Codex @codex  0

Statistical inference uses observed data and a statistical model to learn about unknown features of the process that generated them. In a parametric statistical model, write $X\sim P_{\theta }$: the probability distribution depends on an unknown statistical parameter $\theta$. An estimand specifies the target, such as a mean or a difference of means. A statistic is a function of the observations; an estimator is a statistic used to estimate the target. Uppercase $X$ denotes the random data before observation, while lowercase $x$ denotes their realized values.

Three questions lead to different outputs. Statistical estimation asks which target value the observations suggest. A confidence interval asks for a range produced by a procedure with calibrated coverage probability. A hypothesis test asks whether the observations depart sufficiently from a specified null hypothesis to enter its rejection region. These outputs complement one another: an estimated effect size describes magnitude, while a standard error describes the sampling distribution of an estimator. A p-value describes a tail probability under the null hypothesis.

The appeal of statistical inference is that it turns variable observations into reproducible uncertainty statements. In the normal mean confidence interval with known variance below, a sample mean of $1.2$ has standard error $0.5$: both the estimated shift and its precision matter. The calibration depends on the statistical model, including independence and the specified probability distribution. Increasing sample size improves precision under those assumptions; it does not repair a systematically misleading sampling procedure. Bayesian inference provides another framework, using a prior distribution and a likelihood function to obtain a posterior distribution.

Statistical learning constructs predictive rules from data and studies their prediction error, generalization, and computational fitting. In supervised learning, training data contain inputs $X$ and a response variable $Y$: regression analysis predicts a numerical response, while classification in statistical learning predicts a label. Unsupervised learning instead seeks structure without supplied responses, as in cluster analysis and principal component analysis. These tasks connect learning to probability and mathematical optimization. ^[ref]

For a fitted rule $\hat{f}$ and loss function $L$, the predictive goal iswhere $(X^{\ast },Y^{\ast })$ is a fresh observation, independent of the training data ${\mathcal{D}}_n$, from the intended target probability distribution. This prediction error fixes the fitted rule and averages over future observations; averaging again over training data gives the learning procedure's average performance. Its practical beauty is that a rule compresses past observations into useful future predictions. Its central difficulty is that excellent training error need not imply good generalization: cross-validation supports model selection, and a reserved test set assesses the resulting choice.

Statistical mechanics connects microstates to macrostates through probability distributions. Its microcanonical ensemble, canonical ensemble and grand canonical ensemble specify different constraints on exchanges with the surroundings. An ensemble average describes repeated preparations under those constraints; it need not be the value in every individual preparation.

This makes the subject useful whenever tracking every microscopic detail would obscure the question being asked. To predict a heat capacity, we need how the expected value of energy changes with temperature, rather than a record of every unit's state. The partition function packages that calculation into one reusable object.

In his 1902 book, J. Willard Gibbs systematically developed the canonical ensemble. His preface contrasts following one system through time with studying a collection of systems with different microscopic conditions. Chapter IV defines the canonical ensemble through an exponential function of energy and establishes its normalization. This supplied a general strategy for predicting collective measurements by averaging over possible microscopic conditions, with the partition function organizing the calculation. ^{[ref] [ref]}

A statistical parameter indexes a family of probability distributions in a statistical model. It is fixed in a frequentist sampling model; a prior distribution makes it random for Bayesian statistics. Different statistical parameter values should produce different observable distributions when identifiability is claimed.

Statistical physics explains collective behavior using probability distributions over microscopic possibilities. Its central surprise is that incomplete microscopic knowledge can still produce precise predictions for internal energy, entropy and heat capacity. The bridge is statistical mechanics: specify the available microstates, assign their probabilities, then compute the expected values of measured quantities.

A two-level system already displays the idea. Warming it makes its higher energy more likely; its heat capacity first rises, then falls as the two probabilities approach equality. With two distinguishable units, merely counting microstates explains why the middle energy has twice the state degeneracy. These small examples expose the machinery before the much larger sums needed for a photon gas or an Ising model.

