Convergence to Brownian motion integral

Question

Let $X_i$ be i.i.d with $\mathbb{E}(X_i) = 0$ and $Var(X_i) =1, \, S_n = \sum_{i=1}^n X_i$. I would like to show that $\sum_{i=1}^n \frac{f(S_i/\sqrt{n})}{n}$ converges to $\int_0^1 f(B_t)dt$ in distribution for every continuous and bounded function $f$, where $B_t$ is standard Brownian motion.

I would start with the following application of central limit theorem: $S^n_t \to B_t$ in distribution, where $S^n_t = \frac{1}{\sqrt{n}} \sum_{i=1}^{[nt]}X_i$. Not sure how to proceed further.

I would be grateful for any suggestion or ideas. Thanks.

Answer 1

Recall

Theorem (Donsker, 1952) Let $\{X_i\}$ be a sequence of i.i.d. zero mean square integrable random variable, with variance $1$. Let $S_n^*$ the random piecewise linear function defined on $[0,1]$ such that $S_n^*(kn^{-1})=\frac 1{\sqrt n}\sum_{j=1}^kX_i$. Then $\{S_n^*\}$ converges in law in $C[0,1]$ to $(W_t)_{0\leqslant t\leqslant 1}$.

Then use a Riemann sum approximation.