Quadratic variation in mean-square

HardStochastic Processes~9m

Let $B_t$ be a standard Brownian motion. Partition $[0,t]$ into $2^n$ equal pieces and set $\Delta_{m,n} = B_{tm2^{-n}} - B_{t(m-1)2^{-n}}$ . Evaluate $\mathbb{E}\!\left[\left(\sum_{m=1}^{2^n}\Delta_{m,n}^2 - t\right)^{2}\right]$ as a function of $n$ and $t$ , then give its numerical value at $t = 1$ , $n = 5$ .

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The sum has mean $t$ , so the expectation is its variance = sum of $\operatorname{Var}(\Delta_{m,n}^2)$ (independent increments).
For $\mathcal N(0,\sigma^2)$ , $\operatorname{Var}(\cdot^2)=2\sigma^4$ with $\sigma^2=t2^{-n}$ . Sum $2^n$ terms.

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Asked at: Jane Street, Two Sigma

Quadratic variation in mean-square

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