I truly don't know how to approach the following problem
Consider a sequence of events identically distributed and independent with probability of success $p$. Let $S_i$ be the success of the i-th event. Denote $X_1$ the time at which the first success happened and $X_1+X_2$ the time when the second success happened. Express the events $\{X_1=k, X_2=\ell\}$ for $ k,\ell \in \mathbb{N}$ in terms of $S_i$ and prove that $X_1$ and $X_2$ are independent.
I think that $S_i$ is distributed as a Binomial(N,p), where $N$ is the number of times event was done, and $X_1=\sum_{i=1}^{N} S_1$. Is this correct? Any Hint?
Thanks!
By what I understand from the exercise, $S_i$ is actually a Bernoulli random variable, so $P(S_i = 1)=p$ and $P(S_i = 0)=1-p$, where $1$ is success and $0$ is failure. So, $\sum^n_{i=1} S_i \sim Binomial(n,p)$, but not $S_i$.
Now, for the $X$'s. Note that $X_1$ is the smallest index such that $S_i$ is a success.
$$X_1(w)= \inf\{n : \sum^n_{i=1}S_i(w) \geq 1 \}$$
$X_2$ is similar to $X_1$, and it represents the number of trials until the second success. $$ X_2(w)= \inf\{n : \sum^n_{i=X_1(w)+1}S_i(w) \geq 1 \} $$
The distribution of $X_1 + X_2$ is a Negative binomial, which is given by $$ P(X_1+X_2 = k)= {k+2-1\choose k}p ^2(1-p)^k $$
Since the Negative binomial is identical to the sum of i.i.d Geometric random variables, then $X_1$ and $X_2$ are independent.
One can also show the independence by noting that $$\{X_1=k,X_2=l\} = \{S_1=0,...,S_k=1,S_{k+1}=0,...,S_{k+l}=1\} $$
Since $S$ are independent, it is clear that so are $X_1$ and $X_2$.