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I am a bit confused by SDE's. I am trying to solve the SDE $dX=(c-\mu X )dt+\sigma dB$, with $\mu,\sigma,c$ constants and $X_0=x_0$ deterministic. I already know the solution of $dX=fdt+gdB$ with $X(0)=1$, for $f,g$ continuous functions. How can I apply this to the new SDE, or what is a reasonable approach? :) Thanks everyone!

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Consider a general linear SDE $$dX_t = (a_1 X_t+a_2)dt + (b_1 X_t+b_2)dW_t.\qquad (*),$$ with $X_0 = x_0$ and where $W_t$ denotes standard Brownian motion. Define $$\phi_t = \exp \left \{a_1 t - \frac{b_1^2}{2}t + b_1W_t \right \},$$ then $(*)$ has solution $$X_t = \phi_t \left(x_0+(a_2-b_1b_2)\int_{0}^{t} \phi_s^{-1} ds + b_2 \int_{0}^{t} \phi_s^{-1} dW_s \right).$$ Comparing $(*)$ with $dX_t = (c-\mu X_t)dt + \sigma dW_t$ we observe that $$\begin{cases} a_1 = - \mu \\ a_2 = c \\ b_1 = 0 \\ b_2 = \sigma \end{cases}.$$ To proceed, substitute the above values into the general solution and simplify.

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  • $\begingroup$ how to prove that this solve our SDE? Tanks a lot for your answer :) $\endgroup$ – Liealgebrabach Apr 15 '16 at 11:00
  • $\begingroup$ and where does the $x_0$ show up? $\endgroup$ – Liealgebrabach Apr 15 '16 at 11:04

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