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Assume $0<Z={\sum }_{x}r(x)<\infty$, and work on the support $S=\{x:r(x)>0\}$. In rejection sampling, choose a normalized proposal probability mass function $g$ and a known finite constant $M$ satisfying $r(x)\le Mg(x)$ everywhere. Draw independent random variables $Y\sim g$ and $U\sim \mathrm{Unif}(0,1)$, accepting $Y$ when $U\le r(Y)/(Mg(Y))$. The joint probability of proposing and accepting $x$ is $r(x)/M$, so the overall acceptance probability is $Z/M$ and the accepted value has probability mass function $r(x)/Z$. Fresh trials therefore produce independent and identically distributed random variables with the desired probability distribution. For a fixed positive number $N$ of accepted values, the sample average $N^{-1}{\sum }_{i=1}^{N}f(Y_i)$ is an unbiased estimator of the expectation when that expectation is finite. Its variance is ${\mathrm{Var}}_p(f)/N$ if the second moment is finite. Finding a valid envelope can be difficult, and a small acceptance probability makes the method expensive.

For the Metropolis–Hastings algorithm, choose a proposal transition matrix $q$ and, at the current state $x$, propose $y\sim q(x,\cdot )$. Accept with the Metropolis–Hastings acceptance probabilityA proposal outside $S$ is rejected; an impossible proposal need not be assigned a ratio. Otherwise remain at $x$. This constructs a Markov chain satisfying detailed balance with $p$, without computing $Z$ or a global envelope. Estimate the expectation by averaging $f(X_i)$ over iterations, including repeated states after rejections. Keeping only accepted jumps generally changes the stationary distribution of the recorded process.

The Metropolis–Hastings algorithm produces dependent samples, so the variance of its average includes autocovariances; starting outside stationarity can also cause finite-run bias. With Markov chain irreducibility, suitable recurrence and an integrable $f$, the countable-state Markov chain ergodic theorem gives consistency. Aperiodicity is additionally needed for the usual convergence of one-time probability distributions. In contrast, accepted rejection sampling values need no initial transient. Neither method is universally faster: envelope rejection competes with the proposal quality and mixing time of the Markov chain. Nonnegativity of $f$ alone does not guarantee a finite expectation or variance.

Here $q(x,y)=p(y)$, with fresh proposal randomness independent of the past. For $x,y\in S$, the Metropolis–Hastings acceptance probability isThus $P(x,y)=p(y)$ is independent of $x$, and $X_1,\dots ,X_N$ are independent and identically distributed random variables with probability distribution $p$, even if $X_0$ was chosen arbitrarily in $S$. The method becomes ordinary Monte Carlo method sampling from $p$. The average of $f(X_1),\dots ,f(X_N)$ is an unbiased estimator, with the variance stated in part (a). Including an arbitrary initial state would spoil this finite-sample property. This is the independent stationary proposal lemma.

Neither Markov chain irreducibility nor fast convergence follows from detailed balance. On two states, the identity transition matrix satisfies detailed balance for the uniform probability distribution, but neither state can reach the other.

Even after imposing Markov chain irreducibility and aperiodicity, considerIt satisfies detailed balance for $\pi =(1/2,1/2)$. Its nontrivial eigenvalue is $1-2\epsilon$, and, from state zero, its total variation distance from $\pi$ after $n$ steps is $\frac{1}{2}(1-2\epsilon {)}^n$. This approaches zero arbitrarily slowly as $\epsilon$ decreases. Detailed balance without rapid convergence separates correctness of a stationary distribution from reachability and speed.

Variational inference replaces exact Bayesian inference by an optimization problem over a tractable family $q_{\lambda }(z)$. A guide program samples from this approximating family. Optimizing an evidence lower bound is equivalent to minimizing $D_{\mathrm{KL}}(q_{\lambda }\Vert p(\cdot \mid x))$. Its appeal is that a fitted approximation can be cheap to sample and may scale well when an exact posterior distribution is inaccessible. A restricted family, such as mean-field variational inference, can miss dependence or entire modes; a local optimum can add further error. More optimization cannot remove approximation error caused by an inadequate family.

