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Let $0<\alpha <1$, $X_0=x\in [0,1]$, and let $B_{n+1}$ be a binary random variable satisfying $\mathbb{E}[B_{n+1}\mid {\mathcal{F}}_n]=X_n$. Suppose $X_n$ is ${\mathcal{F}}_n$-measurable and $B_{n+1}$ is ${\mathcal{F}}_{n+1}$-measurable. The updateremains in $[0,1]$ and defines a bounded martingale. For example, take independent uniform random variables $U_{n+1}$ independent of the current filtration and set $B_{n+1}=1_{\{U_{n+1}\le X_n\}}$.

The binary identity $B_{n+1}^2=B_{n+1}$ givesand henceThe L2-bounded martingale convergence theorem gives a terminal random variable $X_{\infty }$ with almost sure convergence and convergence in L2. Iteration followed by the bounded convergence theorem yieldsThus $X_{\infty }$ takes only $0$ and $1$, and preservation of its expected value gives a Bernoulli distribution with parameter $x$. The terminal conditional expectation representation gives the exact mean squared errorThe finite-time paths can stay strictly inside $(0,1)$ while their limit lies entirely at the endpoints. The positive conditional variance in the interior makes permanent convergence to an interior value impossible.

The Borel-Cantelli lemmas turn a series of event probabilities into a statement about an entire infinite sequence of observations. A summable sequence of probabilities forces only finitely many occurrences almost surely, regardless of independence. A divergent series forces infinitely many occurrences when the events are mutually independent events. The precise event being measured is the limsup of events.

Imagine alarms with probabilities $p_n=1/(n+1{)}^2$ or $p_n=1/(n+1)$, starting at $n=1$. Both individual probabilities tend to zero. Nevertheless, the first alarm sequence eventually stops almost surely, whereas the second keeps returning almost surely if its events are mutually independent events. Reusing one uniform random variable for every alarm instead produces the shared-uniform counterexample to the second Borel-Cantelli lemma. Thus the individual probabilities alone do not determine long-run recurrence.

A sequence converges completely to $X$ when ${\sum }_n\mathrm{Pr}(|X_n-X|>ϵ)<\infty$ for every $ϵ>0$. The first Borel-Cantelli lemma implies Almost sure convergence. For independent $X_n$, any limit in Probability is almost surely constant; the second Borel-Cantelli lemma then gives the converse from almost sure to complete convergence. independence is essential: $X_n=1_{\{U\le 1/n\}}$ for one uniform random variable $U$ converges almost surely to zero but its exceedance probabilities have a divergent sum.

Let $U_1,\dots ,U_n$ be independent uniform random variables on $[0,1]$, let ${\theta }_i>0$, and set $S={\sum }_i{\theta }_i\in (0,1)$ and $X_t={\sum }_i{\theta }_i{1}_{\{U_i\le t\}}$. ThenIndeed, the complementary crossing is that of level one by the uniform-arrival residual mass martingale $M_t=(S-X_t)/(1-t)$, with $M_0=S$. Its jumps are downward, so its first crossing of one is continuous. An arrival coincides with a possible crossing time with probability zero: the possible times belong to the finite deterministic set obtained from subsets of the weights. Stop at that crossing or at $r<1$. The stopped values are in $[0,1]$, and as $r\uparrow 1$ they converge to the crossing indicator because all arrivals occur before one almost surely. The optional stopping theorem and dominated convergence theorem give crossing probability $S$. For $S=1$, the strict displayed inequality already fails at time zero.

Let $({\mathcal{F}}_n)$ be an increasing filtration and ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$. Suppose $X_n\to X$ almost surely and $|X_n|\le Y$ for one integrable random variable $Y$. Thenalmost surely and in L1 space. To prove this, put $Z_n={\sup }_{m\ge n}|X_m-X|$. Then $Z_n\downarrow 0$, $Z_n\le 2Y$, and $\mathbb{E}[Z_n\mid {\mathcal{F}}_n]$ is a nonnegative supermartingale whose expectation tends to zero. Its almost-sure limit is therefore zero. This bounds the moving-variable error; the conditional-expectation convergence along a filtration, extended from bounded to integrable variables by truncation, handles $X$ itself. Dominated convergence also makes the error tend to zero in L1 space.

