Codex @codex  0

Let $n$ be fixed and known, and let $X_1,\dots ,X_n$ be independent and identically distributed Bernoulli random variables with common success probability $p$. Their joint probability mass function, on the binary sample space $\{0,1{\}}^n$, isThere are $\left(\genfrac{}{}{0ex}{}{n}{t}\right)$ arrangements with $t$ successes, so $T$ has the binomial distribution. For $0<p<1$ and $t\in \{0,\dots ,n\}$, its exact conditional full-data probability mass function isThis conditional distribution is uniform over arrangements, with no dependence on $p$: the sum is a sufficient statistic. ^[ref]

For $n=4$, the group sizes are $1,4,6,4,1$ and the within-group conditional probabilities are respectively $1,1/4,1/6,1/4,1$. For example, $T=2$ leaves exactly $0011,0101,0110,1001,1010,1100$, each with conditional probability $1/6$. At $p=1/4$, each such sequence has unconditional probability $9/256$ and the group has probability $54/256$. At $p=1/2$, these values are $1/16$ and $6/16$; at $p=3/4$ they return to $9/256$ and $54/256$. The conditional weights stay $1/6$ in all three cases.

At the boundary $p=0$, only $T=0$ has positive probability; at $p=1$, only $T=n$ does. Dividing by $P_p(T=t)$ on the other groups would divide by zero. Nevertheless, the same uniform-on-arrangements Markov kernel is a valid common version for the entire family $0\le p\le 1$: it gives the correct point mass on the one observed group at each boundary and assigns permissible versions on its unobserved groups. Define the joint probability mass function there by the original product of the one-trial Bernoulli probability mass functions, or use the limiting convention $0^0=1$ in the compact formula.

The sample mean $T/n$ is an unbiased estimator of $p$ and retains the sufficient statistic because $n$ is known. By contrast, $X_1$ is also an unbiased estimator of $p$ but, for $n\ge 2$, is not sufficient: the conditional success probability of the independent $X_2$ given $X_1$ is still $p$. Thus unbiasedness does not imply sufficiency.

Sufficiency also does not imply completeness of a statistic. For $n\ge 2$, the full-data sufficient statistic $X$ admits the nonzero function $g(X)=X_1-X_2$ with expectation zero for every $p$. Nor does minimality imply completeness of a statistic: for $n=2$ and the restricted statistical model $p\in \{1/4,3/4\}$, $T$ is a minimal sufficient statistic since the likelihood ratio $P_{3/4}(X=x)/P_{1/4}(X=x)=3^{2t-2}$ distinguishes its three values. Yet $g(0)=3,g(1)=-5,g(2)=3$ has expectation $(27-30+3)/16=0$ at both parameter values and is never zero. Changing the parameter family can change completeness of a statistic even while preserving sufficiency.

Events turn Boolean operations into operations on sets. For a die with sample space $U=\{1,2,3,4,5,6\}$, let $A=\{2,4,6\}$ be the event “even outcome” and $B=\{4,5,6\}$ the event “outcome at least four”. At outcome $u$, use the indicator functions $P=1_A(u)$ and $Q=1_B(u)$. ThenThus logical conjunction becomes set intersection, logical disjunction becomes set union, exclusive or becomes symmetric difference, and logical negation becomes the complement of a set within $U$. Outcome four satisfies both events, so inclusive logical disjunction includes it and exclusive or excludes it. If the die is fair, these four events have probabilities $1/3$, $2/3$, $1/3$, and $1/2$, respectively. No assumption of independent events is needed for the set identities; indeed $A$ and $B$ here are not independent, since $1/3\ne (1/2)(1/2)$.

For an event $A$ in a sample space $\Omega$, the complementary event $A^c=\Omega \setminus A$ occurs exactly when $A$ does not occur. The complement rule of probability follows from the Kolmogorov axioms: $A$ and $A^c$ are disjoint, their union is $\Omega$, and their probabilities therefore add to $\mathbb{P}(\Omega )=1$.

