If $(X_i)$ are independent and $\tau_i=\inf\{k >\tau_{i-1}\mid X_k=1\}$ why $(\tau_i-\tau_{i-1})$ are independents Let $(X_i)$ independent s.t. $\mathbb P=\{X_i=0\}=p$ and $\mathbb P\{X_i=1\}=1-p$. Let $\tau_0=0$ and set $$\tau_i=\inf\{k>\tau_{i-1}\mid X_k=1\}.$$ Set $$T_i=\tau_{i}-\tau_{i-1}.$$ 
I want to prove that $(T_i)_i$ are independents.

proposition I know that if $U_1,...,U_n$ and $V_1,...,V_m$ are independents, then so are $f(U_1,...,U_n)$ and $g(V_1,...,V_m)$ or equivalently that $\sigma (U_1,...,U_n)$ and $\sigma (V_1,...,V_m)$ are independent. 

The fact that the $(T_i)$ are independent looks quite natural, but I would like to avoid to prove that $$\mathbb P\{T_1=k_1,T_2=k_2,...,T_{m}=k_m\}=\mathbb P\{T_1=k_1\}...\mathbb P\{T_m=k_m\}.$$
Q1) Is there a way to use my proposition ? Something as $T_1$ is $\sigma (X_1,...,X_{\tau_1})$ measurable, $T_2$ is $\sigma (X_{\tau_1+1},...,X_{\tau_2})$ measurable... and conclude ?
Q2) Does $\sigma (X_1,...,X_{\tau_k})$ really make sense ? I don't have the impression.
Q3) Is there a way to prove that the $T_i$ are identically distributed ? It looks very natural, but a bit long to prove. Maybe there is a trick ?
Q4) I know that it's not enough to prove that a collection $\{A_i\}_i$ is pairwise independent to prove that the collection is independent, but in this case, wouldn't it be sufficient to prove that $T_i,T_j$ are independent to prove that any finite vector of $T_i$ are independent ? 
 A: First, you need to define a $\sigma$-field associated with $\tau_i$. Let $\mathcal{F}_n:=\sigma(X_1,\ldots,X_n)$. Then we define
$$
\mathcal{F}_{\tau_i}:=\{A\in \mathcal{F}_{\infty}:A\cap \{\tau_i=n\}\in \mathcal{F}_n \text{ for all }n\}.
$$
Note that $\mathcal{F}_{\tau_i}$ is a $\sigma$-field and since $\tau_{i-1}\le \tau_i$, one has $\mathcal{F}_{\tau_{i-1}}\subset \mathcal{F}_{\tau_i}$.
Let $f_i(x):=1\{T_i=x\}$ with $\tau_0\equiv 0$. Then
\begin{align}
\mathsf{E}\left[\prod_{i=1}^n f_i(x_i)\right]&=\mathsf{E}\left[\prod_{i=1}^{n-1} f_i(x_i)\mathsf{E}[f_n(x_n)\mid \mathcal{F}_{\tau_{n-1}}]\right]=\mathsf{E}\left[\prod_{i=1}^{n-1} f_i(x_i)\right]\mathsf{E}f_1(x_n) \\
&=\prod_{i=1}^n \mathsf{E}f_1(x_i)=\prod_{i=1}^n p^{x_i-1}(1-p).
\end{align}
It remains to justify the second equality. Let $A\in \mathcal{F}_{\tau_{i-1}}$. Then
\begin{align}
\mathsf{E}[f_i(x);A\cap \{\tau_{i-1}<\infty\}]&=\sum_{m\ge 1}\mathsf{E}[f_i(x);A\cap \{\tau_{i-1}=m\}] \\
&=\sum_{m\ge 1}\mathsf{E}[\mathsf{E}[1\{X_{m+1}=0,X_{m+2}=0,\ldots,X_{m+x}=1\}\mid \mathcal{F}_m]; \\
&\qquad\qquad A\cap \{\tau_{i-1}=m\}] \\[10pt]
&=\sum_{m\ge 1}\mathsf{E}[\mathsf{E}[f_1(x)];A\cap \{\tau_{i-1}=m\}] \\
&=\mathsf{E}[\mathsf{E}[f_1(x)];A\cap \{\tau_{i-1}<\infty\}].
\end{align}
The last equation shows that $\mathsf{E}[f_i(x)\mid \mathcal{F}_{\tau_{i-1}}]=\mathsf{E}f_1(x)$ a.s. on $\{\tau_{i-1}<\infty\}$. However, it is easy to see that $\mathsf{P}(\tau_i<\infty)=1$ for any $i\ge 1$. Therefore, $\mathsf{E}f_i(x)=\mathsf{E}f_1(x)=p^{x-1}(1-p)$.
