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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.

If $0<\mu =\mathbb{E}{\xi }_1<\infty$, the size-biased distribution of ${\xi }_1$ is the law of $\hat{\xi }$ defined byfor bounded measurable $h$. Equivalently,

The size-biased distribution associated with ${\xi }_1$ is defined byfor every bounded measurable $h$. Thus, if ${\xi }_1$ has density $f$, then ${\hat{\xi }}_1$ has density $\lambda xf(x)$.

Now suppose that ${\xi }_1$ is exponential with rate $\lambda$. The renewal epochs form a Poisson process. Writeso that $L(t)=A_t+R_t$. Independent stationary increments and the exponential waiting-time law show that $R_t$ is exponential with rate $\lambda$, independent of the history up to $t$ and hence of $A_t$. For fixed $a\ge 0$ and $t>a$,Consequently $A_t$ converges in distribution to another rate-$\lambda$ exponential variable, independent of $R_t$. Therefore $L(t)$ converges to the sum of two independent rate-$\lambda$ exponentials, whose density isThe size-biased density of ${\xi }_1$ is likewiseHence the exponential renewal interval limit is

For the geometric distribution, $\mu =1/p$, so the probability generating function of the size-biased distribution isIf $X_1,X_2$ are independent and identically distributed with geometric distribution of parameter $p$, the probability generating function of an independent sum givesAt zero, interpret the expression by its continuous extension. The probability generating functions agree, henceOne can also check the probability mass functions directly: for $k\ge 1$, there are $k$ positive integer pairs with sum $k+1$, each having probability $p^2{q}^{k-1}$. Thus both sides assign probability $k{p}^2{q}^{k-1}$ to $k$.

Since $Y\ge 1$, its finite mean satisfies $1\le \mu <\infty$. Every $q_k=k{p}_k/\mu$ is nonnegative, andThus $(q_k)$ is a probability mass function. It defines the size-biased distribution of $Y$: an outcome of size $k$ receives $k$ times its original relative weight.

Conditioning on the observed time. The native PDF defines ${\tau }_2$ as the first production time and then conditions on that production occurring at ${\tau }_2$. If the intended information is its numerical value ${\tau }_2=t$, the answer depends on $t$. We first compute this conditional probability using a conditional probability density; conditioning on an exact value of a continuous random variable means a density ratio, since $\mathbb{P}({\tau }_2=t)=0$.

Before ${\tau }_2$, no molecules of the second chemical species are present, so the joint-removal chemical reaction cannot occur. Molecules of the first chemical species arrive as a rate-$2$ Poisson process, and each independently starts a clock with a rate-$1/2$ exponential distribution for producing the first molecule of the second chemical species. Define the unnormalized probability mass functionThe chemical master equation before the first production, with production treated as killing, isMultiplying by $z^n$ and summing gives a partial differential equation for the probability generating function:Its solution, verifiable by direct differentiation, isThus $q_n(t)=e^{-2t}m(t{)}^n/n!$, the survival function is $S(t)=F(1,t)=e^{-2t+m(t)}$, and the conditional distribution of $A_1(t)$ given survival to $t$ is a Poisson distribution of mean $m(t)$.

Production in $(t,t+dt)$ from Markov chain state $n$ has joint probability $(n/2)q_n(t)dt+o(dt)$. Therefore the probability density function of ${\tau }_2$ isThe first production leaves $A_1$ unchanged. Taking the density ratio gives a size-biased distribution of the surviving count:In particular, under this reading the requested conditional probability isThe shift by one matters: an observation made exactly when production occurs preferentially selects larger counts, since their production reaction propensity function is larger.

Without observing the numerical time. If the conditioning merely repeats the definition of ${\tau }_2$, the requested probability is instead the unconditional count at that stopping time. Before production, $A_1$ increases by one at rate $2$, while production occurs at rate $n/2$ in Markov chain state $n$. The competing exponential clocks implyThe first factor requires a second arrival before production from count one; the second requires production before a third arrival. The first arrival itself is certain. Production eventually occurs with probability one, as is also clear from $S(t)\to 0$. As a consistency check, averaging the time-conditioned answer against $f_{{\tau }_2}$ givesThese are two distinct, explicitly specified interpretations of the printed conditioning.

Write $F_X(x)=\mathbb{P}(X\le x)$ and $F_Y(x)=\mathbb{P}(Y\le x)$. For $0\le x\le \mu$, the factor $t/\mu$ is at most one throughout $0\le t\le x$, soFor $x>\mu$, compare the tails instead: $t/\mu \ge 1$ throughout $t\ge x$, givingRearranging provesThis is stochastic domination of $X$ by its size-biased distribution. The proof needs only the finite positive mean, not the finite variance assumed in (ii).

First $\mu >0$. If a nonnegative random variable had zero expected value, it would equal zero almost surely, which is impossible for a distribution having a probability density function on $[0,\infty )$: a singleton has probability zero. Thus the given division by $\mu$ is legitimate.

Because $x\ge 0$ and $f(x)\ge 0$, $g(x)=xf(x)/\mu \ge 0$. MoreoverExtend $g$ by zero for negative $x$. It is a probability density function, defining the size-biased distribution of $X$: larger observations receive weight proportional to their value.

Since $Y\ge 1$, its finite expected value satisfies $1\le \mu <\infty$. Every $q_k=k{p}_k/\mu$ is nonnegative andThus these values are a probability mass function. They define the size-biased distribution of $Y$: larger values receive proportionally more weight.

Reweight probabilities proportionally to the nonnegative value being observed, then divide by its positive finite expected value. This produces a size-biased distribution. Values twice as large receive twice the original relative weight; this is a change in the sampling law rather than a deterministic multiplication of the sampled value.

For a positive integer-valued random variable with finite mean $\mu$, the size-biased distribution has probabilities $k\mathbb{P}(Y=k)/\mu$. They sum to one by the definition of expected value. Multiplying the differentiated probability generating function by $s/\mu$ gives the displayed identity for $0\le s<1$, with the endpoint interpreted by its limit.