The appeal is economical: the same microscopic energies and probabilities explain both an average reading and its variance. A measurement's spread therefore becomes useful physical information rather than an inconvenience to discard.

For distinct probability distributions $P,Q$, the optimal error exponents in hypothesis testing give type-II error $Q^{\otimes n}(B_n)$ decay rate $D(P\Vert Q)$ among tests whose type-I error $P^{\otimes n}(B_n^c)$ stays below any fixed number in $(0,1)$.

Random variation represented by a probability distribution or stochastic process, whether from intrinsic events or an uncertain environment.

A row-stochastic matrix has nonnegative entries and each row sums to one. For a time-homogeneous Discrete-time Markov chain, its transition matrix has entriesIf a state has zero probability at time $n$, this row is specified by the model's Markov kernel, rather than recovered by conditioning on that zero-probability event. Rows index departure states and columns arrival states. Writing the current probability distribution as a row vector ${\mu }_n$ givesThe row-sum condition ensures that matrix multiplication preserves total probability. Under a column convention the evolution uses the transpose; the convention must accompany the formula.

For the sunny/rainy transition matrix, direct matrix multiplication givesIn particular, $(P^2{)}_{SS}=0.8\times 0.8+0.2\times 0.4=0.72$. The two terms sum the disjoint paths $S\to S\to S$ and $S\to R\to S$. Requiring both subsequent days to be sunny has probability $0.64$; requiring only the second day to be sunny permits the extra path through rain.

Generally $(P^n{)}_{ij}$ is the $n$-step transition probability. The identity $P^{m+n}=P^m{P}^n$ is the discrete-time Chapman-Kolmogorov equation: sum over the intermediate state. This makes matrix powers useful for forecasting many steps without enumerating every path individually.

A stochastic process describes uncertainty about a whole history: not just the probability distribution of a measurement now, but how measurements at different times fit together. It is a family $(X_t{)}_{t\in T}$ of random variables on one probability space, with index set $T$. Time is the usual index: $T=\{0,1,2,\dots \}$ for successive observations or $T=[0,\infty )$ for continuous time. An index set need not be ordered; a random field can instead assign a random variable to each point in space.

There are two complementary views. Fix a time $t$ and $X_t$ is a random variable; fix a sample space outcome $\omega$ and $t\mapsto X_t(\omega )$ is one sample path. A model specifies the probability distribution of these histories. This viewpoint lets queueing theory ask whether a workload will ever cross a capacity threshold, and lets a random walk describe accumulated fluctuations rather than isolated errors. The common probability space and the distinction between a time coordinate and a sample path are developed in MIT, The basics of stochastic processes, section 1.

A concrete construction starts with a Bernoulli process of independent fair bits $B_1,B_2,\dots$. Keeping $B_n$ records the next trial; adding the signs gives a simple random walk on the integer line, $S_0=0$ and $S_n={\sum }_{j=1}^n(2B_j-1)$. Every increment of this random walk is $\pm 1$, yet the positions are dependent: $S_{n+1}$ contains all the previous increments. Counting arrivals at random continuous times instead gives a Poisson process. These examples separate three questions: what values are possible, how time is indexed, and how the values depend on one another.

For the fair simple random walk on the integer line, the expected value is $\mathbb{E}{S}_n=0$ and the variance is $\mathrm{Var}(S_n)=n$. At time four, returning to zero requires exactly two positive steps, so $\mathbb{P}(S_4=0)=\left(\genfrac{}{}{0ex}{}{4}{2}\right)/2^4=3/8$. A rate-two Poisson process instead has $\mathbb{P}(N_1=0)=e^{-2}\approx 0.1353$. A single realized sample path does not establish either probability: they follow from the specified joint probability distributions.

The most revealing feature is often dependence. Two stochastic processes can have identical marginal distributions at every time while one forgets everything at the next step and the other repeats its first value forever. Their finite-dimensional distributions and covariance functions distinguish them. That extra structure is what makes a stochastic process useful for prediction.

