What is distribution of lengths of gaps between occurrences of ones in Bernoulli process?

Question

Which distribution fits the following data? Data are generated by the process:

$X_t, \, t=1,2,3,\ldots,n$ is equal 1 with probability $p$ and 0 with probability $(1-p)$ for each $t$.

What is the distribution of gaps between occurrences of 1's ?

Example series: 00100110101010001 gives gaps 2,0,1,1,1,3.

How to fit that distribution if I dont know $p$? $p$ is going to be the only free parameter of that distribution, so is $\hat p = \frac{\#1}{n}$, where $\#1$ is the number of ones in the series an unbiased estimator of the free parameter in that distribution ?

Answer 1

The time you have to wait till the next one is a geometric variable $X\sim\mathcal{G}(p)$ with probability parameter $p$, i.e. $$ \mathbb{P}(X=k) = (1-p)^k p \quad k=0,1,2,\ldots $$

Fitting your distribution to the data presumably means estimating $p$ by $\hat p$ and using the pluggin distribution $\mathcal{G}(\hat p)$ for all purposes. If you do not want to run a Bayesian analysis (with prior $\pi(p)=1/\sqrt{p(1-p)}$), estimating $p$ from your proportion of $1$'s along the Bernoulli sequence is indeed an unbiased estimator.