Let {Ltx; (x, t)∈R1×R+1} denote the local time of Brownian motion, and
αt:=∫−∞∞(Ltx)2 dx.
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Let η=N(0, 1) be independent of αt. For each fixed t,
\[\frac{\int_{-\infty}^{\infty}(L_{t}^{x+h}-L_{t}^{x})^{2}\,dx-4ht}{h^{3/2}}\stackrel{\mathcaligr{L}}{\rightarrow}\biggl(\frac{64}{3}\biggr)^{1/2}\sqrt{\alpha_{t}}\eta \]
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as h→0. Equivalently,
\[\frac{\int_{-\infty}^{\infty}(L^{x+1}_{t}-L^{x}_{t})^{2}\,dx-4t}{t^{3/4}}\stackrel{\mathcaligr{L}}{\rightarrow}\biggl(\frac{64}{3}\biggr)^{1/2}\sqrt{\alpha_{1}}\eta \]
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as t→∞.