Monte Carlo method inference instead estimates posterior expectations using simulated values or weighted values. Examples include rejection sampling, importance sampling and Markov chain Monte Carlo. Under their respective support and convergence assumptions, increasing computation can yield consistent estimates without choosing a finite-dimensional posterior family. Their uncertainty includes simulation error; an inefficient proposal or slow mixing time can make that error large. Markov chain Monte Carlo additionally requires convergence and dependence diagnostics.

The central trade-off is optimization within an approximate family versus simulation under a target probability distribution. It is not an absolute distinction between deterministic and random computation: variational inference often uses Monte Carlo method estimates of its gradient, while Monte Carlo method inference may use fitted proposals. Neither method automatically guarantees accurate uncertainty estimates or superiority at a fixed computational budget.

An idealized underlying probability space iswith the product Borel sigma-algebra. Its 27 independent random coordinates are $\mu$, $\sigma$, twenty past standard-normal noises, and five future standard-normal noises. Define $T_i(\omega )=\mu +\sigma {\epsilon }_i$. Sampling an unknown variance does not mean drawing a new variance on each day; a single draw of ${\sigma }^2$ applies to all durations in a trace.

Let $A$ be the event that the first twenty durations satisfy their corresponding observed inequalities. The observations are the twenty late/early labels, conditional on the fixed departure thresholds. They are not exact values of the latent $T_i$. The posterior probability measure on traces isHere $\mathbb{P}(A)>0$: at every interior parameter pair each strict normal-tail probability lies in $(0,1)$, so their finite product has positive prior integral.

The returned outcome isThus the underlying sample space has dimension 27, the successful output space is $\{0,1\}$, and the observations define the conditioning event $A$. The desired output probability distribution is the pushforward measure of ${\mathbb{P}}_A$ under $Y$. Rejection is an implementation device, not a third scientific answer: None is discarded before forming the conditional output law.

The first graph solves the row player's max–min optimization: choose a mixed strategy guaranteeing the greatest possible worst-case expected payoff. The second solves the column player's min–max optimization: choose a mixed strategy placing the smallest upper bound on the row player's expected payoff. Their optimal heights agree, illustrating the minimax theorem.

There is also a direct Nash equilibrium certificate, so the conclusion does not depend on reading coordinates from a plot. At $p^{\ast }=2/7$, both opposing pure columns give row expected payoff $-1/7$; at $q^{\ast }=4/7$, both pure rows give row expected payoff $-1/7$. By linearity, every unilateral deviation, including an arbitrary mixed strategy, leaves the deviator's expected payoff unchanged against the fixed opposing equilibrium mixture. Neither player can improve it. This is a mixed-strategy optimality certificate for a matrix game.

The equilibrium outcome is a probability distribution over cells, rather than one deterministic cell. Independent randomization givesThe resulting row expected payoff is $(16-18-20+15)/49=-1/7$. Finally, each envelope has a unique optimum, so any Nash equilibrium must use the displayed optimal mixed strategies. In particular there is no pure strategy Nash equilibrium.

Use the standard repeated game convention that the first round's joint action is observed before round two. The following original game tree represents simultaneous moves by making row move first in each round while hiding that current move from column. Terminal labels are total payoffs over both rounds, ordered row then column.

Formally, round one starts at a row decision with actions $T,B$. The two ensuing column nodes form one information set $I_2^0$, with available actions $L,R$; column cannot tell which current row action occurred. The four completed first-round histories are $H=\{TL,TR,BL,BR\}$. At each $h\in H$, round two begins at a separate row node. Its two successors, after row's current $T$ or $B$, form a column information set $I_2^h$. Distinct $h$ have distinct information sets: both players know the whole first-round outcome. There are sixteen terminal histories $(h,k)\in H^2$, whose total payoff is $u(h)+u(k)$, as displayed in the tree.

The row player's five information sets are the initial singleton and the four singleton round-two nodes, one per $h$. The column player's five information sets are $I_2^0$ and the four two-node sets $I_2^h$. Thus each player must specify one binary choice at each of five information sets, including histories that its own first choice might prevent:Counting only the four possible two-round action sequences would miss the distinct contingent responses to the opponent's first action.