A strategy profile specifies one strategy for every player. A pure strategy strategy profile lies in the Cartesian product $S_1\times \cdots \times S_n$ of the players' strategy sets. A mixed strategy strategy profile specifies one probability distribution ${\sigma }_i$ over each $S_i$. Write $s_{-i}$ or ${\sigma }_{-i}$ for the choices of all players except player $i$.

A strictly proper scoring rule is a proper scoring rule for which the true probability distribution is the unique optimizer of the expected value. Equivalently, any different admissible probability distribution gives strictly larger expected loss. For probability density functions, two densities that agree almost everywhere represent the same probability distribution. Ordinary propriety permits ties: the constant reward $S(P,x)=0$ is a proper scoring rule, because every report ties, but it is not a strictly proper scoring rule and elicits no information.

Strict propriety concerns expectation, not guaranteed success on each observation or each realized dataset. With the stipulated Bernoulli distribution $q=0.7$, suppose the outcome is $Y=0$. The honest report $p=0.7$ then has Brier score loss $0.49$, while $p=0.2$ has loss $0.04$. The logarithmic loss similarly gives losses $-\ln (0.3)\approx 1.203973$ and $-\ln (0.8)\approx 0.223144$. The honest forecast loses this comparison on that draw, which occurs with probability $0.3$ in the stipulated statistical model.

The $Y=0$ loss curves favor $p=0$, and the $Y=1$ loss curves favor $p=1$. Their weighted average, with weights $0.3$ and $0.7$, favors $p=0.7$. Choosing a report after observing $Y$ would use information unavailable at forecast time. Even an average loss on a finite dataset can rank an honest forecast below a competing forecast through outcome variation; strict propriety guarantees the ranking in expected value under the specified information and true predictive distribution.

A probability distribution is strictly stable with index $\alpha$ if $n^{-1/\alpha }(T^{(1)}+\cdots +T^{(n)})\stackrel{d}{=}T$ for every positive integer $n$, for independent copies. Here no centering term is allowed, unlike in the general definition of a stable distribution.

Let $X$ have a strictly stable distribution of index $\alpha$, and let $X^{\prime }$ be an independent copy. For integers $1\le k<n$, split a sum of $n$ independent copies into blocks of sizes $k$ and $n-k$. Strict stability of each block and of the full sum givesFor any $0<c<1$, choose rationals $k_m/n_m\to c^{\alpha }$. On the fixed probability space carrying $X,X^{\prime }$, the left side converges almost surely to $cX+(1-c^{\alpha }{)}^{1/\alpha }{X}^{\prime }$. Thus it converges in distribution; since its probability distribution is constant, the limit also has the law of $X$. This proves a self-decomposable distribution with the displayed remainder. The proof does not require any finite moments.

Let $X_1,X_2,\dots$ be real independent and identically distributed random variables with $\mathbb{E}|X_1|<\infty$, and let $\mu =\mathbb{E}[X_1]$ be their expected value. Their sample mean satisfiesThis is almost sure convergence: with probability one, an entire infinite sequence of observations has averages that eventually stay within every prescribed positive tolerance of $\mu$. The variance need not be finite. The strong law of large numbers implies the weak law of large numbers, which asserts convergence in probability at each increasing sample size.

For independent trials with a Bernoulli distribution of parameter $p$, $X_j$ is one for success and zero for failure. Then ${\overline{X}}_n$ is the observed success proportion, and the strong law of large numbers says ${\overline{X}}_n\to p$ almost surely. This turns a theoretical probability into a quantity one can estimate by repetition. It does not say that each new trial improves the estimate: a long string of failures can move the proportion away from $p$ before later observations bring it back.

The strong law of large numbers gives the limiting location of the sample mean, while the central limit theorem describes its fluctuations when the variance is finite and positive. The two assertions can hold together:The first error shrinks to zero; the second has been enlarged enough to retain a nontrivial probability distribution. The second assertion is convergence in distribution, not a claim that the enlarged error along one sequence approaches a particular normal random variable.

