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A Bernoulli process is a sequence of independent and identically distributed random variables with a Bernoulli distribution: $\mathbb{P}(B_n=1)=p$ and $\mathbb{P}(B_n=0)=1-p$. It models repeated Bernoulli trials with a fixed success probability and no dependence between trials. The count $K_n={\sum }_{j=1}^n{B}_j$ has a binomial distribution, expected value $np$, and variance $np(1-p)$; the first success time has a geometric distribution on $\{1,2,\dots \}$. The observations $B_n$ and the accumulated count $K_n$ are different stochastic processes: the observations are independent, while consecutive counts share all earlier observations. ^[ref]

Use the zero-based geometric distribution, with $\mathbb{P}(N=k)=p(1-p{)}^k$ for $k\ge 0$ and $0<p<1$. After binomial thinning with retention probability $0\le r\le 1$, the retained count is again zero-based geometrically distributed, with the displayed success probability. Indeed, writing $q=1-p$, its probability generating function isFor $r=0$ this includes the degenerate success parameter $p^{\prime }=1$ and retained count zero.

For a positive-integer geometric distribution $P(N=j)=p{q}^{j-1}$ with $q=1-p$, the capped time has masses $p{q}^{j-1}$ for $1\le j<k$ and $q^{k-1}$ at $k$. The last mass includes all later arrivals, not only $N=k$. Its probability generating function and expected value areThe expectation follows either by differentiating the PGF or summing the tail probabilities up to $k$.

The number of insurance claims in a specified exposure period. Its probability generating function combines with the moment-generating function of an independent claim size through the random-sum transform identity. A Poisson distribution, geometric distribution, or negative binomial distribution provides a possible count model, with different dispersion and zero-claim probabilities.

Take an independent geometric distribution count with $\mathbb{P}(N=k)=p(1-p{)}^k$, $k\ge 0$, and sum its independent positive claims. Its moment-generating function is the displayed composition wherever finite. With $q=1-p$ and severity moments $m_j$, the third cumulant is $(q/p)m_3+3(q/p{)}^2{m}_1{m}_2+2(q/p{)}^3{m}_1^3$, which is strictly positive for nonzero positive claims and $0<p<1$. The law has an atom of mass $p$ at zero. Its positive-count convention would instead add a factor $M_X(t)$ to the numerator, as in the geometric-sum moment-generating function.

Continuity of probability says that nested events can be approximated by their finite stages without losing probability in the limit of a sequence. In a probability space $(\Omega ,\mathcal{F},\mathbb{P})$, let $A_n,B_n\in \mathcal{F}$. ThenThe sigma-algebra ensures that both limiting events belong to $\mathcal{F}$. These are continuity from below of a measure and continuity from above of a measure. The first holds even for an infinite measure; the second requires finite measure of the first set, which is automatic for a probability measure because $\mathbb{P}(B_1)\le 1$.^[ref]

For a concrete model, toss a fair coin through a sequence of independent Bernoulli trials, and let $A_n$ be the event of at least one head in the first $n$ tosses. Its complement $B_n=A_n^c$ is the event of $n$ initial tails. On the sample space of infinite head-or-tail sequences,Every finite waiting horizon has positive tail probability, yet the event of waiting forever has probability zero. That intersection is not empty: it contains the all-tails sequence. Thus a zero-probability event need not be impossible as an outcome, and eventual success almost surely does not supply a deterministic deadline. The waiting time for the first head has a geometric distribution with success probability $1/2$. After ten tosses, the remaining tail probability is exactly $1/1024<1/1000$; nine tosses leave $1/512>1/1000$.

A proof outline exposes what makes countable additivity stronger than finite additivity of a set function. For increasing events, put $D_1=A_1$ and $D_n=A_n\setminus A_{n-1}$ for $n\ge 2$. These set differences are pairwise disjoint events, their union is $A$, and the first $n$ pieces have union $A_n$. ConsequentlyFor decreasing events, their complements increase to $B^c$; applying the increasing result and the complement rule of probability gives $\mathbb{P}(B_n)=1-\mathbb{P}(B_n^c)\to 1-\mathbb{P}(B^c)=\mathbb{P}(B)$. The finite-total-mass condition is essential for this subtraction. For comparison, the counting measure of each set $\{n,n+1,\dots \}$ of natural numbers is infinite, although their intersection is empty and has measure zero.

