Friday 28 August 2026

Part 1: distribution of one spacing

By exchangeability of the $n+1$ spacings (the vector $(D_1,\dots ,D_{n+1})$ is uniformly distributed on the simplex $\{d_i\ge 0,\sum d_i=1\}$), all $D_i$ share one marginal law. Take $D_1=U_{(1)}$, the minimum of $n$ i.i.d. uniforms: $$\mathbb{P}(D_1>t)=\mathbb{P}(\text{all }n\text{ points}>t)=(1-t{)}^n,0\le t\le 1.$$ Differentiating, $D_i$ has density $n(1-t{)}^{n-1}$, i.e. $D_i~\mathrm{B}\mathrm{e}\mathrm{t}\mathrm{a}(1,n)$, with $$𝔼[D_i]=\frac{1}{n+1}.$$

Part 2: the gap containing a fixed point

Write $L(x)=A+B$, where $$A=x-max\{U_j:U_j<x\}(=x\text{ if none}),B=min\{U_j:U_j>x\}-x(=1-x\text{ if none}).$$ The step candidates miss is that $L(x)$ is not a typical spacing: conditioning on containing $x$ length-biases the gap, and there is a boundary correction. The clean route is the tail-integral formula $𝔼[A]={\int }_0^{\infty }\mathbb{P}(A>a)da$.

For $0\le a\le x$, the event $\{A>a\}$ means no sample point falls in the interval $(x-a,x)$ of length $a$: $$\mathbb{P}(A>a)=(1-a{)}^n.$$ Since $A\le x$ always (there is an atom of mass $(1-x{)}^n$ at $A=x$, automatically included by the tail integral), $$𝔼[A]={\int }_0^x(1-a{)}^{n}da=\frac{1-(1-x{)}^{n+1}}{n+1}.$$ Symmetrically, $\mathbb{P}(B>b)=(1-b{)}^n$ for $0\le b\le 1-x$, so $$𝔼[B]={\int }_0^{1-x}(1-b{)}^{n}db=\frac{1-x^{n+1}}{n+1}.$$ Adding, $$𝔼[L(x)]=\frac{2-(1-x{)}^{n+1}-x^{n+1}}{n+1}.$$

Sanity check at $x=\frac{1}{2}$: $𝔼[L]=\frac{2(1-2^{-(n+1)})}{n+1}\approx \frac{2}{n+1}$, roughly twice the mean spacing $\frac{1}{n+1}$ — the inspection paradox: a fixed point is more likely to fall inside a long gap.

Part 3: the gap containing an independent uniform point

Integrate the part-2 result over $x~\mathrm{U}\mathrm{n}\mathrm{i}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}(0,1)$: $$𝔼[L(X)]={\int }_0^1\frac{2-(1-x{)}^{n+1}-x^{n+1}}{n+1}dx=\frac{1}{n+1}(2-\frac{1}{n+2}-\frac{1}{n+2}).$$ $$𝔼[L(X)]=\frac{1}{n+1}·\frac{2(n+1)}{n+2}=\frac{2}{n+2}.$$

Reconciliation (length-biased sampling). Given the spacings, the probability that $X$ lands in gap $i$ is exactly $D_i$. Hence $$𝔼[L(X)]=𝔼[{\sum }_{i=1}^{n+1}{D}_i·D_i]=(n+1)𝔼[D_i^2]=\frac{𝔼[D_i^2]}{𝔼[D_i]},$$ the length-biased mean. For $\mathrm{B}\mathrm{e}\mathrm{t}\mathrm{a}(1,n)$, $𝔼[D_i^2]=\frac{2}{(n+1)(n+2)}$, so $$𝔼[L(X)]=\frac{2/((n+1)(n+2))}{1/(n+1)}=\frac{2}{n+2},$$ agreeing with the integral. The random point sees gaps weighted by their length, so its expected gap is the size-biased mean, strictly larger than the plain mean $\frac{1}{n+1}$ whenever $n\ge 1$.

Answers

1. $D_i~\mathrm{B}\mathrm{e}\mathrm{t}\mathrm{a}(1,n)$, mean ${\displaystyle \frac{1}{n+1}}$.
2. $𝔼[L(x)]={\displaystyle \frac{2-(1-x{)}^{n+1}-x^{n+1}}{n+1}}.$
3. $𝔼[L(X)]={\displaystyle \frac{2}{n+2}}$, the length-biased mean $𝔼[D^2]/𝔼[D]$.

Tempting wrong answer: quoting $\frac{1}{n+1}$ (the plain spacing mean) for either part 2 or part 3 — this ignores that requiring a gap to contain a given point biases toward longer gaps.

Throughout, write $\Pi =\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}(\pi )$ and regard a kernel $P$ as a matrix. The identity that organizes everything: $P$ is $\pi$-reversible (detailed balance) $⟺$ the matrix $\Pi P$ is symmetric $⟺$ $P$ is self-adjoint on $L^2(\pi )$ with $⟨f,g{⟩}_{\pi }={\sum }_x\pi (x)f(x)g(x)$.

Part 1: each coordinate update is reversible

$P_1$ only connects states sharing the same $x_2$, with $$P_1((x_1,x_2)\to (x_1^{\prime },x_2))=\pi (x_1^{\prime }\mid x_2)=\frac{\pi (x_1^{\prime },x_2)}{\pi (x_2)}.$$ Hence $$\pi (x_1,x_2)P_1((x_1,x_2)\to (x_1^{\prime },x_2))=\frac{\pi (x_1,x_2)\pi (x_1^{\prime },x_2)}{\pi (x_2)},$$ which is symmetric under swapping $x_1\leftrightarrow x_1^{\prime }$. For transitions that change $x_2$ both sides are $0$. So $P_1$ satisfies detailed balance; by symmetry so does $P_2$.

