# Show that process satisfy given equation

I have to show that process (1) $$X_t=e^{-bt}X_0+\int_0^te^{-b(t-s)}\sigma dW_s$$ satisfies the following equation (2) $$dX_t=-bX_tdt+\sigma dW_t$$ My attempt:

Multiply both sides of (1) by $e^{bt}$ and that yields (3) $$Y_t=X_0+\int_0^te^{bs}\sigma dW_s$$ where $$Y_t:=e^{bt}X_t$$ Now we have two equalities $$dY_t=d(e^{bt}X_t)=e^{bt}\sigma dW_t$$ Now I think I should use Ito formula for $F(x,t)=e^{bt} x$. Ito formula is given by: \begin{align} df(t,X_t) =\left(\frac{\partial f}{\partial t} + \mu_t \frac{\partial f}{\partial x} + \frac{1}{2}\sigma_t^2\frac{\partial^2f}{\partial x^2}\right)dt+ \sigma_t \frac{\partial f}{\partial x}\,dW_t. \end{align} My problem is that I don't know what are $\mu_t$ and $\sigma_t$ in our case or how to find them? I would like to see some explanation. Thanks a lot!

For any process of the form $$dY_t = \sigma_t \, dW_t + \mu_t \, dt, \tag{1}$$ we have

$$df(t,Y_t) = \left( \frac{\partial f(t,Y_t)}{\partial t} + \mu_t \frac{\partial f(t,Y_t)}{\partial y} + \frac{1}{2} \sigma_t^2 \frac{\partial^2 f(t,Y_t)}{\partial y^2} \right) \, dt + \sigma_t \frac{\partial f(t,Y_t)}{\partial y} \, dW_t. \tag{2}$$

You have already shown that $$dY_t = e^{bt} \sigma \, dW_t,$$ i.e. $\mu_t = 0$, $\sigma_t = \sigma e^{bt}$in $(1)$. Now apply $(2)$ to $f(t,y) := e^{-bt} y$ in order to find an expression for the stochastic differential

$$dX_t = df(t,Y_t) = d(e^{-b t} Y_t)$$

You seem to be able to prove that $\mathrm dY_t=\mathrm e^{bt}\sigma \mathrm dW_t$. Since $X_t=f(t,Y_t)$ where the function $f:(t,y)\mapsto\mathrm e^{-bt}y$ is smooth, the most basic form of Itô's formula yields $$\mathrm dX_t=\partial_yf(t,Y_t)\mathrm dY_t+\partial_tf(t,Y_t)\mathrm dt+\tfrac12\partial^2_{yy}f(t,Y_t)\mathrm d\langle Y,Y\rangle_t,$$ that is, $$\mathrm dX_t=\mathrm e^{-bt}\mathrm dY_t-b\mathrm e^{-bt}Y_t\mathrm dt.$$ Replacing $Y_t$ and $\mathrm dY_t$ by their expressions in terms of $X_t$ and $\mathrm dW_t$ respectively, you are done.

• $d Y_t=0\cdot \mu_t + e^{bt} \sigma dW_t$ so $\sigma_t =e^{bt} \sigma$ and I should put that into my Ito formula to calculate $dF(y,t)$? What is $d<Y,Y>_t$? Feb 2, 2014 at 16:23
• Wait a minute! You are dealing with Itô's formula stuff and you have no idea what the quadratic variation $\langle Y,Y\rangle$ could be?
– Did
Feb 2, 2014 at 16:40
• sorry we didn't have quadratic variation in our classes at all. we have very bad instructor. Feb 2, 2014 at 16:46
• Wow... How did you manage to even state Itô's formula with no notion of quadratic variation?
– Did
Feb 2, 2014 at 16:50
• @Did If you formulate Itô's formula for Itô processes, that's entirely possible. (Actually, in my first course on this topic we didn't introduce the notion of quadratic variation either. At least not properly.)
– saz
Feb 2, 2014 at 17:09