Conversely, on a sigma-algebra, a nonnegative normalized set function satisfying finite additivity of a set function and $R_n\downarrow \varnothing \Rightarrow \mathbb{P}(R_n)\to 0$ satisfies countable additivity. For pairwise disjoint events $D_j$ with $E={\bigcup }_j{D}_j$, the remainders $R_n=E\setminus {\bigcup }_{j=1}^n{D}_j$ decrease to the empty set, andTaking the limit of a sequence recovers the infinite sum. This equivalence explains why continuity of probability is part of the foundations rather than a separate assumption about smooth curves.

One useful application controls infinitely many possible errors. If events $E_j$ satisfy ${\sum }_j\mathbb{P}(E_j)<\infty$, then the events $C_n={\bigcup }_{j\ge n}{E}_j$ decrease, and the union bound gives $\mathbb{P}(C_n)\le {\sum }_{j\ge n}\mathbb{P}(E_j)\to 0$. Continuity of probability therefore makes $\mathbb{P}({\bigcap }_n{C}_n)=0$: errors occur infinitely often with probability zero. This is the Borel-Cantelli first lemma, and it requires no independence. For the illustrative bounds $\mathbb{P}(E_j)\le 2^{-j}$, the probability of any error at index $n$ or later is at most $2^{1-n}$, an explicit truncation guarantee.

The word continuity here concerns nested sets, not the absence of atoms in a probability distribution. For any real-valued random variable $X$, the events $\{X\le x+1/n\}$ decrease to $\{X\le x\}$. Hence its cumulative distribution function is right-continuous even when it has jumps, as for a geometric distribution. The nesting hypothesis matters: an arbitrary sequence of events need not admit either displayed interchange of limit and probability.

Consider a stipulated repeated-attempt model, with independent Bernoulli trials of success probability $p\in (0,1]$. An unsuccessful attempt costs three units of work, and a successful attempt costs one. Let $B_i$ be its success indicator function, $X_i=3-2B_i$ its cost, and $N=\inf \{i\ge 1:B_i=1\}$ the first success. Observing $X_i$ reveals $B_i$, so $N$ is a stopping time for the costs' natural filtration. The geometric distribution gives $\mathbb{P}(N\ge i)=(1-p{)}^{i-1}$ and $\mathbb{E}[N]=1/p$, including the deterministic case $p=1$. ThusAt $p=1/4$, the average is four attempts and ten units of work. An independent check uses the pathwise identity $S_N=3(N-1)+1$: there are $N-1$ failures and one final success. No success probabilities or costs in this example are measured data.

Here $N$ is not independent of the observed costs. In particular, conditioning on $N=n$ forces every preceding cost to be three and the final cost to be one, so $\mathbb{E}[S_N\mid N=n]=3n-2$, generally different from $n\mathbb{E}[X_1]=(5/2)n$ at $p=1/4$. Nevertheless the unconditional identity holds. This distinction explains why conditioning on the total count as though the original cost law remained unchanged is a faulty proof.

For uniform independent draws of $n$ coupon types, the waiting times with $r$ unseen types remaining are independent positive-support geometric distributions of parameter $r/n$. Summing their means gives $n{\sum }_{r=1}^{n}1/r$, and summing their variances gives the displayed formula. Independence follows from fresh draws after each discovery and the fact that the next stage's conditional distribution depends only on the count of remaining types.

The same argument works for any integer base $b\ge 2$. Suppose $N$ has geometric distribution $\mathbb{P}(N=n)=(1-q)q^n$, $n\ge 0$, with $0<q<1$. Its Euclidean division $N=bY+X$ gives $0\le X<b$ and $Y\ge 0$. Their joint probability mass function isThe factors are normalized probability mass functions, so the final digit $X$ and quotient $Y$ are independent. The quotient again has geometric distribution, with success probability $1-q^b$. If $N$ is the integer part of an exponential random variable of rate $\lambda$, take $q=e^{-\lambda }$.

