# Proof

(related to Proposition: Geometric Distribution)

In a Bernoulli experiment, let $$X$$ be the random variable having the realization $$k$$, if and only if we observe the event $$A$$ for the first time at the $$k$$-th repetition of the experiment. The joint event $$X=k$$ is given by

$B_k:=(\underbrace{\overline{A},\ldots,\overline{A}}_{k-1\text{ times}},A).$

Thus, $$B_k$$ is the event of observing $$k-1$$ times the complement event $$\overline{A}$$ and observing the event $$A$$ at the $$k$$-th repetition. By hypothesis, the probability of observing $$A$$ in each repetition is $$p:=p(A)$$ and all repetitions are mutually independent. Therefore, the probability mass function is given by

$p(B_k)=p(X = k)=\begin{cases} p(1-p)^{k-1}&\text{for }k=1,2,3,\ldots \\\\ 0&\text{else.}\end{cases}\quad\quad ( * )$

The geometric distribution (i.e. the probability distribution of the random variable $$X$$) is given by

$\begin{array}{rcll} p(X \le x)&=&0&\text{for }x < 1\\ p(X \le x)&=&\sum_{k=1}^{k=x}p(1-p)^{k-1}&\text{for }1\le x \end{array}$

We have the following cases:

### Case $$p=0$$

Then we have $$p(B_k)=0$$ for all $$k\ge 1$$.

### Case $$p=1$$

It follows from $$( * )$$ that $$p(B_1)=1$$, and $$p(B_k)=0$$ for all $$k\ge 2$$.

### Case $$0 < p < 1$$

Note that the probabilities $$p(B_k)$$ are all positive, i.e. all mutually exclusive events $$B_1,B_2,B_3,\ldots$$ can occur. On the other hand, it can happen that we will observe neither $$B_1$$, nor $$B_2$$, nor $$B_3$$, etc., for (in an extreme case) infinitely many times.

In this extreme case, the event

$\bigcup_{k=1}^\infty B_k$

has the probability following (among others) from geometric progression and the convergence behavior of the sequence $$(x^n)_{n\in\mathbb N}$$ for $$|x| < 1$$:

$\begin{array}{rcll} p(\bigcup_{k=1}^\infty B_k)&=&\sum_{k=1}^\infty p(B_k)&\text{mutually exclusive events}\\ &=&\lim_{n\to\infty}\sum_{k=1}^n p(B_k)&\text{definition of infinite series}\\ &=&\lim_{n\to\infty}\sum_{k=1}^n p(1-p)^{k-1}&\text{by }( * )\\ &=&\lim_{n\to\infty}p\sum_{k=1}^n (1-p)^{k-1}&\text{distributivity law for real numbers}\\ &=&\lim_{n\to\infty}p\sum_{k=0}^{n-1} (1-p)^{k}&\text{change of summing index}\\ &=&\lim_{n\to\infty}p\frac{1-(1-p)^n}{1-(1-p)}&\text{geometric progression}\\ &=&\lim_{n\to\infty}p\frac{1-(1-p)^n}{p}&\text{distributivity law for real numbers}\\ &=&\lim_{n\to\infty}1-(1-p)^n&\text{cancelling }p\\ &=&1-0&\text{convergence of }(x^n)\text{ for }|x| < 1\\ &=&1 \end{array}$

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### References

#### Bibliography

1. Bosch, Karl: "Elementare Einführung in die Wahrscheinlichkeitsrechnung", vieweg Studium, 1995, 6th Edition