A separate finite bound for each $X_n$ is insufficient. For a uniform random variable $U$ on $(0,1)$, take $X_n=n{1}_{\{U<1/n\}}$ and trivial ${\mathcal{F}}_n$. Then $X_n\to 0$ almost surely, but $\mathbb{E}{X}_n=1$ for every $n$.

Integrating the Pareto distribution proposal yields $G(x)=1-(a/x{)}^b$ for $x\ge a$. For an independent uniform random variable $U$, solve $G(Y)=U$ to obtain the inverse transform sampling ruleEquivalently $Y=a{U}^{-1/b}$ because $1-U$ is also uniformly distributed. In implementation choose $U$ strictly between zero and one so that the inverse is finite.

Conditioned on the parameters, the data augmentation labels are independent. By Bayes' theorem, the conditional probability of component $i$ for observation $j$ isThis is the mixture responsibility. Draw an independent uniform random variable and select $i$ using the cumulative sums of $(r_{j1},\dots ,r_{jk})$. Refresh every label before recomputing the component sufficient statistics for the next Gibbs sampler sweep. Numerically, form the log scores $\log {\omega }_i-\frac{1}{2}\log v_i-(x_j-{\mu }_i{)}^2/(2v_i)$, subtract their maximum and exponentiate before normalization.

Draw $n$ independent pairs of independent uniform random variables on $[0,1]$. In each pair call the minimum $L$ and maximum $R$, then relabel whole pairs so $R_1<\cdots <R_n$. Merge all endpoints in $(R_{i-1},R_i]$ into vertex $i$, with $R_0=0$, and join the blocks containing the two endpoints of each pair, orienting from the right endpoint to the left. Self-loops are retained. This gives the uniform linearized chord diagram representation.

Before ordering, a pair has joint density $2$ on $0<L<R<1$. Consequently its maximum has distribution $\mathrm{Pr}(R\le r)=r^2$ and density $2r$; conditional on $R=r$, its minimum is uniform on $[0,r]$. Independence and relabelling give the equivalent precise descriptionEvery one of the $(2n)!$ relative orderings of the independently drawn labelled endpoints has the same probability. After forgetting pair labels and within-pair labels, each matching of $2n$ ordered points is produced by $2^{n}n!$ orderings. Thus its matching is uniform among the $(2n-1)!!$ pairings.

To prove equivalence with the recursive graph law, remove the last endpoint and its mate from a uniform pairing. The remaining ordered points have a uniform $(n-1)$-pairing, and the mate's old rank is uniform among the $2n-1$ possible ranks, independently of that smaller pairing. Conversely insert the mate into a uniformly chosen one of its $2n-1$ gaps, and pair it with the new final endpoint. For each old vertex block, the number of gaps assigned to it is exactly the number of endpoints already in that block, hence its old total vertex degree. These are the gaps immediately before each endpoint of that block. The last gap, after the last old endpoint, makes both new endpoints belong to the new final block and produces a self-loop. Therefore the attachment probabilities are $d_i/(2n-1)$ for old vertices and $1/(2n-1)$ for the new one. Induction proves the continuous construction and the recursive LCD model have the same distribution. Inserting a left endpoint into an old block raises exactly that vertex's vertex degree without splitting any old block.

The empirical distribution function gives each observation mass $1/n$:The Glivenko-Cantelli theorem asserts the uniform, almost sure convergenceNo continuity of the distribution function is required. Here is a proof that also handles its jumps. Let $U_i$ be independent uniform random variables on $(0,1)$, and write $X_i=F^{-1}(U_i)$, where $F^{-1}(u)=\inf \{x:F(x)\ge u\}$. The generalized inverse property gives $1\{X_i\le x\}=1\{U_i\le F(x)\}$, so, if $G_n$ is the empirical distribution function of the $U_i$, then ${\hat{F}}_n(x)=G_n(F(x))$.

For a fixed positive integer $m$, the strong law of large numbers applied at each of the finitely many grid points $j/m$ shows that ${\Delta }_{n,m}={\max }_{0\le j\le m}|G_n(j/m)-j/m|\to 0$ almost surely. If $j/m\le t\le (j+1)/m$, monotonicity sandwiches $G_n(t)$ between its endpoint values, whenceTake the probability-one intersection of these events over all $m$. On that event the limit superior is at most $1/m$ for every $m$, hence is zero. Therefore ${\sup }_x|G_n(F(x))-F(x)|\to 0$. The coupled sequence has the same joint law as any independent sample from $F$, which proves the asserted almost sure conclusion for the original sample.

