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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?

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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 $X_t^n$.

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This does not have continuous paths, though. – Christian Remling Aug 17 '14 at 22:24
OK how about using a smooth cutoff instead of the indicator function? – John Jiang Aug 17 '14 at 23:27
@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? – r_faszanatas Aug 18 '14 at 6:29
@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? – John Jiang Aug 18 '14 at 23:55
@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. – r_faszanatas Aug 19 '14 at 5:13

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