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$dY_t=2Y_tdt+2\sqrt{1+Y_t^2}dW_t$ 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?

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    $\begingroup$ 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
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    $\begingroup$ 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

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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

$$\frac{Y_u}{\sqrt{1+Y_u^2}}=\tanh(U_t)$$

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

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