Detailed balance is preserved by convex combinations: if $\pi (x)P_i(x,y)=\pi (y)P_i(y,x)$ for $i=1,2$, then averaging the two identities gives $\pi (x)P_{\mathrm{r}\mathrm{s}}(x,y)=\pi (y)P_{\mathrm{r}\mathrm{s}}(y,x)$. Thus $P_{\mathrm{r}\mathrm{s}}$ is $\pi$-reversible.

Part 2: the systematic scan breaks detailed balance

First the conditionals. Marginals of $x_2$: $\pi (x_2=0)=0.1+0.3=0.4$, $\pi (x_2=1)=0.6$; marginals of $x_1$: $\pi (x_1=0)=0.3$, $\pi (x_1=1)=0.7$. Then $$\pi (x_1=1\mid x_2=0)=\frac{0.3}{0.4}=0.75,\pi (x_2=1\mid x_1=1)=\frac{0.4}{0.7}=\frac{4}{7},$$ $$\pi (x_1=0\mid x_2=1)=\frac{0.2}{0.6}=\frac{1}{3},\pi (x_2=0\mid x_1=0)=\frac{0.1}{0.3}=\frac{1}{3}.$$

Under $P_{\mathrm{s}\mathrm{y}\mathrm{s}}=P_1{P}_2$ we redraw $x_1$ from $\pi (·\mid x_2^{\text{old}})$, then redraw $x_2$ from $\pi (·\mid x_1^{\text{new}})$, so $$P_{\mathrm{s}\mathrm{y}\mathrm{s}}((x_1,x_2)\to (x_1^{\prime },x_2^{\prime }))=\pi (x_1^{\prime }\mid x_2)\pi (x_2^{\prime }\mid x_1^{\prime }).$$

For $A=(0,0)\to B=(1,1)$: $$P_{\mathrm{s}\mathrm{y}\mathrm{s}}(A\to B)=\pi (x_1=1\mid x_2=0)\pi (x_2=1\mid x_1=1)=0.75·\frac{4}{7}=\frac{3}{7},$$ $$\pi (A)P_{\mathrm{s}\mathrm{y}\mathrm{s}}(A\to B)=0.1·\frac{3}{7}=\frac{0.3}{7}\approx 0.042857.$$

For $B=(1,1)\to A=(0,0)$: $$P_{\mathrm{s}\mathrm{y}\mathrm{s}}(B\to A)=\pi (x_1=0\mid x_2=1)\pi (x_2=0\mid x_1=0)=\frac{1}{3}·\frac{1}{3}=\frac{1}{9},$$ $$\pi (B)P_{\mathrm{s}\mathrm{y}\mathrm{s}}(B\to A)=0.4·\frac{1}{9}=\frac{0.4}{9}\approx 0.044444.$$

Since $0.042857\ne 0.044444$, detailed balance fails: $P_{\mathrm{s}\mathrm{y}\mathrm{s}}$ is not $\pi$-reversible.

Stationarity still holds, however. $\pi$ is a left eigenvector of each factor ($\pi P_1=\pi$ because a full conditional resample leaves the joint invariant, likewise $\pi P_2=\pi$), so $\pi P_{\mathrm{s}\mathrm{y}\mathrm{s}}=\pi P_1{P}_2=\pi P_2=\pi$. This is the point candidates most often miss: a systematic-scan Gibbs sampler has the correct invariant distribution yet is generically not reversible.

Part 3: the commutation criterion

Reversibility is self-adjointness on $L^2(\pi )$. If $P_1,P_2$ are each self-adjoint, then $$(P_1{P}_2{)}^{*}=P_2^{*}{P}_1^{*}=P_2{P}_1,$$ using $(AB{)}^{*}=B^{*}{A}^{*}$ and $P_i^{*}=P_i$. Hence $P_1{P}_2$ is self-adjoint (i.e. $\pi$-reversible) iff $P_1{P}_2=P_2{P}_1$.

In the numeric example the two updates do not commute, which is exactly why detailed balance broke.

A scan order guaranteed reversible: the palindromic sweep $P_1{P}_2{P}_1$ (update $1$, then $2$, then $1$ again). Indeed $$(P_1{P}_2{P}_1{)}^{*}=P_1^{*}{P}_2^{*}{P}_1^{*}=P_1{P}_2{P}_1,$$ so it is self-adjoint regardless of whether $P_1$ and $P_2$ commute. (One must halve the effective work of the repeated block, but reversibility is restored.)

Answer

1. Each $P_i$ satisfies detailed balance; $P_{\mathrm{r}\mathrm{s}}$ is $\pi$-reversible as a convex combination.
2. $\pi (A)P_{\mathrm{s}\mathrm{y}\mathrm{s}}(A\to B)=0.3/7\approx 0.04286$ while $\pi (B)P_{\mathrm{s}\mathrm{y}\mathrm{s}}(B\to A)=0.4/9\approx 0.04444$; unequal, so $P_{\mathrm{s}\mathrm{y}\mathrm{s}}$ is not reversible, though $\pi P_{\mathrm{s}\mathrm{y}\mathrm{s}}=\pi$.
3. $P_1{P}_2$ is $\pi$-reversible iff $P_1{P}_2=P_2{P}_1$; the palindromic scan $P_1{P}_2{P}_1$ is always reversible.

Note. A tempting error is to reason "detailed balance for each block $\Rightarrow$ detailed balance for the sweep." Convex combinations inherit detailed balance (part 1), but compositions do not, because self-adjoint operators need not have self-adjoint products. Reversibility is not required for a valid MCMC sampler — only invariance is — but many CLT and spectral-gap results assume it, which is one practical reason the palindromic (or random) scan is preferred.