Solving $Y_t$ = $e^{-\alpha t}X_t$ using Ito's lemma

Question

G'day I am trying to solve $Y_t$ = $e^{-\alpha t}X_t$, where $dX_t = \alpha X_tdt + \sigma_tdW_t$ using Ito's lemma.

• Defining f(t, X) = $e^{-\alpha t}X$. $\frac{df(t,X)}{dt}$ = $-\alpha e^{-\alpha t}X$; $\frac{df(t,X)}{dX}$ = $e^{-\alpha t}$.

• Using Ito's lemma: $dY_t = -\alpha e^{-\alpha t}X_t(dt) + e^{-\alpha t}(dX_t)$

• Substituting dt and dXt gives:

• = $-\alpha e^{-\alpha t}X_t( \alpha X_tdt) + e^{-\alpha t}(\alpha X_tdt + \sigma_tdW_t$)

• Calculating: $e^{-\alpha t}(\alpha X_tdt + \sigma_tdW_t)$ = $\alpha e^{-\alpha t}dt + \sigma_t e^{-\alpha t}dW_t$

• Moving in to the first part where I think the problem lies in finding the solution:

• $-\alpha e^{-\alpha t}X_t( \alpha X_tdt)$ = $- \alpha^2 e^{-\alpha}X^{2}_tdt$. I think that this step is wrong because no matter what I do, I can't seem to conclude with the given solution:
• = $\sigma_t e^{-\alpha t}dW_t$

Can someone explain the last steps and finish what I started?

Answer 1

I'm not quite following your derivation, but on the fourth line, I don't see how/why you are replacing $dt$ with $\alpha X_{t} dt$ and on the fifth line, when you multiply out the factor, it looks like $X_{t}$ is lost in the first term. I hope this helps.