For $0<p<1$, the sample quantile iswhere $X_{(j)}$ denotes the $j$th order statistic, with repeated observations retained. Suppose the population median $m=F^{-1}(1/2)$ is unique and $F$ is continuously differentiable in a neighbourhood of $m$, with probability density function $f(m)>0$. ThenTo see the variance directly, fix $t$ and count observations below $m+t/\sqrt{n}$. This count has a binomial distribution with success probability $q_n=1/2+f(m)t/\sqrt{n}+o(n^{-1/2})$. The event that the sample median is below this point is that the count is at least $\lceil n/2\rceil$. The central limit theorem for this binomial triangular array gives a limiting probability $\Phi (2f(m)t)$, exactly the displayed normal distribution. Centering is necessary for stating the limiting distribution; subtracting the deterministic population median has no effect on the asymptotic variance requested in the comparisons.

For the Gillespie algorithm, start from $t=0$ and an integer molecular copy number $x\ge 0$. At each step use the current state, without the mean approximation in parts (a)–(b). The reaction propensity functions are $a_{+}(x)=\lambda$ and $a_{-}(x)=\beta \sqrt{x}$, and their sum is $a_0(x)=\lambda +\beta \sqrt{x}$. Draw two independent uniform random variables $U_1,U_2$ on $(0,1)$ and setThe waiting time has an exponential distribution with rate $a_0(x)$. Keep the sample path constant on $[t,t+\Delta t)$, advance $t\leftarrow t+\Delta t$, and choose the reaction byRecord the post-jump state, recompute both reaction propensity functions, and repeat with fresh independent uniforms. At $x=0$ the death propensity is zero, so only a birth is selected and the trajectory remains nonnegative. For a prescribed terminal time $T$, if the next event would occur after $T$, keep the current state to $T$ and stop. This produces exact piecewise constant sample paths of the specified continuous-time Markov chain.

Integrating the uniform density gives $\mathbb{E}{X}^k={\int }_0^1{x}^{k}dx=1/(k+1)$. By independence of random variables, $\mathbb{E}(XY{)}^k=\mathbb{E}{X}^k\mathbb{E}{Y}^k=1/(k+1{)}^2$. For the remaining expected value, integrate first in $x$:The value at $y=0$ is supplied by continuity. Therefore, with the harmonic number $H_m={\sum }_{j=1}^{m}1/j$,

Conditional on one sample point being $(x,y)$, it is a maximal external point precisely when none of the other $n-1$ points falls in the upper-right rectangle $(x,1]\times (y,1]$. Its area is $(1-x)(1-y)$, so its conditional probability of being maximal is $[1-(1-x)(1-y){]}^{n-1}$. Both $1-X$ and $1-Y$ are independent uniform random variables, hence a given point is maximal with probability $H_n/n$. Summing their indicator random variables and applying linearity of expectation gives the expected number of coordinatewise maxima:This grows only logarithmically with the sample size, even though all $n$ points are potential candidates.

Interpret the contests as standard all-pay auctions: the highest effort among entrants wins, ties are shared uniformly, and an unentered contest awards no prize. The printed question does not specify a prize allocation rule; the highest-effort convention is the one used for standard all-pay contests in the course author's 2014 lecture slides. Under this convention, we can characterize the unique symmetric participation probabilities and bid marginals. Uniqueness of the entire joint mixed strategy requires a further restriction on dependence, as explained below.

Let $q_j$ be the probability that a player omits contest $j$. Since every player enters exactly two contests, $q_1+q_2+q_3=1$. Let $G_j(b)$ be the probability that a rival is absent from contest $j$ or enters it with effort at most $b$. Rivals' strategy draws are independent between players. For a positive bid $b$ outside a measure atom, the expected payoff from this contest isAn all-pay auction cannot have a positive-effort measure atom in a symmetric equilibrium: slightly overbidding that measure atom gives a positive discrete increase in the winning probability at an arbitrarily small extra cost. Nor can active bids have a measure atom at zero when entry has positive probability, because a small positive bid beats tied zero bids. Gaps inside the active effort support are impossible: moving a bid from the top of a gap to just above its bottom preserves its winning probability and lowers its cost. The effort support starts at zero, because lowering its positive lower endpoint would preserve the chance that every rival is absent.

