The first passage time for Brownian motion $\tau_{a}=\inf\{t>0:B(t)=a\}$ is a random variable with density $$f_a(t)=\frac{|a|}{(2\pi\,t^3)^{1/2}}\exp(-a^2/(2t))$$ for $t>0$. Show that if $a,b>0$ and $\tau_a$ and $\tau_b$ are independent with densities $f_a$ and $f_b$ respectively, then $\tau_a+\tau_b=\tau_{a+b}$ where $\tau_{a+b}$ has density $f_{a+b}$.
Currently, I have $$P(\tau_{a+b}>t)=P(B(t)<a+b)=P(B(t_2)-B(t_1)+B(t_1)<a+b)$$ where $t_1+t_2=t$ and $B(t_2)-B(t_1)$ and $B(t_1)$ are independent, where $B(t_2)-B(t_1)$ reaches level $b$ and $B(t_1)$ reach level $a$. I am not certain how to use convolution here. Also, am I on the right track? Any comments or suggestions are welcome! Thanks in advanced.
Here is an idea. For a Brownian motion $\{B_t\}_{t\geq 0}$, the processes $\{C_t := B_{\tau_a+t}-B_{\tau_a}\}_{t\geq 0}$ and $\{B_t\}_{t\leq \tau_a}$ are independent, and $C_t$ is once again a Brownian motion, by the strong Markov property. Now, note that, for the same Brownian motion $B_t$, $$\tau_{a+b} = \tau_a + \inf \{t>0: C_t = b\}~.$$ You are done.
Remark: Now, just imagine that you are in an elementary probability course, where you just learned about densities and convolutions, and the professor asks you to prove that $f_{a+b} = f_a * f_b$, where $f_a$ is as defined above. Imagine how cumbersome it would be to do it, via all the Jacobian transformations, etc. But once you know the Brownian motion and its strong Markov property, you just have a two-line proof!