Elapsed times between successes in a Bernoulli sequence are independent and geometrically distributed

Question

3.21(a) mentioned in the solution is just family (Yi) is independent iff $P(\omega) = \prod \rho_i(\omega_i)$.

I am really new to probability theory so I would like to confirm a few things first:

1. This is discrete time space, so elapsed time Li between i-1st and ith success is actually the number of failures taken for the ith success (from the i-1st).

2. In the solution, $n_k$ and $T_k$ are both the total number of trials (or time points) for all k successes to happen.

3. $X_{n_k} = 1 $ for all $1\leq k\leq m$ means out of all m successes each $n_k$the time point is a point of success.

I would also like some hints on part b. To show $(X_n)_{n\geq 1}$ is a Bernoulli sequence, do I just show each Xn is Bernoulli with parameter p?

In this case, I just have to show iid and $P(X_n=1)=p$?

Answer 1

For (a): For each positive integer $k$ and nonnegative integer $n$ we have $$ \mathbb P(L_k>n) = \sum_{m=n+1}^\infty (1-p)^mp = (1-p)^{n+1}, $$ hence $L_k$ has geometric distribution with parameter $p$. Moreover, $L_k$ is $\sigma(X_{T_k}, X_{T_k-1},\ldots,X_{T_k+1})$-measurable, and so the $L_k$ are independent.

For (b): We have \begin{align} \mathbb P(X_n = 1) &= \mathbb E\left[\sum_{k=1}^\infty \mathsf 1_{\{n\}}(T_k)\right]\\ &= \sum_{k=1}^\infty \mathbb E[\mathsf 1_{\{n\}}(T_k)]\\ &= \sum_{k=1}^\infty \mathbb P(T_k=n)\\ &= \sum_{k=1}^\infty \mathbb P\left(\sum_{j=1}^k L_j+k = n \right)\\ &= \sum_{k=1}^n \mathbb P\left(\sum_{j=1}^k L_j = n-k \right)\\ \end{align} Now, $\mathbb P(L_1=k) = (1-p)^kp = \binom{1+k-1}k p^1(1-p)^k$. Suppose $\mathbb P\left(\sum_{j=1}^n L_j = k \right) = \binom{n+k-1}k p^n(1-p)^k$ for some positive integer $n$. Then \begin{align} \mathbb P\left(\sum_{j=1}^{n+1}L_j = k\right) &= \sum_{i=0}^k \mathbb P\left(\sum_{j=1}^n L_j = i\right)\mathbb P(L_{n+1}=k-i)\\ &=\sum_{i=0}^k \binom{n+i-1}i p^n(1-p)^i(1-p)^{k-i}p\\ &=\binom{n+k}k p^{n+1}(1-p)^k, \end{align} so by induction we conclude that $\sum_{j=1}^n L_j$ has negative binomial distribution with parameters $n$ and $p$. It follows that \begin{align} \mathbb P(X_n=1) &= \sum_{k=1}^n\binom{k+(n-k)-1}{n-k}p^k(1-p)^{n-k}\\ &= \sum_{k=1}^n\binom{n-1}{n-k}p^k(1-p)^{n-k}\\ &= p, \end{align} so that $\{X_n\}$ is a Bernoulli process with parameter $p$.