# differential of the geometric brownian motion as a step for the solution of a linear stochastic differential equation

the core of the question is at the bottom, starting at Problem . Though I will give some context : I was trying to verify the solution of the linear SDE (stochastic differential equation) where $X_t$ is a stochastic process in $\mathbb{R^+} \times \Omega\rightarrow \mathbb{R}$ and where $B_t$ is a brownian motion: $$dX_t = (\alpha_t X_t + \beta _t )dt + (\sigma _tX_t + \rho_t) dB_t \tag{1}$$ For $t \in \mathbb{R^+}, \quad X_0 = x_0,\quad \alpha, \beta, \sigma ,\rho :\mathbb{R^+}\times \Omega$.

TLDR; I was planning to do the several steps : use the homogenous equation as a starting point, then have a process in the form of $Y_t=\frac{X_t}{Z_t}$ where $Z_t$ is the solution of the homogenous case, and apply the Ito's rule for a quotient. Then plug back the original values of $dX_t$ and $dZ_t$into the differential of my quotient to have the solution.

The first step was to look for an adequate process $y_t$ such that : $$X_t = y_t.exp\left(\int_0^t(\alpha_u -\frac{1}{2}\sigma_u^2)du \ + \int _0^t \sigma_u dB_u \right) \tag{2}$$ where the exponential part is a solution of the corresponding homogenous SDE: $$dY_t = \alpha_t Y_t \ dt + \sigma_tY_t \ dBt \tag{3}$$

Rearranging $(2)$ we have : $$y_t = \frac{X_t}{exp\left(\int_0^t(\alpha_u -\frac{1}{2}\sigma_u^2)du \ + \int _0^t \sigma_u dB_u \right)} \tag{4}$$ And now comes my question: we should apply Ito's quotient rule and taking advantage of the fact that $dy_t$ is given by equation $(3)$, applying bi-dimensionnal Ito formula

$df(t,X_t,Y_t) = \left(\frac{\partial{f}}{\partial{t}}dt + \frac{\partial{f}}{\partial{x}}dx + \frac{\partial{f}}{\partial{y}}dy + \frac{1}{2}\frac{\partial^2{f}}{\partial{x^2}}(dx)^2 + \frac{1}{2}\frac{\partial^2{f}}{\partial{y^2}} (dy)^2 + \frac{1}{2}\frac{\partial^2{f}}{\partial{x}\partial{y}} dxdy \right)$

for a function $f(t,X_t,Y_t) = \frac{X_t}{Y_t}$ , with $\frac{\partial{f}}{\partial{t}} = 0, \frac{\partial{f}}{\partial{x}} = \frac{1}{y}, \ \frac{\partial^2{f}}{\partial{x^2}} = 0, \ \frac{\partial{f}}{\partial{y}}=\frac{-x}{y^2},$ $\ \frac{\partial^2{f}}{\partial{y^2}} = \frac{2x}{y^3}, \ \frac{\partial^2{f}}{\partial{x}\partial{y}} = \frac{-1}{y^2}$yields :

$$df(t,X_t,Y_t)=d\left(\frac{X_t}{Y_t}\right)= \frac{dX_t.Y_t-X_t.dY_t}{Y_t^2}-\frac{dX_t.dY_t}{Y_t^2}+\frac{X_t(dY_t)^2}{Y_t^3} \tag{5}$$

So for the differential of $y_t$ if I plug $(1)$ and $(3)$ in this equation, we end up with $$dy_t = (\beta_t -\sigma_t)exp \left(-\left(\int_0^t(\alpha_u -\frac{1}{2}\sigma_u^2)du \ + \int _0^t \sigma_u dB_u \right) \right) + \ \rho_t exp\left(\int_0^t(\alpha_u -\frac{1}{2}\sigma_u^2)du \ + \int _0^t \sigma_u dB_u \right) dB_t$$

Integrating it and replacing the value of $y_t$ in equation $(2)$ will give me the solution for the linear SDE (1).

Problem : in my first approach I didn't plug (3), $\ \alpha_t Y_t dt + \sigma_t Y_t dt \$ the homogenous equation in the quotient, I tried to calculate the differential of $y_t$ as : $$d\left[exp\left(\int_0^t(\alpha_u -\frac{1}{2}\sigma_u^2)du \ + \int _0^t \sigma_u dB_u \right)\right]$$ by using the product rule $$de^f = \frac{\partial{f}}{\partial{x}}.e^f.dx$$ which would yield $$\left((\alpha_t -\frac{1}{2}\sigma_t^2 ) dt + \sigma_t dB_t \right) exp\left(\int_0^t(\alpha_u -\frac{1}{2}\sigma_u^2)du \ + \int _0^t \sigma_u dB_u \right)$$ The issue was that in equation (5) i would have an extra term $\frac{-1}{2}\sigma_t^2X_t dt$ impeaching me from integrating and getting a solution for the linear SDE. I have the intuition the problem lies somewhere between using Ito to solve the geometric brownian motion which adds an extra term of order two in comparison of classical calculus and my use of the fondamental theorem of calculus but i can not pinpoint excatly where. Any input would be greatly appreciated, be it related far or close to my issue.

• my intuition comes from an answer of @Did : math.stackexchange.com/questions/756465/itos-lemma-for-integral/… – RandowMalk Aug 19 '17 at 22:13
• It seems you are using a wrong "rule" $$d(e^{F(U_t)})=F'(U_t)e^{F(U_t)}dU_t$$ while, using the shorthand $V_t=F(U_t)$, the careful application of Itô yields $$d(e^{V_t})=e^{V_t}dV_t+\tfrac12e^{V_t}d\langle V\rangle_t$$ where $$dV_t=F'(U_t)dU_t+\tfrac12F''(U_t)dt$$ hence $$d\langle V\rangle_t=F'(U_t)^2d\langle U\rangle_t$$ and finally, $$d(e^{F(U_t)})=F'(U_t)e^{F(U_t)}dU_t+\tfrac12F''(U_t)e^{F(U_t)}dt+\tfrac12F'(U_t)^2e^{F(U_t)}d\langle U\rangle_t$$ – Did Aug 20 '17 at 8:27
• Hi @Did thanks for the amazing answer, so I did read $d\langle V\rangle _t$ as the quadratic variation, doing so i found the solution of the linear SDE. Maybe you can paste your comment in an answer so i can mark it answered. What suits you best. – RandowMalk Aug 20 '17 at 11:56
• *typo: Please replace $$dt$$ in my comment, twice, by $$d\langle U\rangle_t$$ – Did Aug 20 '17 at 12:31