Existence of a Dirichlet process by normalized gamma Poisson random measure

Put $\rho (dt)=t^{-1}{e}^{-t}dt$ on $(0,\mathrm{\infty })$; [A2] and [F22] make this a measure on the Borel sigma-algebra with a nonnegative density relative to Lebesgue measure, and partition the mark axis into $I_0=[1,\mathrm{\infty })$ and $I_m=[1\mathrm{/}(m+1),1\mathrm{/}m)$ for $m\ge 1$. These disjoint strips cover $(0,\mathrm{\infty })$. For $m\ge 1$, $\rho (I_m)\le 1\mathrm{/}m$ because $t^{-1}\le m+1$ and $e^{-t}\le 1$ there, while $\rho (I_m)\ge 1\mathrm{/}(e(m+1))>0$ because $t^{-1}\ge m$, $e^{-t}\ge e^{-1}$, and the interval length is $1\mathrm{/}(m(m+1))$. Also $\rho (I_0)\le {\int }_1^{\mathrm{\infty }}{e}^{-t}dt<\mathrm{\infty }$ and $\rho (I_0)\ge e^{-2}\mathrm{/}2>0$ by restricting to $[1,2]$. Thus every strip has finite positive mass. By [F3] and [F14], $\mu =\alpha \otimes \rho$ is sigma-finite and ${\mu }_m:=\mu {\mathrm{\mid }}_{\mathsf{X}\times I_m}$ has finite positive total ${\lambda }_m=\alpha (\mathsf{X})\rho (I_m)$. Let $Q_m={\mu }_m\mathrm{/}{\lambda }_m$.

Index count coordinates by $m\ge 0$ and mark coordinates by $(m,\mathrm{\ell })$ with $m\ge 0,\mathrm{\ell }\ge 1$. Their spaces are ${\mathbb{N}}_0$ and $\mathsf{X}\times (0,\mathrm{\infty })$, respectively, hence standard Borel by [F1], [F21], and [F23]. Assign every finite set of coordinates the product of the laws $\mathrm{Pois}({\lambda }_m)$ and $Q_m$. These finite-dimensional laws are consistent because coordinate projections of a finite product recover the corresponding marginal products. By [F1] and [A1] they extend to one probability law. The product rectangle formula gives mutual independence of each finite coordinate family by [F20], and [F13] identifies its joint law with the product. Thus it supports mutually independent $N_m\sim \mathrm{Pois}({\lambda }_m)$ and $Z_{m\mathrm{\ell }}=(X_{m\mathrm{\ell }},T_{m\mathrm{\ell }})\sim Q_m$.

Define $N={\sum }_{m\ge 0}{\sum }_{\mathrm{\ell }\ge 1}{\mathbf{1}}_{\{\mathrm{\ell }\le N_m\}}{\delta }_{Z_{m\mathrm{\ell }}}$. Each $N_m$ is finite, so this is a countable sum of Dirac measures and is a counting measure by [F8]. For every measurable $B$, its evaluation is the increasing limit of finite sums of measurable functions of the coordinates; hence $N(B)$ is measurable. Thus $N$ is a random counting measure.

Fix pairwise-disjoint measurable sets $B_1,\dots ,B_r$ with $\mu (B_j)<\mathrm{\infty }$. Within strip $m$, put $q_{jm}=Q_m(B_j)$ and $q_{0m}=1-{\sum }_{j=1}^r{q}_{jm}$. Conditional on $N_m=n$, independent marks assign counts $(k_0,k_1,\dots ,k_r)$ with $n=k_0+\cdots +k_r$. There are $n!\mathrm{/}{\prod }_{j=0}^r{k}_j!$ assignments of the $n$ mark indices to these cells, and each assignment has probability ${\prod }_{j=0}^r{q}_{jm}^{k_j}$, so the count vector has mass $n!{\prod }_{j=0}^r{q}_{jm}^{k_j}\mathrm{/}{\prod }_{j=0}^r{k}_j!$ (with $0^0=1$). Summing over $k_0$ and using the Poisson mass in [F2] gives $\mathrm{Pr}\{N_m(B_j)=k_j,1\le j\le r\}=e^{-{\lambda }_m}({\prod }_{j=1}^r({\lambda }_m{q}_{jm}{)}^{k_j}\mathrm{/}{k}_j!)e^{{\lambda }_m{q}_{0m}}={\prod }_{j=1}^r{e}^{-{\mu }_m(B_j)}{\mu }_m(B_j{)}^{k_j}\mathrm{/}{k}_j!$; here ${\lambda }_m{q}_{jm}={\mu }_m(B_j)$ and ${\lambda }_m(1-q_{0m})={\sum }_j{\mu }_m(B_j)$. Thus the strip counts are independent Poisson variables with means ${\mu }_m(B_j)$. The series over $k_0$ is the exponential series [F17].

