Polya urn proportion martingale

Use $\mathrm{\Omega }=(0,1]$ with the trace Lebesgue sigma-algebra and length probability. Restriction is a measure by [F2] (equivalently restrict its event formula to subsets of $\mathrm{\Omega }$), and [F1] gives total mass one. Define intervals for finite color words recursively: $I_{\mathrm{\varnothing }}=(0,1]$. If $h$ has length $n$, contains $j$ red letters, and $I_h=(a,b]$, put $T_n=r+g+nc$ and $p_h=(r+cj)\mathrm{/}{T}_n$. Then $0<p_h<1$ because both $r+cj$ and $g+c(n-j)$ are positive. Set $I_{hR}=(a,a+p_h(b-a)]$ and $I_{hG}=(a+p_h(b-a),b]$. Both are positive-length half-open intervals Half-open boxes in ${\mathbb{R}}^n$ and their volume, disjoint with union $I_h$. Induction gives a finite partition at every depth, and each point has exactly one compatible word of every finite length. Thus every draw is defined on this one space by the interval containing the point; no infinite-product existence is assumed.

Let ${\mathcal{F}}_n$ consist of all unions of depth-n intervals. Complements and countable unions just select subsets of this finite partition, so it is a sigma-algebra. Refinement makes it a filtration. A word interval is exactly the intersection of the corresponding first n color events; conversely a color event up to n is a union of word intervals. Hence ${\mathcal{F}}_n$ is the draw-history sigma-algebra. On $I_h$ set $R_n=r+cj$ and $U_n=R_n\mathrm{/}{T}_n$. These are finite-valued ${\mathcal{F}}_n$-measurable variables and $0<U_n<1$. If $J_{n+1}$ indicates the next red draw, then ${\int }_{I_h}{J}_{n+1}dP=P(I_{hR})=p_{h}P(I_h)={\int }_{I_h}{U}_{n}dP$. Finite addition proves this identity for every event in ${\mathcal{F}}_n$. Both variables are bounded, so [F3] identifies $E[J_{n+1}\mid {\mathcal{F}}_n]=U_n$.

The pathwise update is $R_{n+1}=R_n+c{J}_{n+1}$. The bounded ${\mathcal{F}}_n$-measurable $R_n$ is its own conditional version, since its event identities are tautologies. Since $T_n+c$ is deterministic and positive, conditional linearity gives $E[U_{n+1}\mid {\mathcal{F}}_n]=(R_n+c{U}_n)\mathrm{/}(T_n+c)=R_n(1+c\mathrm{/}{T}_n)\mathrm{/}(T_n+c)=R_n\mathrm{/}{T}_n=U_n$. Thus the bounded adapted process is a martingale Martingale submartingale and supermartingale. For $r=g=c=1$, $U_0=1\mathrm{/}2$ and the first two possible new proportions are $2\mathrm{/}3$ and $1\mathrm{/}3$, each with probability $1\mathrm{/}2$; their mean is $1\mathrm{/}2$. After the first red draw the next proportions are $3\mathrm{/}4$ and $1\mathrm{/}2$ with probabilities $2\mathrm{/}3$ and $1\mathrm{/}3$, whose mean is $1\mathrm{/}2+1\mathrm{/}6=2\mathrm{/}3$. AC supplies the countable choice assumed in Lebesgue construction and CE existence; the interval recursion itself makes no selections.