Let $(\Omega,\mathcal{F}, \mathbb{P}, (\mathcal{F}_{t})_{t\geq0})$ be a filtered probability space. Furthemore, let $(S_{t}^{1},S_{t}^{2})_{t\geq0}$ be two assets (adapted to filtration, etc). Define $X_{t}=S^{1}_{t}-S^{2}_{t}$. If $X_{t}$ satisfies the SDE:
$dX_{t}=\xi(\zeta-X_{t})dt+\sigma dW_{t}$
($W_{t}$ is a $\mathbb{P}$ Brownian motion)
then what process does $(S_{t}^{1},S^{2}_{t})$ follow (assuming reasonable conditions like nonnegativity)?
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4$\begingroup$ This is a complex question, and there is probably not a unique representation. As @Kermittfrog answer shows, one possible solution is that $S^1$ and $S^2$ follow OU processes themselves. $\endgroup$– Daneel OlivawCommented May 25, 2022 at 12:46
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4$\begingroup$ @Kermittfrog gave probably the answer you expected (+1) but note that your general question allows for silly examples like $\text{d}S_1=\alpha S_1\text{d}t+\sigma_1\text{d}W_1$ and $\text{d}S_2=\left(\alpha S_1-\xi(\zeta-(S_1-S_2))\right)\text{d}t+\sigma_2\text{d}W_2$. You can find infinitely many of these trivial examples unless you narrow down your a problem a little. $\endgroup$– KevinCommented May 25, 2022 at 13:08
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1$\begingroup$ Good point @Kevin! $\endgroup$– KermittfrogCommented May 25, 2022 at 13:25
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1 Answer
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If we allow the mean reversion speeds to be identical, we could assume OU processes for the two components:
Let
$$ \begin{align} dx_1&=\kappa_1(\theta_1-x_1)dt+\sigma_1dW_1\\ dx_2&=\kappa_2(\theta_2-x_2)dt+\sigma_2dW_2 \end{align} $$ with $E(dW_1dW_2)=\rho dt$. Now let $z=x_1-x_2$. Then, if $\kappa_1=\kappa_2=\kappa$,
$$ \begin{align} dz&=dx_1-dx_2\\ &=\kappa_1(\theta_1-x_1)dt+\sigma_1dW_1-\kappa_2(\theta_2-x_2)dt-\sigma_2dW_2\\ &=\kappa(\theta_1-x_1)dt+\sigma_1dW_1-\kappa(\theta_2-x_2)dt-\sigma_2dW_2\\ &=\kappa((\theta_1-\theta_2)-(x_1-x_2))dt+\sigma_1dW_1-\sigma_2dW_2\\ &\equiv\kappa(\theta_z-z)dt+\sigma_zdW_z \end{align} $$
where $\theta_z=\theta_1-\theta_2$ and $\sigma_z^2=\sigma_1^2+\sigma_2^2-2\rho\sigma_1\sigma_2$.
answered May 25, 2022 at 8:42
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$\begingroup$ Excellent response! Thank you. $\endgroup$ Commented May 25, 2022 at 14:54