Brownian local time: can we estimate $\mathbb{E}^B\left[\int_{\mathbb{R}}\left|L_{a}(t)-L_{a}(s)\right|^2da\right]$?

Question

Let $L_a(t),\; (a,t)\in \mathbb R\times [0,T]$ denote the local time of a Brownian motion $B$.

I am interested in the quantity

$$\mathbb{E}^B\left[\int_{\mathbb{R}}\left|L_{a}(t)-L_{a}(s)\right|^2da\right]$$ where without loss of generality we can assume that $t\geq s$.

I've found some estimation of the $L^2$-modulus of continuity of a Brownian local time that states that

$$\mathbb{E}^B\left[\int_{\mathbb{R}}\left|L_{a+h}(t)-L_{a}(t)\right|^2da\right]=4th+O(h^2),$$ but can something be said about the expression above?

Thanks in advance!

Answer 1

Yes, it is possible with the help of Berman's analytic method. Firstly, write occupation formula: $$ \int_0^t e^{i u B(s)} ds = \int_{\mathbb R} e^{iua} L_a(t) da. $$ Secondly, invert the resulting Fourier transform: $$ L_a(t) = \frac{1}{2\pi} \int_{\mathbb R}\int_0^t e^{iu(B(s)-a)}ds \, du. $$ Now it is a matter of Fubini to get $$ \mathbb E\big[\big(L_a(t+h) - L_a(t)\big)^2 \big] = \frac{1}{4\pi^2} \int_{\mathbb R}\int_{\mathbb R} \int_t^{t+h} \int_t^{t+h} \mathbb E\big[e^{i\left(u_1(B(s_1)-a) + u_2(B(s_2)-a)\right)}\big] ds_1 ds_2du_1du_2. $$ This can be found in the literature, but I invite you to estimate this expression, it's not hard at all.

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Some further steps: write this as $2\times \int_t^{t+h} \int_t^{s_2} \dots ds_1ds_2$, estimate the complex exponents by 1, transform $u_1 B(s_1) + u_2 B(s_2) = (u_1 + u_2) B(s_1) + u_2(B(s_2)-B(s_1))$; then the integrand becomes $$ \exp\left\{ -\frac12 \left(s_1(u_1+u_2)^2+ (s_2-s_1)u_2^2 \right)\right\}. $$ Now integrating w.r.t. $u_1$ and $u_2$ (in this order), you will arrive at something like $$ 2\int_t^{t+h}\int_t^{s_2} \frac{ds_1 ds_2}{\sqrt{s_1 (s_2 - s_1)}} = C h^2. $$