Computing distribution of first hitting time of 0 after time 1 for 1-dimensional Brownian motion Let $D = \inf \{t \in [1, \infty): B_t = 0\}$, where $B_t$ is one-dimensional standard brownian motion. $D$ is the first hitting time of $0$ after time $0$ and it is a stopping time since it is the hitting time of a closed set. I want to find the law of $D$. My idea is to partition the event $D \le t$ into events $D \le t \bigcap B_1 = x$ and sum over all $x \in \mathbb{R}$. Abusing notation by writing a summation over the reals:
$$
P(D \le t) = \sum_{x \in \mathbb{R}} P(B_1 = x)P(D \le t | B_1 = x) = \sum_{x \in \mathbb{R}} P(\exists 1 \le s \le t: B_s=0 | B_1 = x)P(B_1 = x) = \sum_{x \in \mathbb{R}}  \left(\sum_{1 \le s \le t} P(B_s-B_1 = -x) \right) P(B_1=x) 
$$
I then wrote this as an integral:
$$
P(D \le t) = \int_{-\infty}^{\infty} \left( \int_{1}^{t} f_{s-1}(-x) ds \right) f_1(x) dx = \int_{-\infty}^{\infty} \left( \int_{1}^{t} \frac{e^{-\frac{x^2}{2(s-1)}}}{\sqrt{2\pi(s-1)}} ds \right) \frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}}dx
$$
However, I am not sure how to proceed since the nested integral seems difficult to compute explicitly. Does anyone have any advice on how to proceed?
 A: Let $X \sim N(0,1)$ and $W$ be a Brownian motion independent from $X$. Define $\tau := \inf\{t\geq0:W_t = X\}$.
It's easy to see that $ \tau +1 \stackrel{\mathcal L} = D$, using $W_t = B_{t+1}-B_1$ and $X = -B_1$.
For any $x\in\mathbb R$, define $\tau_x := \inf\{t\geq0:W_t = x\}$. It's well known that $$\mathbb P(\tau_x \leq t) = 2\mathbb P(W_t\geq \lvert x\rvert) = 2(1-\Phi(\frac {\lvert x\rvert}{\sqrt t}))$$
We have that
$$ \mathbb P(\tau \leq t) = \mathbb E[\chi_{\tau\leq t}] = \mathbb E[\mathbb E[\chi_{\tau\leq t}\mid X]] = \mathbb E[\mathbb E[\chi_{\tau_x\leq t}]_{x=X}] = 2(1-\mathbb E[\Phi(\frac {\lvert X\rvert}{\sqrt t})]).$$
However, $$\mathbb E[\Phi(\frac {\lvert X\rvert}{\sqrt t})] = \int_{\mathbb R}\frac{e^{-\frac{x^2}2}}{\sqrt{2\pi}}\Phi(\frac {\lvert x\rvert}{\sqrt t})dx = \iint_{y\sqrt{t}\leq \lvert x\rvert}\frac{e^{-\frac{x^2+y^2}2}}{2\pi}dxdy = 1-\frac{\arctan{\sqrt{t}}}\pi.$$
(for the last integral, just use the radial symmetry, I hope I calculated the angle correctly).
Therefore $$\mathbb P(D\leq t) = \mathbb P(\tau \leq t-1) = \frac{2\arctan{\sqrt{t-1}}}{\pi}.$$
A: Here is a solution without computing integrals. By symmetry,
\begin{eqnarray} P(D \le t)&=&2P(B_1>0, D \le t) \\
&=&2P(B_1>0, B_t<0)+2P(B_1>0, D \le t, B_t>0) \,.
\end{eqnarray}
By reflecting $B$ at time $D$, we see that the second summand equals the first, so
$$P(D \le t)=4P(B_1>0, B_t<0) =4P(B_1>0, B_1-B_t>B_1) \,.\quad (*) $$
Let $\gamma$ denote the standard Gaussian measure in ${\mathbb R}^2$.$\,$
Since $Y=B_1$ and  $X=\frac{B_1-B_t}{\sqrt{t-1}}$  are independent  standard Gaussian variables, we can rewrite $(*)$ in the form
\begin{eqnarray}
P(D \le t)&=&4P(Y>0, \sqrt{t-1} \cdot X>Y)  
 = \gamma\Bigl\{(x,y): 0<y<\sqrt{t-1} \cdot x \Bigr\} \\
&=&2\frac{\arctan(\sqrt{t-1})}{\pi} \,,
\end{eqnarray}
where we have used the rotation invariance of $\gamma$ in the last step.