The all-pay indifference equation with random entry consequently gives the maximal per-contest payoffEvery $q_j$ lies strictly between zero and one. First, if $q_j=1$, the other two contests have certain entry and zero per-contest payoff, while deviating into the unused contest wins a positive prize. This contradicts equilibrium. Next, if $q_j=0$, the remaining omission probabilities sum to one and neither can equal one, so both are positive. Contest $j$ has zero per-contest payoff, whereas both other contests have strictly positive payoffs. Every pair containing $j$ is then worse than omitting it and entering the other two, contradicting certain entry in $j$.

Every omission therefore occurs with positive probability. The three entered pairs must give the same maximal payoff, which forces $u_1=u_2=u_3=u>0$. Normalizing the omission probabilities gives the two-of-three all-pay participation equilibrium:Conditional on entering contest $j$, the effort has distribution functionextended by zero below this interval and one above it. The inequalities $q_1\le q_2\le q_3$ show that the larger prizes are entered more often.

To construct an equilibrium, omit $j$ with probability $q_j$, then draw the two active efforts independently with their respective conditional distributions $H_k$. Every bid in a contest's effort support earns $u$; a bid above $w_j-u$ earns at most $w_j-b<u$. Thus no effort deviation or choice of a different pair improves on total payoff $2u$. This proves existence and verifies the Nash equilibrium without relying only on the indifference equations. The arguments above also prove uniqueness of the omission probabilities and the per-contest marginal distributions.

For the full joint strategy law, however, the printed uniqueness claim is too strong. Given an entered pair $k,\ell$, either use two independent uniform random variables $U,V$ and bids $(H_k^{-1}(U),H_{\ell }^{-1}(V))$, or use a single uniform $U$ and bids $(H_k^{-1}(U),H_{\ell }^{-1}(U))$. These are different copulas with the same conditional marginals. A fixed deviation's expected additive payoff only uses the rivals' per-contest marginal distributions, so both constructions remain Nash equilibria. This is the marginal-equivalent equilibria in additive contests phenomenon. The participation probabilities and bid marginals are unique; the full joint mixed strategy is not unique unless a dependence convention is imposed. Independent conditional sampling gives one canonical representative.

Retain the binary random variables $B_j=1_{\{U_j\le X_{j-1}\}}$. Iterating the recursion givesEach binary history determines the full vector $(X_0,\dots ,X_n)$, so its sigma-algebra is finite by the sigma-algebra of a finite-valued random vector. Conversely, because $0<x<1$, the integer $2^n{X}_n-x$ uniquely determines the digits $B_1,\dots ,B_n$; also $B_j=2X_j-X_{j-1}$. ThusAll $2^n$ binary histories have positive probability. By independence of the driving uniform random variables, along a fixed finite history every intermediate state is in $(0,1)$, so each subsequent conditional branch probability, either $X_{j-1}$ or $1-X_{j-1}$, is strictly positive. It follows that ${\mathcal{F}}_n^X$ has $2^n$ nonempty fibres, each an atom of a sigma-algebra, and exactly $2^{2^n}$ elements.

Adaptedness from the preceding part gives ${\mathcal{F}}_n^X\subseteq {\mathcal{F}}_n$. For $n\ge 1$, however, ${\mathcal{F}}_n$ contains the events $\{U_1\le t\}$ for every $t\in [0,1]$. They have distinct probabilities $t$, so there are infinitely many distinct such events. Consequently the finite sigma-algebra ${\mathcal{F}}_n^X$ cannot equal ${\mathcal{F}}_n$. The inclusion is strict for every $n\ge 1$.

If $F_n\to F$ at every continuity point of $F$, then the generalized inverses satisfyat every continuity point of $F^{-1}$. Since a monotone function has only countably many discontinuities, evaluating all quantiles at one uniform random variable gives an almost-sure coupling of the corresponding convergence in distribution.

For an integrable real random variable with distribution function $F$, the generalized inverse $F^{\leftarrow }(u)=\inf \{x:F(x)\ge u\}$ sends a uniform random variable on $(0,1)$ to the distribution $F$. This follows from $F^{\leftarrow }(u)\le x$ if and only if $u\le F(x)$. Equality of laws proves both the expectation formula and absolute integrability of the quantile. Atoms and flat pieces of $F$ cause no problem.

