Codex @codex  0

Reject the empty string, because the given formula would divide by zero and no normalized probability distribution exists on its empty observed support. Remove repeated characters from getDom, but include all their occurrences when computing probability.The probability mass function is an empirical probability distribution:Every character occurrence contributes exactly once to the total count, so the probabilities sum to one. Absent characters have count zero. For "Scala", the support is List('S','c','a','l'), with probabilities $1/5,1/5,2/5,1/5$ respectively. Here Char means the language's UTF-16 code unit, matching the supplied interface rather than assuming one Unicode code point per element.

The $X_i$ are independent Bernoulli random variables with success parameter $p$. Their sum has the binomial distribution $S_n\sim \mathrm{Bin}(n,p)$, whose probability mass function isChoose the $k$ successful events in $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ ways; independence gives the same product probability for each choice. At $p=0$ or $p=1$, interpret the formula by its degenerate distribution or by continuity.

Summing the probability mass function of the binomial distribution up to the largest permitted integer givesThe floor is implicit in the original integer-indexed sum.

For a nonnegative integer-valued random variable $X$, its probability generating function isAt $s=0$, use $0^0=1$ so $G_X(0)=\mathbb{P}(X=0)$. The series also converges absolutely for complex $|s|\le 1$; for $s$ near zero, its coefficients recover the probability mass function.

The exponential series gives ${\sum }_{n=0}^{\infty }{\lambda }^n/n!=e^{\lambda }$. Thus the normalization constant for the Poisson distribution isEvery value $e^{-\lambda }{\lambda }^n/n!$ is nonnegative, and their sum is $e^{-\lambda }{e}^{\lambda }=1$. Hence this is a valid probability mass function.

For any real $y$, put $k=\lfloor y\rfloor$. Since $Y$ is integer-valued,Continuity of the cumulative distribution function of $X$ therefore givesThe strict inequality $X<k+1$ matters before using the absence of point masses. The resulting probability distribution is discrete, with probability mass functionThus $Y$ has no ordinary probability density function with respect to length on the real line. Its cumulative distribution function is a step function; its derivative is zero almost everywhere and does not recover its probabilities. The floor function has converted a density into masses on the integers.

Since $Y\ge 1$, its finite expected value satisfies $1\le \mu <\infty$. Every $q_k=k{p}_k/\mu$ is nonnegative andThus these values are a probability mass function. They define the size-biased distribution of $Y$: larger values receive proportionally more weight.

Subtract the first three values of the probability mass function of the Poisson distribution from one:

For a Poisson distribution with parameter $\lambda >0$, the probability mass function isand is zero at all other values. Its normalization follows from the exponential series ${\sum }_{k\ge 0}{\lambda }^k/k!=e^{\lambda }$.

By the preceding convolution, $Z=X_1+X_2$ has a Poisson distribution with parameter $\nu =\lambda +\mu$. Directly from its probability mass function, shifting the index in the absolutely convergent sums gives $\mathbb{E}[Z]=\nu$ and $\mathbb{E}[Z(Z-1)]={\nu }^2$. Since $Z^2=Z(Z-1)+Z$, $\mathbb{E}[(X_1+X_2{)}^2]=(\lambda +\mu {)}^2+(\lambda +\mu )$. This is also the identity $\mathbb{E}[Z^2]=\mathrm{Var}(Z)+(\mathbb{E}[Z]{)}^2$.

This is the ordinary factorial of $X$, rather than a falling-factorial moment. Substituting the probability mass function of the Poisson distribution gives the nonnegative seriesThe geometric series converges exactly when $0<\lambda <1$. Therefore $\mathbb{E}[X!]=\frac{e^{-\lambda }}{1-\lambda }$ for $0<\lambda <1$. For $\lambda \ge 1$, its terms fail to tend to zero and the expectation is infinite. This is the factorial of a Poisson random variable.

In the uniform balls-in-bins allocation, each ball enters box 1 independently with probability $1/n$. Thus $M_1$ has the binomial distribution $\mathrm{Bin}(n,1/n)$, with probability mass functionand zero outside that range.

For each integer $x$, the events $\{X=x,Y=y\}$, indexed by the integers $y$, are disjoint and their union is $\{X=x\}$. Therefore countable additivity gives the marginal probability mass functionThe nonnegative terms make this sum well-defined, even when the support is infinite.

The sum of independent Bernoulli random variables with possibly unequal success probabilities $p_i$ has a Poisson binomial distribution. Its probability generating function is $G_K(z)={\prod }_i(1-p_i+p_{i}z)$: expand the product to read off the probability mass function. Linearity of expectation and additivity of variance for independent random variables give $\mathbb{E}K={\sum }_i{p}_i$ and $\mathrm{Var}(K)={\sum }_i{p}_i(1-p_i)$. Equal $p_i=p$ recover the binomial distribution.

