$dY_t=2Y_tdt+2\sqrt{1+Y_t^2}dW_t$ where $W_t$ is $P-$Brownian motion (Wiener process).

I have defined a new measure $Q$ where the Kernel density (In Girsanov theorem) is $$ \phi_t = \frac{Y_t}{\sqrt{1+Y_t^2}} $$ Now I need to assure that the Novikov condition is satisfied. Hence I need to make sure: $$ E^P [\exp \{ \int_0^t \frac{Y_u^2}{1+Y_u^2}du \}]< \infty. $$ Is it? Is it possible to show that and how can I show that?

  • 1
    $\begingroup$ I don’t understand the first equation. Please correct $\endgroup$
    – Ezy
    Commented Jan 13, 2019 at 12:59
  • $\begingroup$ $dY_t$ instead of $Y_t$. Thanks for the reminder! I have now corrected. $\endgroup$
    – Sanjay
    Commented Jan 13, 2019 at 14:12
  • 4
    $\begingroup$ Note that $0 \le \frac{Y_u^2}{1+Y_u^2} < 1$. Then $\exp\int_0^t \frac{Y_u^2}{1+Y_u^2} du < \exp(t)$, and $E\left( \exp\int_0^t \frac{Y_u^2}{1+Y_u^2} du\right) \le \exp(t)$. $\endgroup$
    – Gordon
    Commented Jan 13, 2019 at 20:23

1 Answer 1


If you make the change of variable $Y_t = \sinh U_t$ and apply Ito then you immediately get

$$dU_t = 2dW_t$$

so the solution of your SDE is $$Y_t = \sinh\left(2W_t + C\right)$$

with $C$ a constant.

Then to answer your question is suffices to notice that


which is bounded therefore your expression is finite since the integrand is bounded.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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