Using Itô Calculus to Differentiate the Ornstein–Uhlenbeck Solution

# Differential of the Ornstein-Uhlenbeck solution


# Differential of the Ornstein-Uhlenbeck solution












I would like to calculate a couples of limits of stochastic integrals.

Given the Ornstein-Uhlenbeck SDE:

$$ dY_t=\alpha(m-Y_t)dt+\beta dW_t $$

the solution is

$$ Y_t=m+(Y_0-m)e^{-\alpha t}+\beta\int_0^t e^{-\alpha(t-s)}dW_s. $$

In the book The Volatility Smile by Derman & Miller at pag. 327 it's written that the differential of the solution is:

$$ dY_t = -\alpha(Y_0-m)e^{-\alpha t}dt+\beta dW_t-\alpha\beta\int_0^te^{-\alpha(t-s)}dW_s. $$

I'm trying to calculate it by starting from the definition of differential of a function: $df(x)=f'(x)dx$. Thus I calculate $Y_{t+\Delta t}-Y_t$ and apply the definition of derivative. Unfortunately, I'm not able to calculate these limits because they involve a stochastic integral:

$$ \lim_{\Delta t \to 0} \frac{\int_t^{t+\Delta t} e^{-\alpha(t+\Delta t -s)}dW_s}{\Delta t} $$

and

$$ \lim_{\Delta t\to 0}\frac{\int_0^t e^{-\alpha(t+\Delta t -s)}-e^{-\alpha(t -s)}dW_s}{\Delta t} $$

How can I proceed? What are the rules/theorems for these calculations? Or am I taking completely the wrong direction?

Please, let me know if more details are needed. Thanks for the help.

Edit I checked the differential proposed by the authors and I don't understand why I get a $dt$ term that is missing in authors' one.

If I consider that $Y_t=f(t,X_t)$ with

$$ f(t,x)=m+(Y_0-m)e^{-\alpha t}+\beta e^{-\alpha t} x $$

and

$$ X_t=\int_0^t e^{\alpha s} dW_s , \quad dX_t=e^{\alpha t} dW_t. $$

I can now apply the Ito-Doeblin formula to obtain that ($f_t$, $f_x$ and $f_{xx}$ denote derivatives with respect to $t$, $x$ and $x^2$):

$$ dY_t = f_t(t,X_t)dt + f_x(t,X_t)dX_t + \frac{1}{2} f_{xx}dX_tdX_t \\= \big[(-\alpha(Y_0-m)e^{-\alpha t}-\alpha\beta \int_0^t e^{-\alpha(t-s)}dW_s\big]dt + \beta dW_t \\= -\alpha(Y_0-m)e^{-\alpha t}dt + \beta dW_t -\alpha\beta \color{red}{dt}\int_0^t e^{-\alpha(t-s)}dW_s $$

What am I doing wrong to obtain the red $dt$? From the comments it seems a notation issue thanks to @Lorenzo Castagno.

Edit 2 I'm trying to follow again the @Lorenzo Castagno tips. If I'm right, the second limit could be solved by invoking something like the Dominated Convergence Theorem, invert the integral with the limit and obtain something close to the author's expression. (What conditions should I check?)

What invalidates my "differential" way to obtain the authors' differential of the solution is that the second limit does not exist. Indeed, if I define a stochastic process $X_t = \int_0^t f(s) dW_s$, in differential form, I have $dX_t=f(t)dW_t$. The limit becomes

$$ \lim_{\Delta t \to 0} (\Delta t)^{-1}(X_{t+\Delta t}-X_{t}) \overset{\color{red}{(1)}}= \frac{dX_t}{dt} \overset{\color{red}{(2)}}= f(t)\frac{dW_t}{dt}. $$

This limit is not defined and "originated" the Ito's calculus to calculate the differential of stochastic functions.

What do you think? Are (1) and (2) equalities right?

## Answer by Rylan (score 2, accepted)

https://quant.stackexchange.com/a/83852

I would like to second Lorenzo Castagno's comment and say it appears to be either shorthand or a typo.

Besides verifying from Itô calculus, which you did, we can take advantage of the fact that we know the solution and the differential. Specifically, by plugging in the solution for $Y_t$ into the differential, we get

$$dY_t = \alpha \Bigg(m - (m + (Y_0 - m)e^{-\alpha t} + \beta\int_0^te^{-\alpha (t-s)}dWs\Bigg)dt + \beta dW_t$$ $$= -\alpha \Bigg((Y_0 - m)e^{-\alpha t} + \beta\int_0^te^{-\alpha (t-s)}dWs\Bigg)dt + \beta dW_t$$ $$= -\alpha (Y_0 - m)e^{-\alpha t} - \alpha \beta\Bigg(\int_0^te^{-\alpha (t-s)}dWs\Bigg)dt + \beta dW_t$$ which is exactly the form you show.

As for the justification of the limits you present, I think you're on the right track that the challenges of working with Brownian motion in are why Itô calculus exists (or at least why we quants bother to learn about it), and it's likely worthwile to go through the derivation of Itô's lemma to see where it disagrees with your differential approach.