For a concrete pair of complete pure strategies, order the second-round histories as $(TL,TR,BL,BR)$ and useThe first entry is the initial action, and the four remaining entries prescribe the action after each first-round history. This induces the unique path $TL$ in round one followed by $BR$ in round two. Hence the total payoff isThe off-path entries are part of each pure strategy, even though this pair does not reach them.

For randomization, a mixed strategy is a probability distribution over the $32$ complete plans, with $31$ free probabilities in a general distribution. A behavioural strategy instead gives one binary-action probability at each of five information sets, requiring five parameters. This game has perfect recall: a player never forgets a previous own action or a first-round observation, and the only concealed action is the opponent's current simultaneous choice. Consequently neither kind of randomization has a greater attainable-payoff advantage: Kuhn's theorem gives realization-equivalent strategies against every fixed opposing strategy.^[ref]

The equivalence can be verified directly here. Given a behavioural strategy, sample all five contingent actions independently before play; the resulting product probability distribution over complete plans is an equivalent mixed strategy. Conversely, let $\mu$ be a mixed strategy for player $i$. Use its marginal probability for the initial action. At history $h$ containing own first action $a_i$, choose second action $b_i$ with conditional probabilitywhenever the denominator is positive; if it is zero, choose arbitrarily at that unreachable history. The opponent's first-round action is simultaneous and cannot depend on the sampled hidden plan beyond the chosen first action. Thus conditioning on $h$ preserves exactly these conditional probabilities of actions, and multiplying the first- and second-round factors reproduces each terminal probability. Correlations between prescriptions at mutually exclusive second-round histories have no further observable effect. Behavioural strategies offer the more compact representation, while mixed strategies describe a single random choice of a full plan. Imperfect information here does not mean imperfect recall.

For sufficiently large $n$, $p_n=\lambda /n$ lies in $(0,1]$. The waiting time has the geometric distribution on $1,2,\dots$, so summing its geometric series giveswhenever $(1-\lambda /n)e^{t/n}<1$. Fix $t<\lambda$. The expansions $e^{t/n}=1+t/n+O(n^{-2})$ show that the numerator is $\lambda /n+O(n^{-2})$ and the denominator is $(\lambda -t)/n+O(n^{-2})$. ThereforeThe convergence theorem for moment-generating functions says that pointwise convergence of finite moment-generating functions on a fixed open interval about zero to the moment-generating function of a probability distribution implies convergence in distribution to that distribution. For example take $|t|<\lambda /2$; all large $n$ satisfy the convergence condition throughout that interval. The limit is the moment-generating function from part (a), proving

For the positive-direction alternative, use the same paired statistic $t_{\mathrm{obs}}=2.607/(4.713/\sqrt{15})$, but reject at level $\delta$ only when $t_{\mathrm{obs}}>t_{14,1-\delta }$. Its one-sided $p$-value isBecause the observed mean difference is positive and the null Student probability distribution is symmetric, the two-sided $p$-value is twice this upper-tail probability, consistent with the supplied $0.05$ and $0.025$.

The two-sided value is the null-model probability of a test statistic at least as far from zero in either direction as the observation. The one-sided value uses only the direction of larger cross-fertilized heights. They are probabilities under the assumed null sampling distribution, not posterior probabilities that the null hypothesis is true.

At the reported precision, the two-sided result is on the conventional 5 percent boundary; rounding can matter for a strict numerical threshold. The one-sided result is significant at5 percent and supports a positive mean difference if that directional alternative was chosen in advance. Choosing the direction after observing the sign would not justify using the smaller one-sided probability as a prespecified test. The paired design controls shared pot conditions, while the inference still relies on the independence and probability distribution assumptions for the differences.

