Let $X_t = e^{-\lambda t} \left(X_0 + \int _0^t e^{\lambda u} dW_u\right)$ where $(W_u)_{u \geqslant 0}$ is a Wiener process, $X_0$ random variable of law $\nu$ and independent of $\int _0^t e^{\lambda u} dW_u$.

I want to show that there is no meaningful distribution of the limit if $\lambda<0$. Using characteristic function and the independence I can show the limiting characteristic function is $0$ always, so it isn't a characteristic function. But does that tell me there can be no such measure?

I know the first term goes to infinity and the variance of the normal distribution of the second term as well, but I would prefer a better argument than that both those measures are respectively not finite or defined.


The usual notion of "distribution of the limit" is weak convergence: a sequence of probability measures $\mu_n$ on $\mathbb{R}$ converges weakly to a probability measure $\mu$ if $\int f\,d\mu_n \to \int f\,d\mu$ for all bounded continuous $f$. In particular, since $f(x) = e^{itx}$ is a bounded continuous function, the chfs of $\mu_n$ must converge to the chf of $\mu$. If you can show the chfs of your $X_t$ do not converge, or converge to something that cannot be a chf, then you have ruled out weak convergence.

(Note you have a slight error: it can't be that the limit of the chfs is zero everywhere, because $\phi(0)=1$ for any chf. But you may be able to show that the limit of the chfs is discontinuous at 0, and this would suffice because chfs must be continuous.)

  • $\begingroup$ Oh yea of course! I should rule out weak convergence, not the existence of a probability measure. Thanks for the tip about 0 - I will double check my calculations. $\endgroup$ – Henrik Dec 3 '12 at 15:50

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