Define $\mathrm{\Gamma }(A)={\sum }_{m\ge 0}{\sum }_{\mathrm{\ell }\ge 1}{\mathbf{1}}_{\{\mathrm{\ell }\le N_m\}}{T}_{m\mathrm{\ell }}{\mathbf{1}}_{\{X_{m\mathrm{\ell }}\in A\}}$ for $A\in \mathcal{A}$. This is a measurable nonnegative weighted sum of Dirac measures, hence a measure by [F8]. For each strip, independence of $N_m$ and its marks [F13], expectation against their law [F11], and $\mathbb{E}{N}_m={\lambda }_m$ from [F2] give $\mathbb{E}{\sum }_{\mathrm{\ell }=1}^{N_m}{T}_{m\mathrm{\ell }}{\mathbf{1}}_{\{X_{m\mathrm{\ell }}\in A\}}={\lambda }_m\int t{\mathbf{1}}_A(x)Q_m(dx,dt)={\int }_{A\times I_m}t\mu (dx,dt)$. Monotone convergence [F16] then gives $\mathbb{E}\mathrm{\Gamma }(A)={\sum }_m{\int }_{A\times I_m}t\mu (dx,dt)=\alpha (A){\int }_0^{\mathrm{\infty }}{e}^{-t}dt=\alpha (A)$ by [F3] and [F6].

Let $A_1,\dots ,A_r$ be pairwise disjoint and choose $\mathrm{\mid }{u}_j\mathrm{\mid }<\delta <1$. On strip $m$, define $g_m(x,t)={\sum }_{j=1}^r{e}^{u_{j}t}{\mathbf{1}}_{A_j}(x)+{\mathbf{1}}_{\mathsf{X}\setminus {\cup }_j{A}_j}(x)$. Conditional on $N_m=n$, the independent marks give the exponential factor $(\int g_{m}d{Q}_m{)}^n$. Summing its Poisson series yields $\mathbb{E}\exp ({\sum }_j{u}_j{\mathrm{\Gamma }}_M(A_j))=\exp ({\sum }_j\alpha (A_j){\int }_{{\cup }_{m\le M}{I}_m}(e^{u_{j}t}-1)\rho (dt))$, where ${\mathrm{\Gamma }}_M$ sums the first $M+1$ strips. The mark integral is finite on each strip because the strips are bounded away from zero and $u_j<1$; the Poisson series is [F17].

For each finite $M$, the sums over $m\le M$ are independent Poisson counts: convolution and [F18] give ${\sum }_{h=0}^k{e}^{-a}{a}^h\mathrm{/}h!e^{-b}{b}^{k-h}\mathrm{/}(k-h)!=e^{-(a+b)}(a+b{)}^k\mathrm{/}k!$, and iteration handles $M$. By [F16], their increasing total counts have $\mathbb{E}N(B_j)={\lim }_{M\to \mathrm{\infty }}{\sum }_{m\le M}{\mu }_m(B_j)=\mu (B_j)<\mathrm{\infty }$. Thus the full counts are finite almost surely and the truncated integer counts eventually stabilize. Their joint point probabilities converge both to the point probabilities of the full count vector (by eventual stabilization) and to the product Poisson masses with means $\mu (B_j)$ (by continuity of the Poisson mass formula in its mean). Hence the full counts are independent Poisson variables with means $\mu (B_j)$. Therefore $N$ satisfies Poisson random measure on a standard Borel space.

Since $\mathbb{E}\mathrm{\Gamma }(\mathsf{X})=\alpha (\mathsf{X})<\mathrm{\infty }$, $\mathrm{\Gamma }(\mathsf{X})<\mathrm{\infty }$ almost surely: if it were infinite on a set of positive probability, its expectation would exceed every finite bound. Also, $\mathrm{\Gamma }(A)$ is measurable for every $A$ as a monotone limit of finite coordinate sums.

For $u<1$, ${\int }_0^{\mathrm{\infty }}(e^{ut}-1)t^{-1}{e}^{-t}dt={\int }_0^u(1-v{)}^{-1}dv=-\log (1-u)$: use $(e^{ut}-1)\mathrm{/}t={\int }_0^u{e}^{vt}dv$, Tonelli on the positive or negative part, and [F6]. In particular $\mathbb{E}{e}^{\delta \mathrm{\Gamma }(\mathsf{X})}=(1-\delta {)}^{-\alpha (\mathsf{X})}<\mathrm{\infty }$, by monotone convergence [F16] on the nonnegative truncations. For general $\mathrm{\mid }{u}_j\mathrm{\mid }<\delta$, the truncated exponentials are dominated by $e^{\delta \mathrm{\Gamma }(\mathsf{X})}$; [F16] therefore gives $\mathbb{E}{e}^{{\sum }_j{u}_j\mathrm{\Gamma }(A_j)}={\prod }_j(1-u_j{)}^{-\alpha (A_j)}$.