For a sample of size $n\ge 2$, defineFor the zero-mean Gaussian sample, the standardized vector $(X_1/\sigma ,\dots ,X_n/\sigma )$ has independent standard normal coordinates. Apply an orthogonal transformation whose first row is $(1,\dots ,1)/\sqrt{n}$. The resulting coordinates $Z_1,\dots ,Z_n$ remain independent standard normals: their joint density depends only on squared Euclidean length and the transformation has unit absolute Jacobian. Its first coordinate is $Z_1=\sqrt{n}\overline{X}/\sigma$, while orthogonal decomposition givesThusThe ratio of the first independent standard normal and the square root of the normalized chi-square variable has a Student probability distribution, giving

For the paired height observations, form one cross-minus-self difference per pot. The sampling units for this test are the fifteen differences, not thirty separate plant heights. Under the usual independent normal model for differences, a null mean of zero gives a $t_{14}$ statistic. The observed statistic isFor the two-sided alternative, reject at significance level $\delta$ if $|t_{\mathrm{obs}}|>t_{14,1-\delta /2}$. Its two-sided $p$-value is $2[1-F_{t_{14}}(|t_{\mathrm{obs}}|)]$. This uses the recorded values without performing the numerical calculation forbidden by the question. The normal difference model is what makes the small-sample Student calibration exact.

Let $T_0^{+}=\inf \{n\ge 1:X_n=0\}$. For $a>0$, the chain is irreducible and its stationary probability distribution makes it positive recurrent. The mean-return-time identity givesThis counts a self-loop at zero as a return after one time unit, as “next zero” requires. Equivalently, with probability $1-a$ that immediate return occurs; with probability $a$ the chain moves to one and then has mean hitting time $1/(b-a)$ back to zero, giving $1+a/(b-a)$. If $a=0$, the return is deterministically at time one and the displayed formula still equals one.

For $i\ge 1$, the stationarity equation iswhich rearranges to the stated recurrence. At zero,Define the adjacent probability current $J_i=a{\pi }_i-b{\pi }_{i+1}$. The recurrence says $J_i=J_{i-1}$, and the boundary equation says $J_0=0$. Hence ${\pi }_{i+1}=(a/b){\pi }_i$ for all $i$. Since $\rho =a/b<1$, normalization givesThe adjacent balance equations also verify stationarity directly. If $a=0$, this formula means the point mass at zero; otherwise the birth-death chain is irreducible and this is its unique stationary probability distribution.

The binomial series gives, for $|z|<14/5$,All coefficients are nonnegative, and $g(1)=2/3+1/3=1$. Thus defineThese numbers sum to $1$, so they define a probability distribution on the nonnegative integers with the desired probability generating function.

Empirical equivalence must be specified relative to a domain and an account of what counts as observable. Two theories may fit all evidence collected so far while predicting different future results; that is ordinary underdetermination by a finite record. A stronger equivalence requires the same observable consequences for every permitted experiment in the stated domain. For stochastic predictions, matching a set of possible outcomes is not enough: the theories must assign the same outcome probability distributions under corresponding preparations and testing assumptions.

A clear physical example at the level of descriptions occurs in electromagnetism. Write the electric field and magnetic field using a scalar potential $\varphi$ and magnetic vector potential $\mathbf{A}$:For a smooth, single-valued function $\chi$, the gauge transformationleaves both fields unchanged: the curl of a gradient vanishes, and the mixed derivatives cancel in the electric-field expression. With the same charges, boundary conditions and force law, these descriptions give the same classical measurements. Agreement is structural, not an accidental fit to a few readings. However, many physicists regard such descriptions as representing one physical situation, so this alone is not an example of two genuinely incompatible physical theories. The smooth global transformation assumption also matters; equality of local fields is not an unrestricted license to identify every globally different potential configuration.

A second example exhibits possible surplus structure. For particles interacting by Newtonian forces depending only on relative positions, a common Galilean transformationleaves relative positions and accelerations unchanged. Accordingly, adding a privileged but completely uncoupled state of uniform rest does not alter any internal mechanical prediction. Rival choices of that unobservable rest state cannot be distinguished by these experiments. This is a deliberately specified completion of the mechanical model: it does not show that rotation or acceleration is undetectable, or that Newtonian mechanics and special relativity are empirically equivalent. The latter give different high-speed predictions. Whether undetectable absolute rest makes a genuinely different theory or a redundant representation is precisely one philosophical issue raised by the example.

