Sum of hitting times and hitting times of sum of Brownian Motion

Question

I have came across a question that asks to prove that for a brownian motion $B$, the first hitting time of $T_a=\inf\{t≥0 : B_t=a\}$ has a $1/2-$stable distribution, in that if we have $n$ independent copies of $T_a$, then: $$\frac{T^{(1)}_a+\ldots+T^{(n)}_a}{n^2}\sim T_a$$ in distribution. So far I tried the following, but I'm doubtful wether it is correct, as I couldn't find any reference for it though it seems to me as something should be easy to find if it was true:

1. If $0≤a≤b$ then $T_a+T_b\sim T_{a+b}$ because by Strong Markov property on $T_a≤T_b$ we have: \begin{equation}\begin{split}P(T_{a+b}≤t)=\int_{0}^t P(T_{a+b}≤t-s |T_a=s)P(T_a=ds)\\ =\int_{0}^t P(T_{b}≤t-s)P(T_a=ds)=P(T_a+T_b≤t)\end{split}\end{equation} where the last equality is just a general expression for the CDF of independent random variables.

2. It follows by induction that $T_{a_1}+\ldots+T_{a_n}\sim T_{a_1+\ldots+a_{n-1}}+T_{a_n}\sim T_{a_1+\ldots+a_n}$. This, coupled with the fact that $\frac{T_{na}}{n^2}\sim T_a$ (see Prove that two stopping times are equal in distribution (Brownian motion)) yields that if $T^{(1)}_a,\ldots,T_a^{(n)}$ are i.i.d. copies of $T_a$ then: $$\frac{T^{(1)}_a+\ldots+T_a^{(n)}}{n^2}\sim \frac{T_{na}}{n^2}\sim T_a$$

Answer 1

Let $\tau_a=\inf\{t:B_t=a\}$, $a>0$. We have $P(\tau_a\leq t)=P(\sup_{s\leq t}B_s\geq a)$. Let $\Phi,\phi$ be respectively the standard normal cdf and pdf. The reflection principle yields $P(\tau_a\leq t)=2P(B_t\geq a)$ so $$P(\tau_a\leq t)=2(1-\Phi(a/\sqrt{t}))\implies f_{\tau_a}(t)=\frac{a}{t^{3/2}}\phi(a/\sqrt{t})$$ This is a Lévy distribution with parameters $(0,a^2)$. We then have $E[e^{i\xi \tau_a}]=e^{-\sqrt{-2ia^2\xi}}$. Now if $(\tau_a^{(k)})_{k\leq n}$ are IID copies of $\tau_a$ we have for any $\xi \in \mathbb{R}$ $$E[e^{i\xi (\tau_a^{(1)}+...+\tau_a^{(n)})/n^2}]=(E[e^{i\xi \tau_a/n^2}])^n=e^{-n\sqrt{-2ia^2\xi/n^2}}=e^{-\sqrt{-2ia^2\xi}}=E[e^{i\xi \tau_a}]$$ and we conclude because the characteristic functions coincide.