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.

  • 2
    $\begingroup$ Do you know Donsker invariance principle? $\endgroup$ – Davide Giraudo Feb 3 '13 at 15:01
  • $\begingroup$ Hi, yes - I do. $\endgroup$ – eugen1806 Feb 3 '13 at 15:25
  • $\begingroup$ Let me try. By Donsker's theorem for continuous bound $f$ one obtains that $\frac{S^*_n(t)}{\sqrt(t)}$ converges weakly to $B_t$ from which follows that $\int_0^1 f\left(\frac{S^*_n(t)}{\sqrt{n}} \right)dt \to \int_0^1 f(B_t)dt$ (1). Here $S^*_n(t)$ is the stochastic process derived and interpolated appropriately from $S_n$ (as in the Donsker's theorem proof). From the other side, the intergal on the left side in (1) can be obtained as a limit of Riemann sums $\lim_{n \to \infty} \frac{1}{n}\sum_{k=0}^nf\left(\frac{S^*_n(k/n)}{\sqrt{n}} \right)$. $\endgroup$ – eugen1806 Feb 4 '13 at 7:43
  • $\begingroup$ Which is equal to $\lim_{n \to \infty}\sum_{k=0}^{n}\frac{1}{n}f\left(\frac{S_k}{\sqrt{n}} \right)$. Davide, is that what you meant? Thanks. $\endgroup$ – eugen1806 Feb 4 '13 at 7:44
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    $\begingroup$ That's what I meant. $\endgroup$ – Davide Giraudo Feb 4 '13 at 10:07


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.


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