Thus there are concrete cases of empirical redundancy, while the existence of equally well-motivated, fully distinct rivals with identical consequences across all physics is a much stronger claim. Historical theories often agree only in limited regimes or after additional assumptions. One must not convert that restricted agreement into an equivalence theorem for every possible test.

The philosophical relevance does not depend on finding a complete historical pair. A constructed rival can expose the logical limits of evidence, although an artificial extension need not deserve the same credibility as an economical account. If two specified models have equal likelihoods for every admissible data set $D$, thenfor finite positive prior odds. The Bayes factor is one; the data do not discriminate between them. Choices about simplicity or interpretation would be additional considerations. Constructive empiricism treats empirical adequacy as the relevant belief commitment in acceptance, while a scientific realist must explain what further warrants belief in the unobservable structure. The equivalence argument identifies a limit of evidential discrimination; it does not establish that every theoretical claim is equally good or that every unobservable claim is false.^[ref]

The covering-law model captures one important feature of scientific explanation: a result should follow from general principles together with the relevant circumstances. In its deductive-nomological model, laws and conditions entail the phenomenon, the laws are needed in the derivation, and the premises must be true and empirically substantive. That is stronger than merely redescribing the event.^[ref]

For example, the ideal gas law, a specified amount of gas and a fixed-volume condition explain why changing its temperature changes its pressure in the model's regime. The derivation identifies a systematic dependence rather than an isolated coincidence. It also makes assumptions visible: a calculation using a law outside its domain is not repaired simply by being logically valid.

Nevertheless, valid subsumption does not by itself determine explanatory direction. A vertical pole of height $h$, a solar elevation $\alpha$ and a shadow of length $l$ satisfy $h=l\tan \alpha$ under suitable geometric assumptions. The same relation allows one to deduce the height from the shadow. In an ordinary intervention on the pole, changing its height changes its shadow; shortening the shadow with a screen does not shorten the pole. The reverse calculation can be excellent evidence for the height without providing the same causal explanation. This is explanatory asymmetry.

Relevance is another difficulty. One can append an irrelevant fact to a sound derivation without invalidating the deduction, but the added fact need not explain anything. Requiring an essential law prevents some vacuous derivations, yet still does not identify which features make a physical difference. A useful explanation should reveal the dependence the question is asking about, not merely contain enough true premises to reach its conclusion.

Nor are all explanations deterministic. A particular radioactive decay can have a good account in terms of a physical decay mechanism and a probability distribution without being entailed by initial conditions. An inductive-statistical model extends the covering-law idea, but requiring a high probability would exclude the explanation of some rare outcomes. Conversely, a high conditional probability produced by irrelevant correlations need not explain a mechanism.

The model is therefore a valuable discipline for organizing some explanations, rather than a sufficient universal analysis. It clarifies derivational structure and testable commitments. An adequate broader account must also address causal relevance, the question's contrast, statistical events and explanations by constraints or unification that are not straightforward causal histories.

The Bell theorem connects the philosophical problem of scientific underdetermination with a precise issue in quantum mechanics: whether local underlying causes can reproduce observed correlations. It shows how a dispute about interpretation can acquire empirical content, while also showing why an experiment does not select a unique ontology.

A shared cause can produce strong correlations without a signal between distant measurements. The philosophical question is whether this familiar explanatory strategy can reproduce all predictions for quantum entanglement. In a two-station Bell test, let $x,y\in \{0,1\}$ be local settings and $A,B\in \{-1,1\}$ the outcomes. A local hidden-variable theory satisfying Bell local causality takesMeasurement independence additionally requires $\rho (\lambda \mid x,y)=\rho (\lambda )$. The same ensemble of underlying states must therefore apply across the four setting pairs.

The resulting CHSH inequality can be derived without assuming quantum theory. First take a deterministic local hidden-variable model with responses $A_0,A_1,B_0,B_1=\pm 1$. For each $\lambda$,because exactly one bracket vanishes. Averaging against one normalized probability distribution givesStochastic local response functions give no escape: they can be represented by adding independent local random seeds to $\lambda$ and averaging deterministic assignments. The essential restriction is the common setting-independent ensemble and local dependence of responses, not a separate insistence that all nature be deterministic.

