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I want to show that the process

$Y(t) = e^t \int_0^t e^{-s}dW(s)$ satisfies the following SDE:

$dX(t) = X(t)dt + dW(t), \ \ t\geq 0 , \quad X(0) = 0$

I think the right approach is to use Ito's formula, but i don't see how. I thought that the argument $dW(s)dt = 0$ is totally wrong in the direct approach. If one would try Ito's formula, $f$ would be $Y(t) = f(t,X(t)) = e^t \int_0^t e^{-s}dX(s)$, where $X(t) = W(t)$, the brownian motion. Now if we try to calculate $dY(t)$ with Ito, we get:

$\frac{\partial f}{\partial t} = tY(t)$ (?)

$\frac{\partial f}{\partial X} = X(t)$

$\frac{\partial^2 f}{\partial X^2} = dX(t)$

Could you give me a hint on how to do this? Thanks!

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1 Answer 1

If we set $$Z(t) := \int_0^t e^{-s} \, dW_s,$$ then $Z_t$ is a Itô process. Moreover, $dZ_t = e^{-t} \, dW_t$. If we apply Itô's formula to

$$f(t,z) := e^t \cdot z,$$

we obtain

$$\begin{align*} \underbrace{f(t,Z_t)}_{Y_t}-\underbrace{f(0,0)}_{0} &= \int_0^t \underbrace{e^s}_{\partial_z f(s,Z_s)} \, dZ_s + \int_0^t (\underbrace{e^s \cdot Z_s}_{\partial_t f(s,Z_s)} + \underbrace{0}_{\partial_z^2 f(s,Z_s)}) \, ds \\ &= \int_0^t \, dW_s + \int_0^t Y_s \, ds \\ \Leftrightarrow dY_t &= dW_t + Y_t \, dt \end{align*}$$

using that $Y_t = f(t,Z_t)$ by definition.

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