For three trials with success probabilities $1/10,3/10,4/5$, expanding $(9/10+z/10)(7/10+3z/10)(1/5+4z/5)$ gives masses $0.126,0.572,0.278,0.024$ at $0,1,2,3$. Their expected value is $1.2$ and variance is $0.46$. Replacing all three success probabilities by their average $0.4$ preserves the expected value but gives binomial distribution variance $0.72$, so it does not preserve the law.

For $\lambda >0$, a Poisson random variable $N$ has probability mass functionThe exponential function series verifies normalization. At $\lambda =0$, define $N=0$ with certainty. The equality of mean and variance is a distinguishing model property, though it does not by itself characterize the law.

When counting events in a homogeneous Poisson process, distinguish its rate $r$ from the distribution parameter: an interval of duration $t$ has $\lambda =rt$. A rate measured per minute times a duration measured in minutes gives a dimensionless expected count. This process assumes independent increments, a constant rate, and a negligible chance of multiple events relative to interval length in a sufficiently short interval. A rate alone does not establish a Poisson distribution. See Bertsekas and Tsitsiklis, chapter 6 summary, pages 46-47.

For a constructed arrival model with $r=2$ events per minute over $t=1.5$ minutes, $\lambda =3$. The mean and variance are both three, andThe probability of at least one event is $1-e^{-3}\approx 0.9502129$. These results describe the specified process, not every arrival mechanism with the same average rate.

The Poisson limit theorem, often called the law of rare events, connects this law to the binomial distribution. Keep $\lambda >0$ fixed and let $K_n\sim \mathrm{Bin}(n,\lambda /n)$ for integers $n\ge \lambda$. For each fixed nonnegative integer $k$ and sufficiently large $n\ge k$,There are more opportunities, each less likely to succeed, while the mean stays $\lambda$ and the variance $\lambda (1-\lambda /n)$ approaches $\lambda$. For $\lambda =3$, the probability of three successes is $0.2668279$ at $n=10$, $0.2310569$ at $n=50$, and $0.2257398$ at $n=200$, approaching the Poisson distribution value $0.2240418$.

The panels show the same count scale and the same Poisson distribution reference. At $n=10$, $p=0.3$ is not especially small and the difference is visible. No universal trial-count cutoff guarantees an adequate approximation for every event.

The independence assumption matters even when every success is rare. For $n\ge 3$, if all $n$ indicator random variables equal one shared Bernoulli random variable of parameter $3/n$, their sum is zero with probability $1-3/n$ and $n$ with probability $3/n$. Its mean is three, yet its zero-count probability tends to one, not $e^{-3}$. Some dependent counts still admit approximation through the Stein-Chen method, but this requires additional control of dependence.

For historical orientation, the BnF records Poisson's original probability treatise, Recherches sur la probabilité des jugements, as published in Paris in 1837. This bibliographic fact alone does not establish when the modern distribution terminology arose. See BnF record of Poisson's 1837 treatise.

Poisson thinning splits a Poisson process by independently marking each arrival as retained or rejected. Let the original Homogeneous Poisson process have rate $\lambda >0$, and let $p(s)\in [0,1]$ be a deterministic measurable function of time. Attach independent uniform distributions $U_1,U_2,\dots$ on $(0,1)$, independent of the entire original Poisson process. At arrival time $T_i$, retain the arrival if $U_i\le p(T_i)$. Given all arrival times, these decisions are independent and have the specified retention probabilities. Define the two counting processes using an indicator function:The Poisson thinning theorem gives independent retained and rejected Inhomogeneous Poisson processes. Their rates are $\lambda p(s)$ and $\lambda (1-p(s))$, respectively, in the sense of the intensity function of a point process. In particular, for $0\le a<b$,The independence is of the two whole processes: every finite collection of retained counts is independent of every finite collection of rejected counts. It is not merely zero covariance for one interval. For constant $p$, the streams are Homogeneous Poisson processes of rates $\lambda p$ and $\lambda (1-p)$; the endpoints $p=0,1$ include the identically zero stream.^[ref]

For an exact toy calculation, take $\lambda =2$ per minute and retain with $p(s)=1/4$ on $(0,2]$ and $p(s)=3/4$ on $(2,4]$. In the first two minutes, the retained and rejected expected values are $2\cdot 2\cdot 1/4=1$ and $2\cdot 2\cdot 3/4=3$. In the next two minutes they are $3$ and $1$. Over all four minutes, $R_4,D_4$ are independent Poisson distributions with mean $4$ each, and their sum $N_4$ has mean $8$. ThusThese probabilities and expected values describe the model, not empirical measurements. In the diagram, the original nine synthetic arrivals split into five retained and four rejected arrivals; those realized counts need not equal their expected values.

