Let $W_i ( i=0, 1, 2, ...)$ be i.i.d. random variables following a standard normal distribution $\mathcal{N}(0,1)$. For a continuously differentiable function $f:[0, 1] \to \mathbb{R}$, consider the following random variable:
$X_n = \sum_{i=0}^{n-1} \sqrt{n} \left( f \left(\frac{i+1}{n} \right) - f\left(\frac{i}{n} \right) \right) W_i$
Well, the equation itself looks a lot like gaussian quadrature, so I thought maybe this has something to do with it...
EDIT1:
I just realized that this is essentially a sum of random variables following a standard normal distribution, so the sum also follows a standard normal distribution.
Then, we have $E[X_n] = 0, Var[X_n] = \sum_{i=0}^{n-1} n \left( f \left(\frac{i+1}{n} \right) - f\left(\frac{i}{n} \right) \right)^2$
So,$X_n$ converges in distribution to a normal distribution, due to the central limit theorem?
EDIT 2: Thank you to those who have commented. I now have for $n \to \infty$ $E[X_n] = 0, Var[X_n] \to \int_0^1f'(x)^2dx$.
Then $X_n$ would intuitively converge to a normal distribution with the above average and variance, but how can I show this more rigorously?
By the mean-value theorem, $$ \frac{1}{n}\sum_{i=0}^{n-1}\left(\frac{f((i+1)/n)-f(i/n)}{1/n}\right)^2 = \frac{1}{n}\sum_{i=0}^{n-1}\left(f'(x_{n,i})\right)^2, $$ where $x_{n,i}\in [i/n,(i+1)/n]$. The RHS converges to $V\equiv\int_0^1 (f'(x))^2\,dx$ and, therefore,
$$ \varphi_{X_n}(t)=\exp\!\left(-\frac{t^2}{2}\times\frac{1}{n}\sum_{i=0}^{n-1}\left(\frac{f((i+1)/n)-f(i/n)}{1/n}\right)^2\right)\to e^{-\frac{t^2V}{2}}, $$ which is the ch.f of a zero-mean normal random variable with variance $V$.