Ok I'm trying to understand the second arcsine law which states:
Let $g_t:=\sup\{s\leq t:W_s=0\}$, then $$\mathbb{P}(g_t\leq s)=\frac{2}{\pi}\arcsin \left(\sqrt{\frac{s}{t}}\right )$$
This won't be anywhere near a rigorous proof, I just need clarification of whether my reasoning is correct.
Ok so, if we condition on $W_s=x$ we have $$\mathbb{P}(g_t> s)=\int_{-\infty}^{\infty}\mathbb{P}(g_t>s| W_s=x)\cdot \mathbb{P}(W_s=x)$$ $$=\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi s}}e^{-x^2/2s}\cdot \mathbb{P}(g_t>s| W_s=x)$$
Now the rest is the part I'm unsure of. We wish to calculate the probability that the Brownian motion returns to $0$ before time $t$ given that $W_s=x$. Hence as we conditioned on $W_s=x$, and Brownian motion has the strong Markov property we have that $$W_{t+s}-W_s\sim N(0,t)$$ which implies that the conditional density $$W_{t+s}\sim N(x,t-s)$$ This now determines the motion after $s$, given $W_s=x$.
Therefore if $x>0$ then we require $\mathbb{P}(W_{t+s}<0)=\Phi\left ( \frac{0-x}{\sqrt{t-s}} \right )=\Phi^c\left ( \frac{x}{\sqrt{t-s}}\right )$
And if $x<0$ we require $\mathbb{P}(W_{t+s}>0)=\Phi^c\left (\frac{-x}{\sqrt{t-s}}\right )$
Either way this is $\Phi^c\left (\frac{|x|}{\sqrt{t-s}}\right )$ and so $$\mathbb{P}(g_t>s|W_s=x)=2\Phi^c\left (\frac{|x|}{\sqrt{t-s}}\right )$$.
Plugging this into the integral and evaluating it gives the required arcsine law.
$\textbf{My problem in understanding is whether $s$ is the first hitting time to $x$}$. If it is not, then does this argument still hold? Also is this a viable method of showing the proof, provided one can evaluate the integral?