion.

Combining (4) and (5) we get $$ X_t=\sigma\,e^{-\beta t}\int_0^te^{\beta s}\,dW_s $$ which is seen to solve the familiar SDE $$ dX_t=-\beta X_t\,dt+\sigma\,dW_t\,. $$ Few remarks:

  • It is easy to see that $X_t$ is Gauss-Markov but not stationary. Its covariance function is $$\tag{6} {\rm Cov}\big[X_t,X_s\big]=\sigma^2e^{-\beta(t+s)}\frac{e^{-2\beta(t\wedge s)}-1}{2\beta}=\sigma^2\frac{e^{-\beta|t-s|}-e^{-\beta(t+s)}}{2\beta} $$ (cf. [2]).

  • The Brownian bridge $$ Z_t=W_t-\frac{tW_T}{T}\,,\quad 0\le t\le T\,, $$ is continuous, Gaussian and Markov with respect to its own filtration [3] and has covariance function $$\tag{7} {\rm Cov}\big[Z_t,Z_s\big]=t\wedge s-\frac{ts}{T}\,. $$ But it is not stationary and therefore not an O.U. process $Y$ in the sense of Doob [1]. The Brownian bridge does also not satisfy the Langevin SDE (2) because its covariance function cannot be achieved by (6). (For $\beta\to 0$ the expression in (6) converges to $\sigma^2(t\wedge s)$, not to (7).)

[1] Doob, J.L. (April 1942). The Brownian Movement and Stochastic Equations. Annals of Mathematics. JSTOR. 43 (2): 351–369. doi:10.2307/1968873. ISSN 0003-486X. JSTOR 1968873.

[2] Wikipedia Ornstein Uhlenbeck Process.

[3] Borisov, I.S. On a Criterion for Gaussian Random Processes to Be Markovian. https://doi.org/10.1137/1127097

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  • $\begingroup$ The Brownian bridge is Markov (with regard to its own filtration) since its covariance function factorises into a function of the minimum and a function of the maximum. If you do not know this result, see the paper of Borisov "on a criterion for gaussian random processes to be markovian" $\endgroup$
    – W. Volante
    Commented Mar 28, 2022 at 19:39
  • $\begingroup$ Thanks for this. Unfortunately I see only a fragment of this paper but I believe you are right. The resolution is that Brownian bridge is not stationary, therefore not OU. $\endgroup$
    – Kurt G.
    Commented Mar 28, 2022 at 19:47
  • $\begingroup$ It does not solve the question. The claim in the book is wrong as highlighted in the last paragraph of my question. So what is the correct claim? What type of SDE do Gauss-Markov process satisfy (if it exists)? $\endgroup$
    – W. Volante
    Commented Mar 28, 2022 at 19:58
  • $\begingroup$ @W.Volante . Thanks for asking that interesting question. I had to update the answer to erase some false statements I made about Wikipedia and go into further details. Your question I tried to answer as briefly as possible in the first sentence. What remains is the question if the result of Doob was ever extended to non stationary Gauss-Markov processes. I don't know. $\endgroup$
    – Kurt G.
    Commented Mar 29, 2022 at 2:24
  • $\begingroup$ I cannot accept your answer because it is precisely what you do not know that I am asking about, but +1 for the amazing presentation. $\endgroup$
    – W. Volante
    Commented Mar 30, 2022 at 11:47
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Every non-degenerated Gauss-Markov (GM) process with continuous mean and covariance is a time-dependent Ornstein-Uhlenbeck process, meaning that they solve the SDE $$ \mathrm{d}X_t = \theta(t)(\alpha(t) - X_t)\,\mathrm{d}t + \sigma(t)\,\mathrm{d}B_t, $$ Proving this statement comes easily by directly computing the drift and volatility terms as $$ \lim_{h\downarrow 0}h^{-1}\mathbb{E}_{t, x}\left[X_{t+h} - x\right], \quad \lim_{h\downarrow 0}h^{-1}\mathbb{E}_{t, x}\left[\left(X_{t+h} - x\right)^2\right], $$ where $\mathbb{E}_{t, x}\left[\cdot\right] = \mathbb{E}\left[\cdot | X_t = x\right]$. You will also need to use the fact that conditioning a Gaussian process to take a deterministic value at some point leaves the resulting process Gaussian with a well-known formula for the covariance and mean. Also, the characterization of GM processes via the factorizability of their covariances will come in handy.

I'm not working out the details here, but rather cite this paper, which also treats GM bridges.

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