A discrete probability distribution is a probability distribution concentrated on a finite or countable set of possible values $S$: a random variable $X$ with this law satisfies $\mathbb{P}(X\in S)=1$. Its probability mass function assigns a probability to each value:Thus the probability of an event is obtained by adding masses, rather than integrating a probability density function. The possible values need not be equally spaced, or even integers. When the relevant moments are finite, its expected value and variance are

For a constructed example, assign values $0,2,5$ masses $0.5,0.3,0.2$. They sum to one; $\mathbb{P}(X\ge 2)=0.3+0.2=0.5$, $\mathbb{E}X=0(0.5)+2(0.3)+5(0.2)=1.6$, and $\mathbb{E}[X^2]=4(0.3)+25(0.2)=6.2$. Hence the variance is $6.2-1.6^2=3.64$. A mean can lie between possible values: no outcome here equals $1.6$.

The useful organizing question is what an experiment records. One yes/no indicator random variable gives a Bernoulli distribution; a success count in a fixed number of independent attempts gives a binomial distribution; the attempt number of the first success gives a geometric distribution. A Poisson distribution describes an event count under different assumptions. Naming the quantity first helps prevent selecting a familiar formula for the wrong experiment.

A time-homogeneous Discrete-time Markov chain is a sequence of random variables $Y_0,Y_1,\dots$ indexed by nonnegative integers, with the displayed conditional probabilities. Here ${\mathcal{F}}_n$ is its natural filtration and $P$ is a row-stochastic transition matrix. The Markov property removes dependence on earlier observations once $Y_n$ is specified; the time-homogeneous Markov property makes $P$ independent of $n$.

Consider states $S$ (sunny) and $R$ (rainy), with the constructed transition matrixRows give today's state and columns tomorrow's, in order $S,R$. These numbers define an illustrative model, with one step interpreted as one day; they are not estimates of real weather. Let $q_n=\mathbb{P}(Y_n=S)$. The law of total probability gives the recurrence relationSubtracting $2/3$ makes the deviation multiply by $0.4$ at each step. ThusStarting sunny gives $q_0=1$, $q_1=0.8$, $q_2=0.72$ and $q_3=0.688$; starting rainy gives $q_0=0$ and $q_2=0.56$. The two predictions approach the same stationary distribution while retaining a shrinking trace of the initial probability distribution.

An expected hitting time answers a different question. Starting sunny, let $T_R=\inf \{n\ge 0:Y_n=R\}$. Every day before the first rain remains sunny, soThis geometric distribution has finite expected value. A first-step calculation gives $h_S=1+0.8h_S+0.2h_R$, with $h_R=0$, hence $h_S=5$ days. The same result follows from the geometric series ${\mathbb{E}}_S{T}_R={\sum }_{n\ge 0}0.8^n=5$. This is a mean waiting time, not a prediction that rain first occurs exactly five days later.

A probability row vector $\pi$ is a stationary distribution when $\pi P=\pi$. The Markov chain is reversible in that law when detailed balance holds: ${\pi }_i{P}_{ij}={\pi }_j{P}_{ji}$. Multiplying those equations along a finite path gives equal forward and reversed stationary path probabilities.

Precision comparisons using Fisher information must specify what one observation records. A Bernoulli distribution with success probability $\theta$ gives information $I_{\mathrm{Bern}}=1/[\theta (1-\theta )]$ per trial. A geometric distribution on $1,2,\dots$ records the number of trials until the first success. Its log-likelihood for one waiting time $X$ is $\log \theta +(X-1)\log (1-\theta )$, soAt equal numbers of observations, geometric waiting times therefore provide $1/\theta$ times as much information, and the asymptotic variance of their maximum-likelihood estimator is $\theta$ times the Bernoulli variance. But each waiting time uses an expected $1/\theta$ trials. Information per expected underlying trial is consequently $\theta I_{\mathrm{geom}}=I_{\mathrm{Bern}}$. This separates an observation count from the cost of collecting the observations.

