Brownian positive occupation time has the arcsine law

By [F2], $A_t$ is a random variable with values in $[0,t]$; by [F3] applied with $c=t$ one has $A_t{=}^{d}t{A}_1$, because the time change maps the set of positive times for $B$ to the corresponding set for the scaled motion, and hence $A_t\mathrm{/}t$ has the law of $A_1$.

The probability measure $\mu$ on $[0,1]$ with density ${\pi }^{-1}{x}^{-1\mathrm{/}2}(1-x{)}^{-1\mathrm{/}2}$ satisfies ${\int }_0^1(\alpha +\beta x{)}^{-1}\mu (dx)=1\mathrm{/}\sqrt{\alpha (\alpha +\beta )}$ for all $\alpha ,\beta >0$: substituting $x={\sin }^2\theta$ turns the integral into $\frac{2}{\pi }{\int }_0^{\pi \mathrm{/}2}(\alpha +\beta {\sin }^2\theta {)}^{-1}d\theta$, then $s=\tan \theta$ turns it into $\frac{2}{\pi }{\int }_0^{\mathrm{\infty }}dt\mathrm{/}(\alpha +(\alpha +\beta )t^2)$, and finally $w=t\sqrt{(\alpha +\beta )\mathrm{/}\alpha }$ and [F5] give $\frac{2}{\pi }\cdot \frac{1}{\sqrt{\alpha (\alpha +\beta )}}\cdot \frac{\pi }{2}$.

For $\alpha ,\beta >0$, ${\int }_0^{\mathrm{\infty }}{e}^{-\alpha t}E[e^{-\beta A_t}]dt=E[(\alpha +\beta A_1{)}^{-1}]$: [step 1.1] gives $E[e^{-\beta A_t}]=E[e^{-\beta t{A}_1}]$, so the left side is ${\int }_0^{\mathrm{\infty }}E[e^{-(\alpha +\beta A_1)t}]dt$, and [F4] equals it to $E[{\int }_0^{\mathrm{\infty }}{e}^{-(\alpha +\beta A_1)t}dt]=E[1\mathrm{/}(\alpha +\beta A_1)]$, the integrand being nonnegative and $\alpha +\beta A_1\ge \alpha >0$.

For $V=\alpha +\beta 1_{\{y>0\}}$ and $t>0$ one has ${\int }_0^{t}V({\widehat{B}}_r)dr=\alpha t+\beta A_t$ up to the single point $r=0$, which is Lebesgue-null; hence the function $u$ of [F1] satisfies $u(0)={\int }_0^{\mathrm{\infty }}{e}^{-\alpha t}E[e^{-\beta A_t}]dt$, and [F1] with [step 2.1] yields $E[1\mathrm{/}(\alpha +\beta A_1)]=1\mathrm{/}\sqrt{\alpha (\alpha +\beta )}$ for all $\alpha ,\beta >0$.

The law of $A_1$ and $\mu$ have the same moments: fixing $\alpha =1$ and expanding $1\mathrm{/}(1+\beta x)={\sum }_{k\ge 0}(-\beta x{)}^k$ for $x\in [0,1]$ and $0<\beta <1$, uniformly on the square, [F7] shows that $E[1\mathrm{/}(1+\beta A_1)]={\sum }_k(-\beta {)}^{k}E[A_1^k]$ and $\int (1+\beta x{)}^{-1}\mu (dx)={\sum }_k(-\beta {)}^k\int x^k\mu (dx)$ for every such $\beta$; since [step 3.1] and [step 1.2] make the two sides equal for all $\beta \in (0,1)$, subtracting the two power series gives ${\sum }_k(-\beta {)}^k(E[A_1^k]-\int x^k\mu (dx))=0$ on an interval, so every coefficient vanishes and all moments agree.

Consequently $E[f(A_1)]=\int fd\mu$ for every continuous $f:[0,1]\to \mathbb{R}$: given $\epsilon >0$, [F6] supplies a polynomial $p$ with $\mathrm{\parallel }f-p{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon$, and $\mathrm{\mid }E[f(A_1)]-\int fd\mu \mathrm{\mid }\le 2\epsilon +\mathrm{\mid }E[p(A_1)]-\int pd\mu \mathrm{\mid }=2\epsilon$ by [step 4.1]; taking continuous $f_n\downarrow 1_{[0,x]}$ and applying [F7] to both sides gives $P(A_1\le x)=\mu ([0,x])=\frac{2}{\pi }\arcsin \sqrt{x}$ for $0\le x\le 1$.

By [step 1.1], $P(A_t\mathrm{/}t\le x)=P(A_1\le x)=\frac{2}{\pi }\arcsin \sqrt{x}$ for every $t>0$ and $0\le x\le 1$, which is the displayed distribution function.

Differentiating the distribution function on $(0,1)$ gives $\frac{d}{dx}\frac{2}{\pi }\arcsin \sqrt{x}=\frac{2}{\pi }\cdot \frac{1}{2\sqrt{x(1-x)}}=\frac{1}{\pi \sqrt{x(1-x)}}$; this density is integrable on $(0,1)$ (substitute $x={\sin }^2\theta$), so it is the density of $A_t\mathrm{/}t$ and both endpoints carry zero mass.

The boundary cases are covered: $x=0$ gives $P(A_1=0)=0=\frac{2}{\pi }\arcsin 0$ and $x=1$ gives $1=\frac{2}{\pi }\arcsin 1=\frac{2}{\pi }\cdot \frac{\pi }{2}$; the value $\theta =\pi \mathrm{/}2$ of the substitution is the endpoint of the principal branch of [F5]; the parameters satisfy $\alpha ,\beta >0$ in [step 3.1] and $0<\beta <1$ in [step 4.1]; the occupation time is taken over the half-line $\{y>0\}$ so the single instant $s=0$ is excluded by a null set; and AC enters only through [F8].