Can someone guide me on how to solve the SDE
\begin{equation} dX_{t} = \gamma(a-\beta X_{t})dt + \delta X_{t}dW_{t} \end{equation} where $a,\beta,\gamma,\delta$ are all positive constants? I tried to approach like an Ornstein-Uhlenbeck SDE but I don't think it is the right way.
On
In interest rate modeling, this is known as the Brennan and Schwartz model. My understanding is that, there is no close form solution, and numerical methods are used in practice.
On
Since it seems like what you wrote is a linear SDE of the form $dX_t=(a+b.X_t)dt+(c+d.X_t)dW_t$, I think a solution is available (no ?) Define $$\epsilon _t := \exp\left[\int_0^t b-\frac{d^2}{2} ds + \int_0^td. dW_s\right]$$, then $$X_t=\epsilon _t.\left[X_0+\int_0^t\frac{a-c.d}{\epsilon _s}ds + \int_0^t\frac{c}{\epsilon _s}dW_s \right]$$ Also since in that case the coefficients are constant then I think you can work out the integrals. In your case the substitutions are : $$ a \Rightarrow a\gamma \\ b \Rightarrow -\gamma\beta \\ c \Rightarrow 0 \\ d \Rightarrow \delta $$
Correct me if I am wrong though
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This linear SDE can be solve by an integrating factor, treating it formally just like an ODE. Firstly, $\gamma$ is unnecessary and can be absorbed into $\alpha$ and $\beta$. We thus need to solve (I use physics notation):
$$ \frac{dX}{dt} = (\alpha - \beta X) + c X f(t), $$ with $\left\langle f(t) \right \rangle = 0$ and $\left\langle f(t) f(t') \right \rangle = \delta(t-t')$. This may be rearranged to get $$\frac{dX}{dt} + [\beta-c f(t)]X = \alpha $$ This equation permits the simple integrating factor $$ \exp\left(\beta t-c \int^t f(t')dt' \right),$$ giving $$ \frac{d}{dt}\left[ X \exp\left(\beta t-c \int^t_0 f(t')dt' \right) \right] = \alpha \exp\left(\beta t-c \int^t_0 f(t')dt' \right) $$ Integrate from $0$ to $t$ to find $$X(t) = \left[ X_0 \exp\left(-\beta t+c \int_0^t f(t')dt' \right) + \int_0^t \alpha \exp\left(-\beta (t-t')+c \int^{t}_{t'} f(t'')dt'' \right) dt' \right]. $$ This is the formal solution of your equation.
Another Idea can be: Take $g(x,t)=e^{\gamma\beta t}\times X$,apply It$\hat o$ dif. rule. $$dg=\frac{\partial g}{\partial t}dt+\frac{\partial g}{\partial x}dX+\frac 12\frac{\partial^2 g}{\partial X^2}(dX)^2\\ dg=\gamma\beta e^{\gamma\beta t}Xdt+e^{\gamma\beta t}dX+\frac12(0)\\$$now put $dX_{t} = \gamma(a-\beta X_{t})dt + \delta X_{t}dW_{t} $so $$dg=\gamma\beta e^{\gamma\beta t}Xdt+e^{\gamma\beta t}dX\\ dg=\gamma\beta e^{\gamma\beta t}Xdt+e^{\gamma\beta t}\big(\gamma(a-\beta X_{t})dt + \delta X_{t}dW_{t})\big)\\dg=e^{\gamma\beta t}\gamma adt+e^{\gamma\beta t}\delta X_{t}dW_{t}$$left hand side is complete differential ,so apply integration to both sides $$\int dg=\int e^{\gamma\beta t}\gamma adt +\int e^{\gamma\beta t}\delta X_{t}dW_{t}\\g(x,t)-g(x_0,0)=\gamma a\int_0^t e^{\gamma\beta s}ds+\delta \int_0^t e^{\gamma\beta s} X_{s}dW_{s}\\ e^{\gamma\beta t}\times X-X_0=\gamma a\int_0^t e^{\gamma\beta s}ds+\delta \int_0^t e^{\gamma\beta s} X_{s}dW_{s}\\ e^{\gamma\beta t}\times X-X_0=\gamma a(\frac{1}{\gamma\beta})(e^{\gamma\beta t}-1)+\delta \int_0^t e^{\gamma\beta s} X_{s}dW_{s}\\ e^{\gamma\beta t}\times X=X_0+(\frac{a}{\beta})(e^{\gamma\beta t}-1)+\delta \int_0^t e^{\gamma\beta s} X_{s}dW_{s}\\\times e^{-\gamma\beta t} \\ X=X_0e^{-\gamma\beta t}+\frac{a}{\beta}(1-e^{-\gamma\beta t})+\delta e^{-\gamma\beta t}\int_0^t e^{\gamma\beta s} X_{s}dW_{s}$$