Solving a General Linear SDE by Removing Its Itô Integral
Summary
The document studies a linear stochastic differential equation with both additive and state-dependent drift and diffusion. It starts from an integrating-factor solution, writing the process as a product of a stochastic exponential and a second process that contains a stochastic integral. The author seeks a form that removes that remaining Itô integral, proposing an expression involving the ratio of the diffusion coefficients and its time derivative.
The post suggests checking this expression against two special cases from cited literature: a model with no additive diffusion and a displaced lognormal-type specification. It works through the latter using Itô’s formula and reports that the proposed specialization reproduces the stated closed-form solution. However, the general formula is presented as a question rather than independently established, and the document supplies no rigorous conditions on the coefficients or a general proof. Readers should treat it as a derivation to verify, not a confirmed result.
Key ideas
- An integrating factor expresses the solution of a linear SDE using a stochastic exponential.
- The remaining Itô integral can sometimes be transformed using a function of the integrating factor.
- The proposed general expression uses the ratio of diffusion coefficients and its time derivative.
- A displaced diffusion example is checked with Itô’s formula, but the general identity is not proved in the document.
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# Simplifying the Ito integral term in general linear SDE
# Simplifying the Ito integral term in general linear SDE
I am looking to simplify the Ito integral term in the solution for the general linear stochastic differential equation. Let the process $X_u$ follow $$\text{d}X_u=(a_u+b_uX_u)\text{d}u+(\gamma_u+\sigma_uX_u)\text{d}W_u.$$ The solution has the form $$X_u=Y_uZ_u\\ Z_u=e^{\int_t^u\sigma_s\text{d}W_s+\int_t^u(b_s-\frac{1}{2}\sigma_s^2)\text{d}s}\\ Y_u=x+\int_t^u\frac{a_v-\sigma_v\gamma_v}{Z_v}\text{d}v+\int_t^u\frac{\gamma_v}{Z_v}\text{d}W_v$$ (i.e., $Z_u$ satifies $\text{d}Z_u=b_uZ_u\text{d}u+\sigma_uZ_u\text{d}W_v$. From various sources (https://citeseerx.ist.psu.edu/document?repid=rep1&type=pdf&doi=7d38b03cfc62a15bdfd755c793d4e70a821725cc equation 11) I know that with $\gamma_v=0$ the solution is of the form $$X_u=xe^{\int_t^u\sigma_s\text{d}W_s+\int_t^u(a_s-\frac{1}{2}\sigma_s^2)\text{d}s}+\int_t^ua_ve^{\int_v^u\sigma_s\text{d}W_s+\int_v^u(a_s-\frac{1}{2}\sigma_s^2)\text{d}s}\text{d}v.$$
I also know (https://www.researchgate.net/profile/Fabio-Mercurio-2/publication/266035719_Displaced_and_Mixture_Diffusions_for_Analytically-Tractable_Smile_Models/links/56e6ccc008ae65dd4cc1d323/Displaced-and-Mixture-Diffusions-for-Analytically-Tractable-Smile-Models.pdf equations 13, 14) that for $a_v=0, b_v=\mu, \sigma_v=\beta_v, \gamma_v=-\alpha\beta_ve^{\mu v}$, which gives $$X_u=Y_uZ_u\\ Z_u=e^{\int_0^t\beta_s\text{d}W_s+\int_0^t(\mu-\frac{1}{2}\beta_s^2)\text{d}s}\\ Y_u=x+\int_0^t\frac{\alpha\beta_ve^{\mu v}}{Z_v}\text{d}v-\int_0^t\frac{\alpha\beta_ve^{\mu v}}{Z_v}\text{d}W_v,$$ which I solved by postulating a function $f(t, x)=g(t)/x$ and using Ito's formula $$\text{d}f(t, Z_t)=\frac{g'(t)}{Z_t}\text{d}t-\frac{g(t)}{Z_t^2}\text{d}Z_t+\frac{g(t)}{Z_t^3}\text{d}Z_t\text{d}Z_t\\ =\frac{g'(t)}{Z_t}\text{d}t-\frac{g(t)}{Z_t}(\mu\text{d}t+\beta_t\text{d}W_t)+\frac{g(t)}{Z_t}\beta_t^2\text{d}t\\ =\frac{1}{Z_t}(g'(t)-g(t)\mu+g(t)\beta_t^2)\text{d}t-\frac{g(t)}{Z_t}\beta_t\text{d}W_t$$ to match $$=\frac{\alpha \beta_t^2 e^{\mu t}}{Z_t}\text{d}t-\frac{\alpha \beta_t e^{\mu t}}{Z_t}\text{d}W_t.$$ This is done by letting $g(t)=-\alpha e^{\mu t}.$ Using $f(t, Z_t)=-\alpha e^{\mu t}/Z_t$, gives $$X_t=(X_0-\alpha )e^{\int_0^t\beta_s\text{d}W_s+\int_0^t(\mu-\frac{1}{2}\beta_s^2)\text{d}s}-\alpha e^{\mu t}.$$ Which is good. I want to reduce the general solution to the SDE without the Ito integral, skipping my workings (let me know if you want me to paste it here) I get $$X_u=(x+\frac{\gamma_t}{\sigma_t})Z_u+\int_t^u\frac{Z_u}{Z_v}\left(a_v-\frac{b_v\gamma_v}{\sigma_v}\right)\text{d}v -\frac{\gamma_u}{\sigma_u} +\int_t^u\left(\frac{\gamma_v}{\sigma_v}\right)'\frac{Z_u}{Z_v}\text{d}v $$ Using $a_v=0, b_v=\mu, \sigma_v=\beta_v, \gamma_v=-\alpha\beta_ve^{\mu v}$ and letting $u\mapsto t$ and $t\mapsto 0$ gives the required solution to Brigo and also gives the required answer ($\gamma_v=0$) from Swishchuck. Is this this correct? And If so, is this the most I can reduce it? Thanks in advance.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.