This rule is useful when an event is easier to describe through its failure. For a nonnegative integer-valued random variable $K$, $\mathbb{P}(K\ge 1)=1-\mathbb{P}(K=0)$ turns an at-least-one calculation into a single value of the probability mass function. In sampling without replacement, the hypergeometric distribution gives the chance of missing all marked objects, so its complement gives the chance of finding at least one.

A constant random variable takes the same fixed real number $c$ at every outcome in its sample space. Its probability distribution puts mass one at $c$. An almost surely constant random variable may differ on an event of probability zero, while having the same probability distribution.

For finitely many events, or a countable family with finite sum of individual probabilities, equality in the uncapped union bound holds exactly when every pairwise intersection has probability zero. Equivalently, at most one event occurs almost surely. Pairwise disjoint events give equality, but literal disjointness is stronger than necessary: shared outcomes of probability zero do not change either side. For the capped bound, if the sum exceeds one, equality instead means that the union is an almost sure event; overlap can then have positive probability.

The cap and the disjoint case are both sharp. If the individual probabilities sum to at most one, put the events on separate pieces of a sample space to attain their sum. The twelve-outcome disjoint example does this with two events of probability $1/3$. If the sum exceeds one, a covering union can attain the cap: on the same twelve-outcome sample space, $A=\{1,\dots ,8\}$ and $B=\{5,\dots ,12\}$ have sum $4/3$ and exact union probability one. These are sharpness examples, not an assertion that every overlapping family attains the bound. Positive-probability disjoint events are also not independent events, because their joint probability is zero while the product of their individual probabilities is positive.

A failure-probability budget assigns nonnegative numbers ${\delta }_i$ to a finite or countable family of failure events $F_i$ in a common probability space, with verified bounds $\mathbb{P}(F_i)\le {\delta }_i$. The union bound guarantees that the probability of any failure is at most $\min \{1,{\sum }_i{\delta }_i\}$. By De Morgan's laws and the complement rule of probability, the probability that every requirement succeeds obeysThus ${\sum }_i{\delta }_i\le \delta$ guarantees simultaneous success with probability at least $1-\delta$, even when all failure events depend on the same observations. The numbers ${\delta }_i$ are upper allowances; adding them is not a calculation of the exact failure probability.

For three requirements, give each a failure-probability budget ${\delta }_i=1/20$. The total budget is $3/20$, so simultaneous success has probability at least $17/20$. On a sample space of twenty equally likely outcomes, let $F_1=\{1\}$, $F_2=\{2\}$, and $F_3=\{3\}$. These disjoint events attain the budget and the success bound. If instead all three failure events equal $\{1\}$, their individual probabilities are unchanged, but exact failure has probability $1/20$ and simultaneous success has probability $19/20$. This second model shows how overlap can leave the guarantee conservative.

For $n$ requirements, an equal allocation ${\delta }_i=\delta /n$ is convenient; unequal allocations also work when some requirements are harder to satisfy. For infinitely many requirements indexed by $i\ge 1$, the allocation ${\delta }_i=\delta /2^i$ uses total budget $\delta$ by the geometric series. A fixed positive allowance for every requirement instead gives an infinite sum and no positive simultaneous-success guarantee from the union bound. The individual probability bounds must cover all the failure events being combined: omitting possible failure modes leaves the resulting guarantee about the listed requirements only. The same addition of error allowances underlies the Bonferroni correction in simultaneous hypothesis tests.

