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For $a\ge 1/2$, at most one component exceeds $a$, so $1-F_{\theta }(a)=\theta {\int }_a^1{x}^{-1}(1-x{)}^{\theta -1}dx$. Below one half several components can exceed the threshold. Put $N_a=\#\{i:p_i>a\}$ and use the finite identity $1_{\{N_a=0\}}={\sum }_m(-1{)}^m(N_a{)}_m/m!$, with the factorial moment densities of Poisson-Dirichlet components. It gives the full probability distribution as finite inclusion-exclusion over $x_i>a$, ${\sum }_i{x}_i<1$. In contrast, a size-biased component is beta-distributed.

For $\theta >0$, the one-parameter Poisson-Dirichlet distribution is a probability distribution on decreasing random mass sequences. A precise construction uses a Poisson random measure $\Xi$ on $(0,\infty )$ with intensityArrange its points as $J_1>J_2>\cdots$ and put $T={\sum }_i{J}_i$. The intensity is diffuse, so there are almost surely no ties. There are infinitely many points because the intensity has infinite mass near zero, but their sum is finite almost surely: the Campbell first-moment formula gives $\mathbb{E}T=\int u{\rho }_{\theta }(du)=\theta <\infty$. Also $T>0$ almost surely, since the probability of having no points is zero. DefineIts law is $PD(\theta )$, also written $PD(0,\theta )$. In particular, $p_1$ is the largest component; the ordering is part of the definition.

The Laplace functional of a Poisson random measure determines the total:For the integral identity, differentiate the integral with respect to $s$, obtaining $1/(1+s)$, and use its value zero at $s=0$. Thus $T$ has the Gamma distribution with shape $\theta$ and rate one, and density $t^{\theta -1}{e}^{-t}/\Gamma (\theta )$.

We use the Mecke formula for a Poisson random measure in its point-removal formHere the process inside the expectation on the right has the original law. To justify this identity, first take finite intensity. Conditional on a Poisson count $n$, its points are independent samples from normalized intensity. Expanding the expectation as a sum over $n$ and singling out one of the $n$ points replaces $n/n!$ by $1/(n-1)!$; the remaining points have the original Poisson law. For sigma-finite intensity, restrict the selected point to a finite-intensity set, condition on the independent outside process, and then exhaust the space. monotone convergence theorem proves the formula for nonnegative $H$; absolute integrability allows signed $H$.

Apply it with $H(u,\eta )=\phi (u/(u+\int v\eta (dv)))$. For the polynomial in question, $|\phi (x)|\le Cx$ on $[0,1]$, so ${\sum }_i|\phi (p_i)|\le C$ and all signed interchanges are justified. Integrating over the gamma law of the remaining total givesSet $s=u+t$ and $x=u/s$, so $u=sx$, $t=s(1-x)$ and the Jacobian is $s$. The integrand factors as $\phi (x)x^{-1}(1-x{)}^{\theta -1}{s}^{\theta -1}{e}^{-s}$. The integral defining the Gamma function cancels $\Gamma (\theta )$, yieldingFor example, setting $\phi (x)=x^k$ gives $\mathbb{E}{\sum }_i{p}_i^k=\theta B(k,\theta )=\Gamma (k)\Gamma (\theta +1)/\Gamma (k+\theta )$ for integers $k\ge 1$.

The same nonnegative calculation identifies the mean measure of Poisson-Dirichlet components for arbitrary nonnegative test functions. Write $N_a=\#\{i:p_i>a\}$. If $1/2\le a<1$, at most one component exceeds $a$, so $N_a=1_{\{p_1>a\}}$. Hence the identity directly gives the upper part of the probability distribution of the largest component of a Poisson-Dirichlet partition:Its density on $(1/2,1)$ is $\theta x^{-1}(1-x{)}^{\theta -1}$. For $\theta =1$, this tail is $-\log a$, and $\mathbb{P}(p_1>1/2)=\log 2$. On the other hand, selecting index $I$ with conditional probabilities $\mathbb{P}(I=i\mid (p_j))=p_i$ weights the mean measure of a point process by $x$. Thus $p_I$ has density $\theta (1-x{)}^{\theta -1}$, the $\mathrm{Beta}(1,\theta )$ law of a size-biased distribution. That beta law describes the selected component.

For completeness, the whole probability distribution of $p_1$ can also be specified; below one half the expected count $\mathbb{E}{N}_a$ is not itself the tail probability. Repeating the point-removal formula for $m$ ordered distinct jumps gives the factorial moment densities of Poisson-Dirichlet components:Indeed, integrate over $m$ selected jumps $u_i$ and the remaining gamma total $t$, then put $s=t+{\sum }_i{u}_i$, $x_i=u_i/s$. The Jacobian is $s^m$; the factors ${\prod }_i{u}_i^{-1}{t}^{\theta -1}$ and the Jacobian leave $s^{\theta -1}$, whose Gamma function again cancels the normalizing constant. Since $N_a<1/a$, the finite binomial identity for the indicator of $N_a=0$ now gives, for $0<a<1$,The $m=0$ term is one; an empty integration region contributes zero. This is inclusion-exclusion, since $1_{\{N_a=0\}}={\sum }_m(-1{)}^m(N_a{)}_m/m!$. Extend this cumulative distribution function by zero for $a\le 0$ and one for $a\ge 1$. For $a\ge 1/2$ only the terms $m=0,1$ can contribute, recovering the simpler formula above.