Donsker's theorem

For independent identically distributed increments of mean zero and variance one, the diffusively rescaled, linearly interpolated random walk converges weakly in $C[0,1]$ to Brownian motion.

For centered unit-variance independent and identically distributed increments with $m_4=\mathbb{E}{\xi }^4<\infty$, an interpolated increment is $N^{-1/2}{\sum }_j{c}_j{\xi }_j$. Here $c_j$ is the length of the overlap of $(Ns,Nt)$ with the $j$th unit cell, so $0\le c_j\le 1$ and $\sum c_j=N(t-s)$. Independence gives its fourth moment as $N^{-2}[3(\sum c_j^2{)}^2+(m_4-3)\sum c_j^4]$. This is at most $\max (3,m_4)N^{-2}(\sum c_j^2{)}^2$, and $\sum c_j^2\le N(t-s)$ gives the bound, including intervals crossing a mesh point. A uniform fourth-moment continuity criterion yields tightness of probability measures on $C_0[0,1]$.

For a random walk with centered independent identically distributed increments of variance one, the maximum of its polygonal interpolation is the maximum at its grid vertices. The maximum functional is Lipschitz for the supremum norm, so the continuous mapping theorem applied to the Donsker invariance principle proves the displayed convergence in distribution. The Brownian reflection principle then gives upper-tail limits $2(1-\Phi (a))$ at all $a>0$, since the limiting maximum has no atom there.

For a random walk with independent identically distributed steps of mean zero and variance one, the Donsker invariance principle and the continuous mapping theorem for integration give this limit. The integral of the polygonal interpolation differs from the right-endpoint sum by $S_n/(2n^{3/2})$, whose squared expectation is $1/(4n^2)$. The Slutsky theorem removes that error. The limiting variable is the time-one value of integrated Brownian motion, with law $N(0,1/3)$.