Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 3 down vote accepted

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.)

share|cite|improve this answer
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. – Henrik Dec 3 '12 at 15:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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