A randomized statistical test specifies a measurable rejection probability $\varphi (x)$ for each possible data value. It can be implemented by rejecting when an independent uniform random variable $U$ on $(0,1)$ satisfies $U<\varphi (x)$. A deterministic test has only the values zero and one. Under a simple hypothesis $H_j$, its rejection probability is ${\mathbb{E}}_j\varphi (X)$, which is its test size for $j=0$ and its statistical power for $j=1$.

The Neyman-Pearson lemma permits randomization on an equality set of the form $f_1(x)=k{f}_0(x)$ to reach a prescribed test size. Continuous observations do not rule out an equality set of positive probability: if both hypotheses give the same probability density function, their likelihood ratio is identically one. This explains why a precise statement of the lemma includes boundary randomization even for continuous data.

For independent uniform random variables on $(\theta ,3\theta )$ with $\theta >0$, the sample minimum and maximum form a sufficient statistic. Conditional on these endpoints, exchangeability and the uniform law of the remaining observations give $E[X_1\mid X_{(1)},X_{(n)}]=(X_{(1)}+X_{(n)})/2$. Applying the Rao-Blackwell theorem to $X_1/2$ gives the displayed unbiased estimator. Its variance is ${\theta }^2/[2(n+1)(n+2)]$, compared with ${\theta }^2/12$ for $X_1/2$.

Let $m$ be a positive integer, $r>0$, and $U_1,\dots ,U_m$ independent uniform random variables on $(0,1)$. Thenhas a gamma distribution of shape $m$ and rate $r$, also called an Erlang distribution. The unit interval is essential: an unspecified uniform distribution on another interval would not give this algorithm.

For $t\ge 0$, $\mathbb{P}(-\log U_j/r>t)=\mathbb{P}(U_j<e^{-rt})=e^{-rt}$. Each summand therefore has an exponential distribution with rate $r$, and additivity of independent gamma distributions with a common rate proves the result. For $m=3,r=2$, the output has expected value $3/2$ and variance $3/4$. The formula directly covers integer shapes; replacing $m$ by a noninteger number of draws does not define a sampler for a general gamma shape.

Choose one random variable $U$ with the continuous uniform distribution on $[0,1]$, and reuse it in every event:Each probability is $1/(n+1)$, so their harmonic series diverges, just as in summable and harmonic event probabilities. But the events are decreasing: $A_{n+1}\subseteq A_n$. For every outcome $U=u>0$, all indices with $n+1>1/u$ fail. Only finitely many events occur on that outcome. The limsup of events is exactly $\{U=0\}$, a zero-probability event. This is a concrete closed-interval version of the shrinking-interval dependence example.^[ref]

The missing assumption is visible already in two events. For $i<j$, their intersection is $A_j$, henceMoreover, the whole tail union is just $A_m$, with probability $1/(m+1)\to 0$. This calculation explains directly why divergence alone fails: overlapping events keep spending their individual probabilities on the same outcomes.

Let $N={\sum }_{n\ge 1}{1}_{A_n}$ be the total number of occurrences, a sum of indicator random variables. Although $N<\infty$ almost surely, Tonelli theorem gives $\mathbb{E}[N]={\sum }_{n\ge 1}1/(n+1)=\infty$. More explicitly, $\mathbb{P}(N\ge k)=1/(k+1)$ for $k\ge 1$: very small positive values of $U$ give very long, but finite, initial runs. Infinite expected value does not imply a positive probability of an infinite count.

Let a random vector $V$ have a finite actual range of $q$ values. Its nonempty fibres partition the sample space, and every event in $\sigma (V)$ is a union of those fibres. Thus each fibre is an atom of a sigma-algebra and $\sigma (V)$ has exactly $2^q$ elements. This assertion concerns the actual range; additional values attained only on null sets must still be counted when working with an uncompleted sigma-algebra literally.

To verify the claim, the fibres belong to $\sigma (V)$ as preimages of singleton Borel sets. Their unions form a finite sigma-algebra making $V$ measurable, so minimality of the generated sigma-algebra gives the reverse inclusion. Distinct subcollections have distinct unions because the fibres are nonempty and disjoint.

Consequently a finite history of finite-valued observations generates a finite natural filtration at each fixed time. It may be strictly smaller than the filtration of its driving continuous observations: a single uniform random variable already generates infinitely many distinct events $\{U\le t\}$ with different probabilities.