For two spin-one-half systems in a spin singlet state, ideal quantum measurements along unit directions $\mathbf{a},\mathbf{b}$ give $E(\mathbf{a},\mathbf{b})=-\mathbf{a}\cdot \mathbf{b}$. ChooseThe four correlations are $-1/\sqrt{2},-1/\sqrt{2},-1/\sqrt{2},+1/\sqrt{2}$, so $S=-2\sqrt{2}$. Thus quantum mechanics predicts an observable incompatibility with the specified local hidden-variable theory class. This is stronger than saying that two interpretations happen to fit the observations collected so far.

Actual Bell tests require more than the ideal calculation. Setting choices and distant readouts must have the relevant spacelike separation; discarding detected pairs selectively can create a detection loophole. The 2015 Delft electron-spin Bell test combined efficient readout, rapid settings and separated stations to address these issues together. The reported statistic was $S=2.42\pm 0.20$, with a test of the local-model hypothesis accounting for device memory.^[ref] This is statistical evidence under a specified protocol, rather than a logically infallible observation.

The philosophical conclusion must be equally precise. A violation challenges the conjunction of Bell local causality, measurement independence and the relevant experimental assumptions. It does not disprove every possible hidden-variable account, or establish that the observer's mind creates the outcome. Different theories can reject different premises while agreeing on the observable correlations. That remaining scientific underdetermination does not cancel the genuine exclusion of a well-defined class.

Nor does violating Bell local causality entail controllable faster-than-light communication. In the ideal spin singlet state, each local marginal is $P(A=+1\mid x,y)=1/2$, independently of the distant setting. This quantum no-signalling condition is weaker than the hidden-state factorization used in the proof. Conflating them turns a precise result into an unwarranted conclusion about signalling.

The example therefore illustrates both the power and the limit of empirical testing in philosophy of science. Mathematical clarification can transform an apparently metaphysical disagreement into rival observational constraints. Evidence can discriminate those constraints without determining every question about the reality represented by a successful theory.

Convergence in distribution means that the cumulative distribution functions satisfyat every $x$ where the cumulative distribution function of $X$ is continuous. Values at its discontinuities need not converge. This definition uses only the probability distributions, so the random variables need not be defined on the same probability space.

Every $X_n$ has the same exponential distribution, so $X_n\stackrel{d}{\to }X$ for $X\sim \mathrm{Exp}(1)$. Part (ii) gives $Z_n-X_n\stackrel{p}{\to }0$. Applying Slutsky's theorem to $Z_n=X_n+(Z_n-X_n)$ yieldsThe limiting probability distribution does not depend on $c$. This is precisely the attraction to the exponential invariant law of a random affine recursion.

Let $X$ have the Pareto distribution with probability density function $f_X(x)=3x^{-4}$ for $x\ge 1$, and zero otherwise. Its normalization of a probability distribution and first two moments follow fromSet $Y=X-3/2$. ThenThe probability distribution of $Y$ is not a symmetric distribution: $Y\ge -1/2$ almost surely, whereas $\mathbb{P}(Y>1/2)=\mathbb{P}(X>2)=1/8>0$ and $\mathbb{P}(Y<-1/2)=0$. For $X\ge 3$ we have $|Y|\ge X/2$, soThus $\boxed{Y=X-3/2}$ has every property required.

First justify that the moments being used exist. For every $r>0$,The bound $|y{|}^k\le k!r^{-k}{e}^{r|y|}$ gives $\mathbb{E}|Y{|}^k<\infty$ for all $k$.

If $Y$ has a symmetric distribution, then $Y$ and $-Y$ have the same probability distribution, so for odd $k$,which forces $\mathbb{E}{Y}^k=0$.

Conversely, suppose every odd moment vanishes. For fixed $t\in \mathbb{R}$, the absolute values of the partial Taylor series for $e^{tY}$ are bounded by $e^{|t||Y|}$, whose expected value is finite by the preceding inequality. The dominated convergence theorem therefore permits termwise expectation:The uniqueness theorem for moment-generating functions implies equality of the probability distributions of $Y$ and $-Y$. Hence