slightly related:Is there a specific term for this SDE?

Using $f(x_t, t ) = x_te^{at}$, then $df(x_t,t) = ax_te^{at} + e^{at}dx_t$

We can plug in the original question, $dX_t = aX_tdt + bdW_t$ which results in

$df(x_t,t) = be^{-at}dW_t$ then I get, $x_te^{-at} = \int_0^t{x_te^{-rt}dW_t}$

From here, I am unsure of how to get to $x_t = x_0e^{at}+be^{at}\int_0^t{e^{-at}dW_s}$ which is the final solution.

I also want to use the $x_t$ to find $E[x_t^2]$. I know that $E[x_t] = x_0e^{at}$ but don't know how to to do it when x is squared.


By application of Ito's lemma, we have $$d(e^{-at}X_t)=-a e^{at}X_t dt+e^{at}dX_t+\underbrace{d[e^{at},X_t]}_0 =e^{at}b dW_t$$ thus $$X_t=e^{-at}X_0+b\int_{0}^{t} e^{-a(t-s)}dW_s.$$ As a result $$\text{Var}(X_t)=b^2\text{Var}\left(\int_{0}^{t} e^{-a(t-s)}dW_t\right)=b^2\mathbb{E}\left[\left(\int_{0}^{t} e^{-a(t-s)}dW_s\right)^2\right]$$ By application of Ito's isometry formula, $$\text{Var}(X_t)=b^2\mathbb{E}\left[\int_{0}^{t} e^{-2a(t-s)}ds\right]=b^2\int_{0}^{t} e^{-2a(t-s)}ds=\frac{b^2}{2a}(1-e^{-2at})$$ Finally we have $$\mathbb{E}[X_t^2]=Var(X_t)+\mathbb{E}[X_t]^2$$

| cite | improve this answer | |
  • $\begingroup$ So using my $E[x_t]^2$ equation, $E[X_t^2] = \frac{b^2}{2a}(1-e^{2at}) + (X_0e^{at})^2$? $\endgroup$ – mathematician123493 Dec 16 '16 at 12:19
  • $\begingroup$ $$E[X_t]=X_0e^{-at}$$ $\endgroup$ – Behrouz Maleki Dec 16 '16 at 12:21
  • $\begingroup$ ah I see thank you so much! I could understand this so much easier than I probably would have if I were on my own. $\endgroup$ – mathematician123493 Dec 16 '16 at 12:22
  • $\begingroup$ How did you use Ito's formula in the first line to get $d(e^{-at}X_t) = e^{at}bdW_t$? I keep getting $d(e^{-at}X_t) = e^{-at}bdW_t$ $\endgroup$ – user2139 May 16 '18 at 0:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.