Suppose you have a sequence of non-negative stochastic processes $(X^n)_{t \in \mathbb{R}}$, $n \geq 1$, with continuous paths and continuous in $t$ such that $$\int_{-\infty}^{\infty} X^n_t \, \mathrm{d} t =\int_{m}^{M} X_t^n \, \mathrm{d}t < \infty,$$ where $M = \sup_{0 \leq s \leq T} B_s$ and $m = \inf_{0 \leq s \leq T} B_s$, for a standard Brownian motion $B$ and some fixed $T>0$. Let $Y_t$ be a continuous random variable which also satisfies that $$\int_{\infty}^{\infty} Y_t \, \mathrm{d} t = \int_m^M Y_t \, \mathrm{d} t.$$

Is it true (maybe under some additional assumptions) that $$\int_{-x}^x X^n_t \, \mathrm{d} t \xrightarrow{d} \int_{-x}^{x} Y_t \,\mathrm{d}t $$ for any $x>0$ implies the same convergence in distribution with $x$ replaced by $\infty$?

EDIT: How about the case where $X_t^n = X_t$ for $n \geq 1$ where $X_t$ is non-negative and continuous? Does the conclusion hold?


Unless there is something I didn't understand about your Brownian motion statement, the result is not true in general. Consider even a deterministic example, where $X_t^n = \frac{1}{2n} 1_{|t| < n}$. Then $X_t^0$ is the constant function at $0$, so its integral is always $0$. However $\int_{-\infty}^\infty X_t^n dt = 1$, which clearly doesn't converge to $0$ in any sense.

One possible additional assumption you are looking for is uniform integrability of $X_t^n$.

  • $\begingroup$ This does not have continuous paths, though. $\endgroup$ – Christian Remling Aug 17 '14 at 22:24
  • $\begingroup$ OK how about using a smooth cutoff instead of the indicator function? $\endgroup$ – John Jiang Aug 17 '14 at 23:27
  • $\begingroup$ @JohnJiang Basically, your counterexample (with a smooth cutoff) should work, thank you! As the stochastic interval $[m,M]$ is on average finite you might have to increase the blowup rate so that $P \left\{ \int_m^M X_t^n \, \mathrm{d} t > 0 \right\} > p > 0$ for all $n$. However, my original question is connected to a problem I'm trying to solve and I was able to lose the $n$, i.e. one can assume that all $X_t^n$ are equal to the same, continuous and positive random variable $X_t$. Still I'm not able to figure out an answer. Do you have an idea? $\endgroup$ – r_faszanatas Aug 18 '14 at 6:29
  • $\begingroup$ @r_faszanatas: under your new assumption, the distributional convergence condition holds always. Also the conclusion holds trivially, since $\int_{-\infty}^\infty X_t^n dt \equiv \int_{-\infty}^\infty X_t^0 dt$. Is there anything I am missing? $\endgroup$ – John Jiang Aug 18 '14 at 23:55
  • $\begingroup$ @John Jiang Actually I chose very bad notation. Note that $X^n_t=X_t$ only for $n \geq 1$ (not for $n=0$). The random variable $X^0_t$ might still be different. $X^0_t$ should have been called $Y_t$, I was just too lazy to repeat the assumptions. I have corrected this now. $\endgroup$ – r_faszanatas Aug 19 '14 at 5:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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