# Analytical solution of OU process with scaled, time-transformed Wiener process?

From wikipedia, the solution of OU processes can be written as

$$x_t = x_0 e^{-\theta t} + \mu (1-e^{-\theta t}) + \frac{\sigma}{\sqrt{2\theta}}W_{1-e^{-2\theta t}}.$$ The driving brownian motion is scaled and time-transformed. If I scale it back, I get (can I do it?) $$x_t = x_0 e^{-\theta t} + \mu (1-e^{-\theta t}) + \frac{\sigma}{\sqrt{2\theta}}\sqrt{\frac{1-e^{-2\theta t}}{t}}W_{t}.$$

My problem is that if I use Ito formula to find the coefficient of $$dW_t$$, I get $$dX_t = fdt + \frac{\sigma}{\sqrt{2\theta}}\sqrt{\frac{1-e^{-2\theta t}}{t}}dW_{t}$$,

where f is just some function that is not of interest. The coefficient of $$dW_t$$ is a function of time but is not $$\sigma$$ of the SDE of OU processes. What is wrong here?

• Not $W_{1-2\theta t}$ . It is $W_{1-e^{-2\theta t}}$. Mar 29, 2022 at 8:43
• @KurtG. Yes, I corrected it. Mar 29, 2022 at 12:03

In Wikipedia they write that the well-known OU process $$\tag{1} x_t=x_0\,e^{-\theta t}+\mu(1-e^{-\theta t})+\sigma\int_0^te^{-\theta(t-s)}\,dW_s$$ can be written as $$\tag{2} x_t=x_0\,e^{-\theta t}+\mu(1-e^{-\theta t})+\frac{\sigma}{\sqrt{2\theta}}\,W_{1-e^{-2\theta t}}\,.$$

• (a) $$\quad$$ When the same Brownian motion is used in (1) and (2) the resulting OU processes $$x_t$$ constructed by these equations do not agree pathwise. It is easy to see that their expectation and variance agree. Therefore, since they are Gaussian, their distribution agrees.

• (b) $$\quad$$ To put this differently, if we want the two OU processes to agree pathwise then the BMs in (1) and (2) cannot be the same. This was nicely shown by user6247850 in a comment: The Brownian term in (1) depends on the entire path $$[0,t]\ni s\mapsto W_s$$ whilst the Brownian term in (2) depends only on the path $$[0,1]\ni s\mapsto W_s$$ (take $$s=1-e^{-2\theta t}$$ which is in $$[0,1]$$ for all $$t$$).

• (c) $$\quad$$ There is an infinitude of scalings $$h(t)$$ and time changes $$T_t$$ such that, with the same BM as in (1), the OU process $$\tag{3} x_t=x_0\,e^{-\theta t}+\mu(1-e^{-\theta t})+\sigma\,h(t)\,W_{T_t}$$ has the same distribution as in (1).

• (d) $$\quad$$ Using the quadratic variation (5) below there is a standard way (e.g. ) of time changing another BM $$B_t$$ such that the OU process $$\tag{4} x_t=x_0\,e^{-\theta t}+\mu(1-e^{-\theta t})+\sigma\,e^{-\theta t}B_{\langle M\rangle_t}$$ agrees pathwise with (1).

To see (a) simply compare the variances: From (1) or (2) it is easy to see that $$x_t$$ has variance $${\rm Var}[x_t]=\sigma^2\int_0^te^{-2\theta(t-s)}\,ds=\sigma^2e^{-2\theta t}\frac{e^{2\theta t}-1}{2\theta}=\sigma^2\frac{1-e^{-2\theta t}}{2\theta}\,.$$ To see (c) observe that the variance of (3) is $${\rm Var}[x_t]=\sigma^2\,h^2(t)\,T_t\,.$$ In (2) we had $$T_t=1-e^{-2\theta t}$$ and $$h^2(t)=1/(2\theta)$$.

In your question you had $$T_t=t$$ and $$h^2(t)=(1-e^{-2\theta t})/(2\theta t)\,.$$

These are obviously not the only possibilities. There is an infinitude of function pairs $$h(t)$$ and $$T_t$$ that yield the correct variance. The pairs only have to obey the equation $$h^2(t)\,T_t=\frac{1-e^{-2\theta t}}{2\theta}\,.$$ To see (d) we follow the standard steps for time-changed Brownian motion . With the BM from (1) $$M_t=\int_0^te^{\theta s}\,dW_s$$ is a martingale with quadratic variation $$\tag{5} \langle M\rangle_t=\int_0^te^{2\theta s}\,ds=\frac{e^{2\theta t}-1}{2\theta}\,.$$ Writing $$T_t=\frac{\ln(1+2\theta t)}{2\theta}$$ for the inverse function of $$\langle M\rangle_t$$ it turns out that $$B_t=M_{T_t}$$ is a martingale with quadratic variation $$\langle M\rangle_{T_t}=t\,,$$ in other words, $$B_t$$ is a Brownian motion. To put it differently, $$M_t=B_{\langle M\rangle_t}$$ is a time-changed Brownian motion. This gives (4)

 I. Karatzas, S. Shreve, Brownian Motion and Stochastic Calculus.

• It would be good if you could also explain why the two BM are different, or why scaling by $\sqrt{\frac{1-e^{-2\theta t}}{t}}$ produces an error. Thank you! Mar 29, 2022 at 13:48
• The Brownian motions are different because, if using the formula $x_t = x_0 e^{-\theta t} + \mu (1-e^{-\theta t}) + \frac{\sigma}{\sqrt{2 \theta}} W_{1-e^{-2\theta t}}$, then $x_t$ only depends on $(W_s)_{s \in [0,1)}$. On the other hand, writing $x_t$ in terms of a Brownian integral with respect to $\tilde W_t$ shows $x_t$ depends on $(\tilde W_s)_{s \in [0,\infty)}$. That scaling is essentially doing a time change, which is why it seems to give an error. Mar 29, 2022 at 14:11
• Edited. In fact this raised more questions that it answered. Interesting though. Do you mind to try your luck after going through my steps from K&S ? Mar 29, 2022 at 14:19
• @user6247850 . +1 . It looks like (2) and Vincent's ansatz only agree with (1) in distribution while (3) agrees pathwise. What do you think ? Mar 29, 2022 at 14:26
• @KurtG. I think I agree. It looks like the next step would be to define, for every $t$, a new Brownian motion by $\tilde B_s := c B_{s/c^2}$ for $c := \sqrt{2 \theta}e^{-\theta t}$, so $B_s = \frac{1}{c} \tilde B_{s c^2}$. Then, at $s = \langle M \rangle_t$, the $\sigma e^{-\theta t}$ term becomes $\frac{\sigma}{\sqrt{2\theta}} \tilde B_{1-e^{-2\theta t}}$. Mar 29, 2022 at 14:55