Suppose a contest uses $N_0$ trials initially and ends there with probability $1-\alpha$. Conditional on an initial tie, independent further rounds each use $\ell$ trials and resolve the contest with a fixed probability $0<r\le 1$. The number $J$ of further rounds, conditional on that tie, has a geometric distribution: $\mathrm{Pr}(J=j\mid \text{tie})=(1-r{)}^{j-1}r$ for $j\ge 1$. Thus the total number $N$ of trials has an atom $1-\alpha$ at $N_0$ and the displayed probabilities thereafter. The geometric series sums to $\alpha$, so termination has probability one, and the expected value is $\mathbb{E}[N]=N_0+\ell \alpha /r$. The displayed distribution requires independence of the future rounds from the initial tie.

Independently retain each event of a zero-based geometric distribution count with probability $\alpha$. Conditional on the original count, the retained count is binomial, so its probability generating function is $p/[1-(1-p)(1-\alpha +\alpha z)]$. This is again a zero-based geometric distribution, with success parameter $p_{\alpha }$ and mean $\alpha (1-p)/p$. In excess of loss reinsurance, $\alpha$ is the probability that a claim exceeds the retention.

Consider successive independent Bernoulli trials with the same success probability $0<p\le 1$. A geometric distribution models waiting for their first success. Two conventions coexist, so always say what is counted. Let $T$ count all trials up to and including the successful trial, and let $F=T-1$ count preceding failures. Their probability mass functions areThe factors describe failures followed by one success; independence permits multiplication. Summing the geometric series verifies that each law has total mass one. Their expected values differ by one, while shifting by a constant leaves the variance unchanged:The positive convention and these moments are also given by Penn State STAT 414.^[ref]

For a constructed retry mechanism with independent attempts and $p=0.3$, success on attempt four has probability $\mathbb{P}(T=4)=0.7^3(0.3)=0.1029$. More than four attempts are needed precisely when the first four fail: $\mathbb{P}(T>4)=0.7^4=0.2401$. The expected values are $\mathbb{E}T=10/3$ and $\mathbb{E}F=7/3$, and the common variance is $70/9\approx 7.7778$. The large spread explains why the average waiting count is not a deadline.

The two probability mass functions have identical heights with different labels: $T=4$ and $F=3$ describe the same event. For $0<p<1$, consecutive masses have ratio $1-p$, explaining the name geometric distribution. The support of a probability distribution is the positive integers for $T$ and the nonnegative integers for $F$. At $p=1$, it collapses to $T=1$ and $F=0$, both with variance zero; interpret an empty product, including $(1-p{)}^0$, as one. These counts concern attempts; converting them into elapsed time requires specifying the time spent per attempt.

The survival function makes the counting convention especially visible. Put $q=1-p$. For nonnegative integers $n$, and for $k\ge 1$,The event $T>n$ requires $n$ failures, whereas $T\ge n$ requires only $n-1$ when $n\ge 1$. For the constructed $p=0.3$ mechanism, $\mathbb{P}(T\ge 4)=0.343$ differs from $\mathbb{P}(T>4)=0.2401$. The geometric success deadline uses the latter tail to control the chance of still waiting.

The striking memorylessness of the geometric distribution says that, for nonnegative integers $m,r$ with $\mathbb{P}(T>m)>0$,After five failures, the conditional probability of requiring more than two additional attempts is still $0.7^2=0.49$. The analogous zero-based identity uses $\mathbb{P}(F\ge m+r\mid F\ge m)=\mathbb{P}(F\ge r)$; keeping the inequalities straight avoids an off-by-one error. This property follows from fresh independent trials with unchanged $p$; learning or changing conditions can destroy it. At $p=0$, success never occurs, so there is no proper finite waiting-count law.

