I am trying to understand the central limit theorem established for integrals.

Specifically, let $\left( X_n \right)$ be a sequence of random variables. I understand that under a set of conditions, for the sample mean $\bar{X}_N := \frac{1}{N} \sum_{i=1}^N X_i$, we have $\sqrt{N} \left( \frac{\bar{X}_N - \mu}{\sigma} \right) \overset{d}\to N\left(0,1\right)$.

$\textbf{My question is:}$ if, instead of looking at the sample mean $\bar{X}_N$, we look at the integral $\int_0^1 X(t) dt$, where $X(t)$ is some stochastic process on the interval $\left[0,1\right]$, such that $\int_0^1 X(t) dt$ is well-defined, then will $\frac{\int_0^1 X(t) dt - \mathbb{E}\left[\int_0^1 X(t) dt\right]}{\sqrt{\text{Var}\left(\int_0^1 X(t) dt\right)}}$ (or something in this format) converge to standard normal?

Thank you!

  • $\begingroup$ Converges when which quantity converges ? The expression you wrote is a number not depending on any parameter. $\endgroup$ – Stéphane Laurent Aug 10 '17 at 20:16
  • $\begingroup$ Thank you for your reply! This is actually where I am confused. What I am thinking is: suppose the integral itself is a random variable, would it be correct if I view it as the limit of a summation approximation and re-write it in the format of $\frac{1}{T}\sum_{j=1}^T X_j$ on interval $\left[0,1\right]$ as $T \to \infty$? If I can do so, would it make the expression a normal distribution directly, since it is already "at the limit"? $\endgroup$ – Tian Aug 10 '17 at 20:56
  • $\begingroup$ You don't have any parameter that goes to anything in that final expression, hence there's no sequence to converge to anything - it's just a number, as Stephane pointed out. If it's the limit of <something>, it's still just a number, not a sequence. $\endgroup$ – jbowman Aug 10 '17 at 21:11
  • $\begingroup$ Thank you for your reply. My concern here is that, in the case where the integral itself is a random variable, as opposed to a number, there is no "convergence in distribution"; in other words, it is "$\sim N(0,1)$" as opposed to "\overset{d}\to N(0,1)". I can't figure out whether/why this thought is correct or incorrect. $\endgroup$ – Tian Aug 10 '17 at 21:39
  • $\begingroup$ instead of "\overset{d}\to N(0,1)", I meant to say "$\overset{d}\to N(0,1)$" $\endgroup$ – Tian Aug 10 '17 at 22:23

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