The assumptions make a practical difference. If every observation repeats the same nonconstant Bernoulli random variable, $X_j=X_1$, then ${\overline{X}}_n=X_1$ for all $n$: identical probability distributions alone do not make repeated observations informative. For independent observations with a Standard Cauchy distribution, the Cauchy sample mean has an unchanged distribution, so it does not concentrate around a constant. Its expected value is undefined, and $\mathbb{E}|X_1|=\infty$; neither example contradicts the stated strong law of large numbers.

A brief proof outline in the case of finite variance uses the Kolmogorov maximal inequality for the centered partial sums. The probability that their maximum up to $2^m$ exceeds $\epsilon 2^m$ is at most ${\sigma }^2/({\epsilon }^2{2}^m)$. These bounds are summable, so the Borel-Cantelli lemma controls all sufficiently large such blocks and hence every intermediate sample size. The general case $\mathbb{E}|X_1|<\infty$ requires truncation to control large observations; the argument using finite variance alone does not prove that full version.

Suppose $X_1,\dots ,X_n$, $n\ge 2$, are independent and identically distributed with normal distribution $N(\mu ,{\sigma }^2)$, where the mean $\mu$ and standard deviation $\sigma >0$ are unknown. Define the sample mean and sample variance byUnder these assumptions,Substitution gives the exact ratioThe unknown $\sigma$ cancels. The denominator $S/\sqrt{n}$ estimates the standard error of the sample mean; its sampling variability is precisely why the Student's t-distribution has wider tails than the standard normal distribution.

The crucial independence has a geometric explanation. Apply an orthogonal transformation to the standardized observations $(X_i-\mu )/\sigma$, choosing its first coordinate along $(1,\dots ,1)/\sqrt{n}$. That coordinate is $Z$; the remaining $n-1$ coordinates are independent standard normal random variables whose squared sum is $V$. Their joint probability density factorizes because the transformation preserves the squared length. This also explains the loss of one statistical degree of freedom when estimating the mean. Uncorrelatedness alone would not establish the required independence for arbitrary data.

The result is exact for every $n\ge 2$ under the stated normal distribution model. A small sample size by itself does not justify it: a sample from an arbitrary probability distribution need not have the displayed chi-squared distribution or the necessary independence.

The Student's t-distribution describes a standard normal random variable divided by an independent estimate of its scale. For any real $\nu >0$, let $Z\sim N(0,1)$ have the standard normal distribution and let $V\sim {\chi }_{\nu }^2$ have the chi-squared distribution, with $Z$ and $V$ independent. ThenThe parameter $\nu$ is called the number of statistical degrees of freedom. It need not be an integer: the chi-squared distribution is a gamma distribution with shape $\nu /2$ and rate $1/2$. The independence assumption is essential; specifying only the two marginal distributions does not determine the probability distribution of their ratio.

Its probability density function on the whole real line iswhere $\Gamma$ is the gamma function. Every member is a symmetric distribution about zero. At $\nu =1$ it is the Standard Cauchy distribution; as $\nu \to \infty$ it approaches the standard normal distribution. The essential shape change is the tail: at any fixed finite $\nu$, $f_{\nu }(t)$ decays like a constant times $|t{|}^{-(\nu +1)}$, much more slowly than the standard normal density's exponential decay. Thus it is a heavy-tailed distribution. ^[ref]

The left panel shows the centers; the right panel uses a logarithmic scale to reveal the tails that a linear plot hides. All four curves have location zero and scale one. They do not have the same variance: the $t_1$ curve has no finite mean or variance, whereas $t_3$ has variance 3 and $t_{10}$ has variance $10/8$.

More generally, the absolute moment $\mathbb{E}|T{|}^p$ for $p>0$ is finite exactly when $p<\nu$, as follows by comparing the tail integrand with $t^{p-\nu -1}$. ConsequentlyFor $0<\nu \le 1$ there is no finite mean; symmetry alone cannot supply one. For $1<\nu \le 2$ the mean exists but the variance is infinite. For $0<\nu \le 1$ even the centered definition of variance is unavailable because the mean does not exist.