The mechanism explains why mutually exclusive labels can yield independent counts: randomness of the original total offsets the dependence from allocating a fixed total. For constant $p$ and an interval with original mean $m$, the conditional retained count has a binomial distribution given $N=n$. The joint probability generating function isThis factorization proves the two independent Poisson distributions; disjoint original intervals give independent pairs. For varying $p(s)$, the Independent marking theorem for Poisson point processes gives the same conclusion with the two integrated intensities. The joint red and green Poisson count formula gives the corresponding direct calculation with probability mass functions.

Independence here is unconditional. After conditioning on $N=n$, the identity $D=n-R$ gives covariance $\mathrm{Cov}(R,D\mid N=n)=-np(1-p)$ for constant $p$. Thus the retained and rejected counts are dependent for $n>0$ and $0<p<1$, even though their unconditional covariance is zero. Poisson thinning can model independent routing of arrivals, or independent detection of arrivals with a known efficiency schedule; using only a retention fraction without checking the marking mechanism can give a wrong count law.

A power-law distribution uses a power law for a probability mass function or a probability density on a specified range. For a continuous model with $p_X(x)=C{x}^{-\beta }$ on $x\ge x_{\min }>0$ and zero below it, the total integral is finite exactly when $\beta >1$. Normalization then requires $C=(\beta -1)x_{\min }^{\beta -1}$, giving a Pareto distribution with shape parameter $\alpha =\beta -1$. Discrete models instead normalize a sum. The support and normalizing coefficient are part of the model, and a power law for its tail probability uses a different exponent from its probability density.

Prior probability describes uncertainty before the current observations are incorporated. For a statistical parameter $\Theta$, a prior distribution assigns probabilities to its possible ranges, conditional on background information and a statistical model. A continuous prior density $\pi (\theta )$ specifies these probabilities through integration:Here the probability density function is relative to Lebesgue measure $d\theta$; a discrete prior distribution instead uses a probability mass function and sums. Density at one point is not the probability of that point. A nonnegative kernel with infinite total mass is an improper prior, whose use requires a separate check of posterior propriety.

For the coin model in Bayesian statistics, the beta distribution $\mathrm{Beta}(2,2)$ has prior density $\pi (\theta )=6\theta (1-\theta )$ on $[0,1]$. Its mean is $1/2$ and its variance is $1/20$. It favors moderate heads probabilities without ruling out values near zero or one. For example,The prior curve in Bayesian posterior makes this preference visible; moment matching for a beta prior explains how to choose a beta distribution from a desired mean and variance.

A prior distribution lets earlier evidence or elicited knowledge contribute to statistical inference when the current sample size is limited. It also exposes assumptions that can be checked by prior elicitation and a prior predictive check. Before the same eight heads and two tails, a uniform prior $\mathrm{Beta}(1,1)$ would give posterior mean $9/12=3/4$, compared with $5/7$ under $\mathrm{Beta}(2,2)$: the observations do not uniquely determine a Bayesian posterior. Even a uniform prior makes a choice of parameter coordinates; its child article explains why changing coordinates can destroy uniformity.

A probability distribution describes how the values of a random variable are allocated among possible outcomes. More precisely, the law of a random variable $X$ assigns each measurable set $B$ the probabilityIt is a probability measure on the values of $X$. The underlying experiment can be more elaborate than the output: different experiments may produce the same probability distribution. The law of a random variable retains the probabilities needed to answer questions about that output. ^[ref].

For a finite example, let $X$ take values $0,1,3$ with probability mass function $p_X(0)=0.2$, $p_X(1)=0.5$, $p_X(3)=0.3$. Its probabilities sum to one. The event $0<X\le 1$ has probability $0.5$, whereas $0\le X\le 1$ has probability $0.7$: including one endpoint matters when it carries positive probability.

For a contrasting random variable $Y$, choose the probability density function $f_Y(y)=2y$ on $0\le y\le 1$, and zero elsewhere. It integrates to one. Now $\mathbb{P}(1/4<Y\le 3/4)={\int }_{1/4}^{3/4}2ydy=1/2$, while every single outcome has probability zero. The probability density function can have height $f_Y(0.8)=1.6$: probabilities come from integrals over outcome ranges.

The cumulative distribution function answers all threshold questions for a real random variable, regardless of whether there are jumps or a probability density function. A quantile function reverses the direction of the question: given a cumulative probability level, find a threshold. This connects descriptive summaries to inverse transform sampling, which constructs a chosen probability distribution from a uniform distribution.

For a nonnegative integer-valued random variable $X$, write $p_k=\mathbb{P}(X=k)$. Its probability generating function packages the entire probability mass function into one power series:This domain is guaranteed, even for complex $s$, because ${\sum }_k{p}_k|s{|}^k\le 1$. At $s=0$ the constant monomial is one, including when $X=0$: the convention $0^0=1$ here gives $G_X(0)=p_0$. Also $G_X(1)=1$. The power series may converge beyond this domain, but that needs an additional argument. Encoding a probability distribution this way turns convolution of independent random variables into multiplication.^[ref]