Hitting time and its distribution ÷I'm reading an italian book about casual process (Probabilità e modelli aleatori of Enzo Orsingher). At pag 105 there's the probability of the stopping time $T_\beta$.
$$P\{T_\beta \leq t\}=P\{\max_{0\leq s\leq t} B(s) \geq \beta\}=$$

$$=2P\{B(t) \geq \beta\} = \sqrt{\frac{2}{\pi}} \int_{\frac{\beta}{\sqrt{t}}}^{\infty} \exp\{-\frac{w^2}{2}\}dw$$
Because the book used the last result :
$$P\{T_\beta \leq t\}=P\{\max_{0\leq s\leq t} B(s) \geq \beta\}= 2P\{B(t) \geq \beta\}$$

Now I don't undersatnd why $T_\beta$ has the inverse gaussian density
$$f_{\beta}(t)= \frac{|\beta|}{\sqrt{2\pi t^3}}\exp\left\{-\frac{\beta^2}{2t}\right\}$$
for $t &gt;0$.
Where did it find?
$P(B(t) &gt; \beta) = P(B(t) &gt; \beta | T_\beta &gt; t ) \cdot P(T_\beta &gt; t ) 
+ P(B(t) &gt; \beta | T_\beta \leq t ) \cdot P(T_\beta \leq t )  \\
= 0 + \frac{1}{2} P(T_\beta \leq t ) \\
\Longrightarrow P(T_\beta \leq t ) = 2  P(B(t) &gt; \beta) $.

$ P(B(t) &gt; \beta | T_\beta \leq t ) = 1/2$ because the Brownian motion is symmetric conditioned on the hitting point (independent increase). 

So, $P(T_\beta \leq t ) = 2  P(B(t) &gt; \beta) = 2 - 2 P(B(t) \leq \beta) \\
= 2 - 2 P( \frac{ B(t)}{ \sqrt{t} } \leq \frac{ \beta }{ \sqrt{t} } ) 
= 2 - 2 \Phi (\frac{ \beta }{ \sqrt{t} } ) $

where $\Phi$ is the CDF of standard Gaussian distribution. 
So $f_{T_\beta} (t) = \frac{d}{d t}( 2 - 2 \Phi (\frac{ \beta }{ \sqrt{t} } ) ) $.
$P(B(t) &gt; \beta) = P(B(t) &gt; \beta | T_\beta &gt; t ) \cdot P(T_\beta &gt; t ) 
+ P(B(t) &gt; \beta | T_\beta \leq t ) \cdot P(T_\beta \leq t )  \\
= 0 + \frac{1}{2} P(T_\beta \leq t ) \\
\Longrightarrow P(T_\beta \leq t ) = 2  P(B(t) &gt; \beta) $.

$ P(B(t) &gt; \beta | T_\beta \leq t ) = 1/2$ because the Brownian motion is symmetric conditioned on the hitting point (independent increase). 

So, $P(T_\beta \leq t ) = 2  P(B(t) &gt; \beta) = 2 - 2 P(B(t) \leq \beta) \\
= 2 - 2 P( \frac{ B(t)}{ \sqrt{t} } \leq \frac{ \beta }{ \sqrt{t} } ) 
= 2 - 2 \Phi (\frac{ \beta }{ \sqrt{t} } ) $

where $\Phi$ is the CDF of standard Gaussian distribution. 
So $f_{T_\beta} (t) = \frac{d}{d t}( 2 - 2 \Phi (\frac{ \beta }{ \sqrt{t} } ) ) $.