Let $J_{+}=\{j:\alpha (A_j)>0\}$. For $j\in J_{+}$, [F4], [F5], and the Euler integral [F19], with $t=(1-u)y$, give $\mathbb{E}{e}^{uY}=\mathrm{\Gamma }(a{)}^{-1}{\int }_0^{\mathrm{\infty }}{y}^{a-1}{e}^{-(1-u)y}dy=(1-u{)}^{-a}$ for $Y\sim \mathrm{Gamma}(a,1)$ and $u<1$. Thus the vector $(\mathrm{\Gamma }(A_j){)}_{j\in J_{+}}$ and the product of the corresponding Gamma laws have the same moment-generating transform near zero by step 4.3; [F3] supplies that finite product law, and [F7] identifies the distributions when $J_{+}$ is nonempty; if $J_{+}$ is empty this clause is vacuous. For each $j\mathrm{\notin }{J}_{+}$, step 3.3 gives $\mathbb{E}\mathrm{\Gamma }(A_j)=0$. If $\mathrm{\Gamma }(A_j)>0$ with positive probability, some event $\{\mathrm{\Gamma }(A_j)\ge 1\mathrm{/}q\}$ has positive probability, forcing its nonnegative expectation to be positive; hence $\mathrm{\Gamma }(A_j)=0$ almost surely. Since there are finitely many cells, these zero coordinates are simultaneously zero almost surely. Their events in any measurable rectangle have probability zero or one, so adjoining these constants preserves the product law and independence.

For a partition with $r\ge 2$ positive cells and parameters $a_1,\dots ,a_r$, the independent Gamma increments have product density ${\prod }_j{z}_j^{a_j-1}{e}^{-z_j}\mathrm{/}\mathrm{\Gamma }(a_j)$. Change variables to $s={\sum }_j{z}_j$ and $p_j=z_j\mathrm{/}s$ for $j<r$, with $p_r=1-{\sum }_{j<r}{p}_j$. Its Jacobian has absolute value $s^{r-1}$: factor $s$ from each of the $r-1$ proportion columns, then add the first $r-1$ rows of the remaining matrix to its last row, whose first entry becomes ${\sum }_j{p}_j=1$ and whose other entries become zero. The inverse map is $s={\sum }_j{z}_j$, $p_j=z_j\mathrm{/}s$, so this is a $C^1$ diffeomorphism on the positive orthant and open simplex. The joint density factors as the Gamma$({\sum }_j{a}_j,1)$ density in $s$ times $\mathrm{\Gamma }({\sum }_j{a}_j){\prod }_j{p}_j^{a_j-1}\mathrm{/}{\prod }_j\mathrm{\Gamma }(a_j)$ on the simplex. Integrating the normalized Gamma factor in $s$, which has total mass one by [F4], leaves the second factor as the marginal density of $p$. By [F9] this is the Dirichlet law. With one positive cell the normalized mass is one, and zero-base cells have zero mass by step 5.1.

The total $\mathrm{\Gamma }(\mathsf{X})$ is Gamma$(\alpha (\mathsf{X}),1)$ by step 5.1, so it is strictly positive almost surely; step 4.2 gives finiteness. Since $\alpha (\mathsf{X})>0$, the space $\mathsf{X}$ is nonempty; choose $x_0\in \mathsf{X}$ and set $G(A)=\mathrm{\Gamma }(A)\mathrm{/}\mathrm{\Gamma }(\mathsf{X})$ on that probability-one event, assigning ${\delta }_{x_0}$ on its null complement. The event is measurable and each evaluation $G(A)$ is measurable, so [F12] makes $G$ a random probability measure with the partition laws from step 6.1.

Finite evaluation cylinders form a generating pi-system for the evaluation sigma-algebra. For any $A_1,\dots ,A_s\in \mathcal{A}$, the nonempty intersections of $A_i$ and $A_i^c$ form a finite partition, and each $G(A_i)$ is the sum of the corresponding cell masses. Hence its joint law is fixed by step 6.1. Two probability laws with those partition laws therefore agree on the generating pi-system; [F10] gives equality on the entire evaluation sigma-algebra. By [F15], this is uniqueness of $DP(\alpha )$.