Finite Dirichlet gamma representation and weighted Bernstein bound

The product Gamma law exists by [F3]. In its density make the substitution $z_j=s{\omega }_j$, ${\omega }_m=1-{\sum }_{j<m}{\omega }_j$. Its Jacobian is $s^{m-1}$: factor $s$ from the proportion columns of the derivative matrix and add its first $m-1$ rows to the last; the remaining last row has first entry one and all other entries zero. The resulting density factors as $s^{B-1}{e}^{-s}\mathrm{/}\mathrm{\Gamma }(B)$ times the simplex density in [F2]. Integrating either normalized factor proves both distributional assertions. For $m=1$, ${\omega }_1=1$ directly.

Substituting $y=(1-u)z$ in the Euler Gamma density gives $\mathbb{E}{e}^{u{Z}_j}=(1-u{)}^{-{\beta }_j}$ for $u<1$. Its first two derivatives at zero yield mean ${\beta }_j$ and variance ${\beta }_j$; differentiation is justified by an exponential majorant with any fixed parameter in $(0,1)$. Independence makes the joint transform of disjoint group sums the product ${\prod }_h(1-u_h{)}^{-{\sum }_{j\in I_h}{\beta }_j}$, which is the transform of the corresponding product Gamma law. By [F5], these are the same laws. In particular $S$ has the asserted mean and variance, also following from step 1.1.

For $\mathrm{\mid }\lambda \mathrm{\mid }b<1$, independence and step 2.1 give $\log \mathbb{E}{e}^{\lambda T(q)}={\sum }_j{\beta }_j[-\log (1-\lambda q_j)-\lambda q_j]\le {\lambda }^{2}V\mathrm{/}[2(1-\mathrm{\mid }\lambda \mathrm{\mid }b)]$. Indeed the convergent logarithm series gives $-\log (1-u)-u\le {\sum }_{r\ge 2}\mathrm{\mid }u{\mathrm{\mid }}^r\mathrm{/}r\le u^2\mathrm{/}[2(1-\mathrm{\mid }u\mathrm{\mid })]$ for $\mathrm{\mid }u\mathrm{\mid }<1$. If $V>0$, set $\lambda =x\mathrm{/}(V+bx)$; exponential Markov then bounds the upper tail by $\exp [-x^2\mathrm{/}(2(V+bx))]$. Apply the same calculation to $-q$ and sum the bounds. If $V=0$, positivity of all ${\beta }_j$ forces every $q_j=0$, so $T(q)=0$.