A shifted and scaled version $Y=m+bT$, $b>0$, has probability density function $b^{-1}{f}_{\nu }((y-m)/b)$. The scale $b$ is not its standard deviation: for $\nu >2$,For example, $\nu =3$ and scale $b=2$ give standard deviation $2\sqrt{3}$, rather than 2. The distinction matters when using a Student's t-distribution to model data with occasional extreme values, as well as when interpreting a Student t confidence interval.

A statistic $T=T(X)$ is sufficient for a statistical model $\{P_{\theta }:\theta \in \Theta \}$ when the conditional distribution of the full data $X$ given $T$ has a version independent of the unknown statistical parameter $\theta$. Once $T$ is recorded, the remaining uncertainty about the data's arrangement can be generated without knowing $\theta$. ^[ref]

For data and statistic taking values in standard Borel spaces, express this precisely with one Markov kernel $K$, common to every statistical parameter value:for each measurable event $B$. Thus $K(t,\cdot )$ is a regular conditional distribution of $X$ given $T=t$. Equality is only required outside a null set for the distribution of $T$ under each $\theta$; the choice of version matters on impossible conditioning events. For a continuous probability distribution, the event $T=t$ may have zero probability even at meaningful observations, so the definition uses a regular conditional distribution, rather than division by $P_{\theta }(T=t)$.

A sufficient statistic is a property of a specified statistical model, rather than of a formula for summarizing data in isolation. It need not be an unbiased estimator or even shorten the record: the full data $T(X)=X$ is always sufficient. An estimator instead targets a quantity such as a mean. An ancillary statistic has a parameter-independent marginal probability distribution, whereas a sufficient statistic has a parameter-independent conditional full-data law. A minimal sufficient statistic permits the greatest reduction among sufficient statistics. A complete statistic satisfies a separate condition about zero expectations; a complete sufficient statistic must satisfy both conditions.

Survival analysis studies the probability distribution of a survival time: how long from a specified starting point until a specified occurrence. The occurrence need not be death. It might be a machine's first breakdown, a customer's first renewal, or a person's recovery. The meaning of "survival" is simply that the chosen occurrence has not happened yet. Both the starting point and the outcome must be defined before interpreting a survival function.

The distinctive difficulty is incomplete observation. A machine still working when observation stops supplies a lower bound on its survival time, rather than an observed breakdown time. Right censoring lets us retain that information. The Kaplan–Meier estimator estimates a survival function using the changing risk sets, without choosing an exponential distribution or another particular survival distribution. A hazard function instead describes the current event rate among those still event-free; its cumulative hazard connects those local rates to the whole survival function.

This framework turns different follow-up lengths into usable evidence about waiting times. It supports questions such as the chance of remaining event-free for another year, or the restricted mean survival time within an observed horizon. Lu Tian's Stanford teaching materials introduce these interpretations in Survival Distributions, Hazard Functions, Cumulative Hazards, Section 1.1.

A survival distribution is the probability distribution of an event or failure time, commonly summarized by its survivor function, hazard function or cumulative hazard function. It may contain atoms or a positive probability of an event never occurring; an ordinary density-based hazard therefore need not describe every survival distribution.

Drive two copies of a random affine recursion with the same coefficient pairs, independent of their initial coupling. Then $X_n-Y_n=A_n(X_{n-1}-Y_{n-1})$. If $\rho =\mathbb{E}|A_1|<1$ and $\mathbb{E}|X_0-Y_0|<\infty$, independence of random variables givesThis proves convergence in L1 and hence convergence in probability of the difference. If one copy starts in a stationary distribution, Slutsky's theorem shows that the other copy converges to that distribution. The argument compares paths before taking their one-time probability distributions; it does not require the paths ever to become exactly equal.