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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 $N(0,1/3)$.

Let $W_n$ be the interpolation in part (a). The operationis continuous, since $|I(f)-I(g)|\le \Vert f-g{\Vert }_{\infty }$. By the Donsker invariance principle and the continuous mapping theorem,The exact trapezoidal integral of the linear interpolation isTherefore the statistic in question differs from $I(W_n)$ by $S_n/(2n^{3/2})$. Using independence, zero means, and unit variances givesThis error tends to zero in $L^2$, hence in probability. The Slutsky theorem now proves the integrated random-walk limit:The limiting law can also be made explicit. The time integral is a Gaussian random variable, as a mean-square limit of linear combinations of a Gaussian process. It is centered, and the covariance identity $\mathbb{E}(B_s{B}_t)=\min (s,t)$ givesThus the terminal value of integrated Brownian motion here has law $N(0,1/3)$.