Assume the full-data sample space and the range of the statistic $T$ are standard Borel spaces, $T$ is a measurable function, and the statistical model $\{P_{\theta }\}$ is dominated by one common sigma-finite measure $\nu$. Write $f_{\theta }=d{P}_{\theta }/d\nu$ for its densities. Then $T$ is a sufficient statistic if and only if there exist nonnegative measurable functions $g_{\theta }$ and $h$, with $h$ independent of $\theta$, such thatThe statistical parameter can enter $g_{\theta }$ only through $\theta$ and the observed $T(x)$. ^[ref]

For discrete data, take counting measure; for ordinary continuous data, take Lebesgue measure. Independence, identical sampling laws, smooth dependence on $\theta$, and a fixed positive support of a probability distribution are not requirements of this theorem. Parameter-dependent support must nevertheless be included in $f_{\theta }$, usually through an indicator function, and must fit the displayed factorization. The sufficient statistic for a symmetric uniform sample illustrates exactly this: its indicator function depends on the unknown scale through the observed absolute maximum. The null sets in the theorem concern versions of densities; the factors are not generally themselves normalized probability distributions.

For a discrete sample space, the cancellation behind the theorem is particularly transparent. On a group with positive probability, factorization givesConversely, multiply a common conditional probability mass function by $P_{\theta }(T=T(x))$ to obtain the required factorization. For continuous data, a sum over a group is not a density integral over the original Lebesgue measure: the group may have measure zero. Use the regular conditional distribution formulation and the Fisher-Neyman factorization theorem instead of that discrete calculation.

The Kolmogorov axioms require a probability measure $\mathbb{P}$ to be nonnegative, normalized by $\mathbb{P}(\Omega )=1$, and countably additive on pairwise disjoint events.

For a countable measurable partition $(B_i)$ with positive probabilities,

Pairwise independent events describe a subtle kind of uncertainty: learning that one event occurred leaves each other event's probability unchanged, yet learning that two occurred can reveal a third completely. Checking every pair therefore answers a weaker question than checking the whole collection. The parity construction of pairwise independent bits makes this distinction visible with just four equally likely outcomes.

A family of events $A_i$ is pairwise independent whenIf $\mathbb{P}(A_j)>0$, this is equivalent to the conditional probability identity $\mathbb{P}(A_i\mid A_j)=\mathbb{P}(A_i)$. The product definition also handles events of probability zero. Independent events, meaning mutually independent events, require the same product rule for every finite subfamily, including triples. For three events, the three pair equations and the triple equation are four separate requirements. ^[ref].

For a worked example, take independent fair bits $U,V$: the sample space is $\{00,01,10,11\}$ with probability $1/4$ per outcome. Set $X=U$, $Y=V$, and $Z=U\oplus V$, using exclusive or, which is one precisely when its inputs differ. The events $A=\{X=1\}$, $B=\{Y=1\}$, and $C=\{Z=1\}$ each contain two outcomes. Their pair intersections areThus each event has probability $1/2$ and each pair intersection has probability $1/4=(1/2{)}^2$: these are pairwise independent events. But $X=Y=1$ forces $Z=0$, soThe three are not independent events. This is the complementary-third-event version of the two-toss example in Tsitsiklis and Jaillet, MIT Introduction to Probability, Lecture 3, slides 7–8.

Another useful diagnostic preserves a special common-intersection case. Suppose three events have common probability $s$ and every pair and triple intersect in the same event, of probability $q$. Pairwise independence requires $q=s^2$, while mutual independence additionally requires $q=s^3$. Both hold only for $s=0$ or $s=1$. For a concrete nontrivial instance, assign probability $1/4$ to each of four outcomes, let all three events contain the first outcome, and give each event one different additional outcome. Then $s=1/2$ and $q=1/4$: the pair equations hold, while the triple equation fails. Unlike the exclusive or example above, this triple intersection is nonempty.

The probability of an event is its value under a probability measure, a real number between zero and one. It compares the mass assigned to that event with the unit mass of the whole sample space. For a finite sample space with equally likely outcomes, $\mathbb{P}(A)=|A|/|\Omega |$. For unequal outcome masses, add their individual probabilities instead.

For example, selecting a coin with head probability $1/2$ with chance $4/5$, or a coin with head probability $3/4$ with chance $1/5$, gives head probability $(4/5)(1/2)+(1/5)(3/4)=11/20$. This is the law of total probability: averaging over the hidden selection is part of calculating the outcome's probability.