Checking Novikov’s Condition with a Bounded Diffusion Integrand

Summary

The note considers whether an exponential expectation involving a diffusion process is finite, as needed to justify a change of measure using Girsanov’s theorem. The proposed approach transforms the process with the inverse hyperbolic sine, which turns the stated stochastic differential equation into a Brownian motion with constant scale. The transformed process can then be used to express the original process through a hyperbolic sine.

The key finiteness argument is simpler than solving for the process: the change-of-measure kernel is the original process divided by the square root of one plus its square, a ratio bounded in absolute value by one. Its square is therefore bounded, so its time integral over a finite horizon is bounded, and the exponential has finite expectation. The answer does not specify the time horizon or initial condition, and its transformation calculation is presented without working through the Itô steps. The boundedness argument itself is sufficient for the stated integrability question.

Key ideas

• A bounded change-of-measure kernel has a bounded squared integrand.
• Over a finite time interval, the integral of that squared kernel is bounded.
• The exponential of a bounded quantity has finite expectation.
• An inverse hyperbolic sine transformation can simplify the given diffusion.

Tags

# How to check if $ E [\exp \{ \int_0^t \frac{Y_u^2}{1+Y_u^2}du \}]< \infty $


# How to check if $ E [\exp \{ \int_0^t \frac{Y_u^2}{1+Y_u^2}du \}]< \infty $












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

## Answer by Ezy (score 6, accepted)

https://quant.stackexchange.com/a/43472

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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