Show that the two solutions of the SDE are equivalent

Question

I have a process: $$dr_t = (W_t^1 - ar_t)dt +\sigma dW_t^2$$ where $W_t^1$ and $W_t^2$ are brownian motions with instantaneous correlation coefficient $\rho$. I want to show that the solution of this process can be written in a form: $$r_t = r_0 e^{-at}+ \int_0^tk(t,s)dW_s^1 + \sigma\int_0^th(t,s)dW_s^2$$. I started with the substitution $\tilde{r}_t = e^{at}r_t$ which led me to the solution. $$r_t = r_0 e^{-at}+ \int_0^t e^{a(s-t)}W_s^1ds + \sigma\int_0^te^{a(s-t)}dW_s^2 $$ which differs from the asked solution in the first integrand ($ds$ instead of $dW_s^1$). I thought I could use Fubini theorem but I am not quite sure how. So essentially what I would like to know is how to show that these two solutions are equivalent. Thanks for any suggestions.

Answer 1

This is a slightly extended and corrected (thanks @DaneelOlivaw) version of my comment.

Consider a process $h(t) W_t$, where $h$ is a function of time only. Using the Ito product rule, this can be expressed in integral form as

\begin{equation} h(t) W_t = \int_0^t h'(s) W_s \mathrm{d}s + \int_0^t h(s) \mathrm{d}W_s, \end{equation}

where we used that $h(0) W_0 = 0$. Matching terms we find that

\begin{equation} h'(s) = e^{a (s - t)} \end{equation}

and thus

\begin{equation} h(s) = \frac{1}{a} e^{a (s - t)}. \end{equation}

Rearranging yields

\begin{eqnarray} \int_0^t e^{a (s - t)} W_s \mathrm{d}s & = & \frac{1}{a} W_t - \frac{1}{a} \int_0^t e^{a (s - t)} \mathrm{d}W_s\\ & = & \frac{1}{a} \int_0^t \left( 1 - e^{a (s - t)} \right) \mathrm{d}W_s. \end{eqnarray}

Answer 2

"I thought I could use Fubini theorem but I am not quite sure how."

Note that we can write:

$$ \begin{align} \int_0^t e^{a(s-t)}W^1_s\text{d}s & = \int_0^t e^{a(s-t)}\left(\int_0^s\text{d}W^1_u\right)\text{d}s \\[6pt] & = \int_0^t \int_0^se^{a(s-t)}\text{d}W^1_u\text{d}s \end{align}$$

In the Appendix of Heath et al. (1992) you can find a nice statement of the stochastic Fubini theorem, which they themselves take from Ikeda and Watanabe (1981).

Let $s \leq T$ with $T \in \mathbb{R}^+$, we then define the continuous function $\phi(s)=e^{a(s-t)}$ for $s \in [0;T]$ $-$ note that in our case the function only depends on one variable thus for $u \in [0;T]$ we have $\phi(s,u)\triangleq\phi(s)$ which makes things easier. The function $\phi(s)$ is well-defined for any value of $t \in \mathbb{R}$. Therefore:

1. The function $\phi(s)$ is deterministic thus mesurable w.r.t. the filtration generated by $W^1$;
2. $\phi(s)$ is clearly square-integrable over $[0;T]$;
3. For any $\tau \in [0;T]$: $\int_0^T (\int_0^{\tau}\phi(s)\text{d}W^1_u)\text{d}s=\int_0^T \phi(s)(\int_0^{\tau}\text{d}W^1_u)\text{d}s=W_{\tau}^1\int_0^T \phi(s)\text{d}s$ $=cW_{\tau}^1$ for some $c \in \mathbb{R}$, which is continuous in $\tau$ by property of Brownian Motion.

The hypothesis of Lemma 0.1 of Heath et al. are fulfilled (p.98), thus applying Corollary 2 (p.99):

$$\begin{align} \int_0^t e^{a(s-t)}W^1_s\text{d}s & = \int_0^t \int_0^se^{a(s-t)}\text{d}W^1_u\text{d}s \\[6pt] & = \int_0^t \left(\int_u^te^{a(s-t)}\text{d}s\right)\text{d}W^1_u \\[6pt] & = \int_0^t \left(\frac{1-e^{a(u-t)}}{a}\right)\text{d}W^1_u \end{align}$$

@LocalVolatility's comment gives an alternative derivation avoiding Fubini, although I believe the $1/a$ must be outside the parenthesis.