Let the increments define an upward skip-free random walk, and suppose $\varphi (\lambda )=\mathbb{E}{e}^{\lambda X_1}$ is finite for all $\lambda \ge 0$. If ${\lambda }_0>0$ satisfies $\varphi ({\lambda }_0)=1$ and ${\varphi }^{\prime }({\lambda }_0)>0$, the exponential tilting at ${\lambda }_0$ gives integrable increments of positive mean. The strong law of large numbers makes every nonnegative integer level attainable under this tilted law. Upward jumps are at most one, so the value at a first level hit is exactly that level. The stopped likelihood ratio under exponential tilting therefore gives hitting probability $e^{-{\lambda }_{0}k}$ for integer $k\ge 0$. Subtracting consecutive tail probabilities yields the displayed geometric distribution on $j=0,1,\dots$. For real thresholds use $\lceil k\rceil$ in place of $k$.

For independent positive-support geometric random variables with success probabilities $p_i$, all must exceed $k$ for their minimum to exceed $k$. The resulting survival function is ${\prod }_i(1-p_i{)}^k$. This is the positive-support geometric distribution with the displayed success probability. Independence is needed for multiplying the survival probabilities.

Integrating the Poisson distribution probability generating function over the exponential distribution of the epoch length givesThus $Y_j$ has a geometric distribution on $\{0,1,\dots \}$, with expectation $\theta /(j-1)$ and variance $\theta /(j-1)+{\theta }^2/(j-1{)}^2$. The support convention matters: the count includes zero mutations.

A geometric series adds the terms of a geometric progression:Its ordinary infinite sum is the limit of a sequence of partial sums $S_N={\sum }_{j=0}^{N-1}a{r}^j$, rather than a new operation that adds infinitely many terms at once. For $a,r$ that are real numbers or complex numbers, if $|r|<1$, the series is absolutely convergent andWhen $a\ne 0$, this condition on the modulus of $r$ is also necessary: for $|r|\ge 1$ the terms fail to tend to zero. The all-zero series $a=0$ converges for every $r$. The finite geometric series identity and $r^N\to 0$ give the result with almost no extra algebra. ^[ref]

With $a=1$ and $r=1/2$, the partial sums approach $2$ from below. Changing the common ratio to $-1/2$ makes the partial sums approach $2/3$ from alternating sides: their increments still shrink. At $r=-1$, those increments never shrink, and the partial sums remain $0,1,0,1,\dots$. Substituting $r=-1$ into $1/(1-r)$ does not supply the ordinary sum of that divergent series.

The formula is useful because many apparently different problems share a fixed multiplier. A repeating decimal such as $0.\overline{27}$ isA geometric distribution counting failures before the first success normalizes by the same identity: for $0<q<1$, the probabilities $(1-q)q^n$ on $n=0,1,\dots$ add to one. The comparison test for series uses a bound by a decaying geometric progression to establish absolute convergence without an exact formula for the series being studied. The iterated contraction gives a summable orbit result uses precisely this idea for repeated applications of a map.

The remainder of a geometric series gives more than an eventual limit: it states exactly how much an approximation omits. A common ratio close to one still gives a convergent series, but can require many terms for a useful approximation.

How many attempts give at least a chosen success probability $0<\alpha <1$? For a positive-integer geometric distribution with $0<p<1$, completion by attempt $N$ means $T\le N$, soThe logarithm $\log q$ is negative, which reverses the inequality when dividing. The least integer meeting the target is therefore obtained with the ceiling function:For our constructed independent retry mechanism with $p=0.3$ and $\alpha =0.95$,To verify minimality directly, eight attempts give $\mathbb{P}(T\le 8)=1-0.7^8=0.94235199<0.95$, whereas nine give $\mathbb{P}(T\le 9)=1-0.7^9=0.959646393\ge 0.95$. Under the zero-based convention, the same completion event is $F\le 8$, with failure to complete given by $F\ge 9$; saying $F\le 9$ would allow ten attempts. A 95% deadline controls a tail probability, while the expected value $10/3$ describes an average.

At $p=1$ success occurs on the first attempt, so $N_{\alpha }=1$; the logarithm formula is inapplicable. At $p=0$ no finite deadline achieves a positive target. For $0<p<1$, no finite deadline guarantees success with probability one because $q^N>0$. A retry cap at $N$ produces the capped geometric waiting time $\min (T,N)$ as the number of attempts actually spent; reaching the cap can mean either success on the last allowed attempt or failure of all $N$ attempts.