Application of Ito's Lemma to integral expression

Question

I have a problem applying Ito's lemma.

I know that if: $dX_t= \mu_t \, dt + \sigma_t \, dB_t$ then for $f(t,x)$:

$df(t,X_t) =\left(\frac{\partial f}{\partial t} + \mu_t \frac{\partial f}{\partial x} + \frac{1}{2}\sigma_t^2\frac{\partial^2f}{\partial x^2}\right)dt+ \sigma_t \frac{\partial f}{\partial x}\,dB_t$

Now I want to apply Ito's theorem to

$g:=∫_0^t \theta(u) \exp(-\alpha(t-u) ) dB(u)$ (background of this expression can be found here)

How, if possible, can I do this?

The integral expression confuses me.. I have absolutely no clue how to do it based on what I know (see above)..

Answer 1

$$g(x,t) = ∫_0^t \theta(u) \exp(-\alpha(t-u) ) dB(u) \\ = e^{-\alpha t} ∫_0^t \theta(u) \exp(\alpha u ) dB(u) $$ Now use the fact that $t \to e^{-\alpha t}$ is of finite variation and the Ito formula for a product:

If $X,Y$ are two Ito processes then so is $XY$: $$ dX_t = a_t dt + \sigma_t dB_t \\ dX_t = b_t dt + \sigma'_t dB_t \\ d(XY)_ t = X_t dY_t + Y_t dX_t + \sigma_t\sigma'_t dt $$

$$ dg(x,t) = d(e^{-\alpha t})∫_0^t \theta(u) \exp(\alpha u ) dB(u) + e^{-\alpha t} d\left[∫_0^t \theta(u) \exp(\alpha u ) dB(u)\right] \\ = -\alpha e^{-\alpha t} dt ∫_0^t \theta(u) \exp(\alpha u ) dB(u) + e^{-\alpha t} \theta(t) \exp(\alpha t ) dB(t) + 0 \\ = -\alpha e^{-\alpha t} dt ∫_0^t \theta(u) \exp(\alpha u ) dB(u) + \theta(t) dB(t) $$