# How can I show that this is a solution to the given stochastic differential equation?

Let $$(B_t)_{t\geq 0}$$ be a brownian motion on a probability space $$(\Omega, \mathcal{F}, \Bbb{P})$$ and consider the stochastic differential equation $$dX_t=\left(\sqrt{1+X_t^2} +\frac{1}{2} X_t\right)~dt+\sqrt{1+X_t^2}~dB_t$$I want to show that $$X_t:=\sinh(\sinh^{-1}(x)+t+B_t)$$ is a solution for the differential equation from above.

My idea was to compute both sides and show that they are equal. But I am a bit confused with the $$dX_t,dB_t,dt$$. On the right hand side I got that $$\left(\sqrt{1+X_t^2} +\frac{1}{2} X_t\right)=\cosh\left(sinh^{-1}(x)+t+B_t\right)+\frac{1}{2} \sinh\left(\sin^{-1}(x)+t+B_t\right)$$ and $$\sqrt{1+X_t^2}=\cosh\left(sinh^{-1}(x)+t+B_t\right)$$ on the left hand side what does $$dX_t$$ mean, does it mean $$dX_t=\cosh(\sin^{-1}(x) +t+B_t)(1+dB_t)$$ If yes I don't see how to continue on the right hand side. Could someone help me?

• Are you familiar with Ito's formula? May 13 at 18:39
• @user6247850 yes I am May 13 at 18:41
• Did you try applying it to determine $dX_t$? May 13 at 18:57
• @user6247850 I don't managed it May 13 at 19:35

Since $$X_t = \phi(B_t,t)$$, with $$\phi(z,t) = \sinh\left(\sinh^{-1}(x)+t+z\right)$$, we find thanks to Itô's lemma : $$\begin{array}{rcl} \mathrm{d}X_t &=& \displaystyle \left(\dot{\phi}(B_t,t) + \frac{1}{2}\phi''(B_t,t)\right)\mathrm{d}t + \phi'(B_t,t)\mathrm{d}B_t \\ &=& \displaystyle \left(\cosh\left(\sinh^{-1}(x)+t+B_t\right) + \frac{1}{2}\sinh\left(\sinh^{-1}(x)+t+B_t\right)\right)\mathrm{d}t \\ && + \cosh\left(\sinh^{-1}(x)+t+B_t\right)\mathrm{d}B_t \end{array}$$ where $$\dot{\phi} = \partial_t\phi$$ and $$\phi' = \partial_z\phi$$. This last expression matches what you have already found for the right-hand side and the work is done.
• But how do you get this $dt$ and $dB_t$ behind? Because I only know the ito formula as $$f(t,B_t)=f(0,X_0)+\int_0^t f(s,B_s) ~ds+\int_0^t f(s,B_s) dX_s+\frac{1}{2} \int_0^t f(s,B_s) d[B,B]_s=f(t,B_t)=f(0,X_0)+\int_0^t f(s,B_s) ~ds+\int_0^t f(s,B_s) dX_s+\frac{1}{2} \int_0^t f(s,B_s) ds$$ if I'm not mistaken May 14 at 18:46