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I am studying Sethuraman's paper on the Dirichlet process and having difficulty showing a lemma. He states: Let $\boldsymbol\gamma=(\gamma_1,...\gamma_k)$ and $\gamma=\sum_j \gamma_j$ and let $\beta_j=\frac{\gamma_j}{\gamma}$, $j=1,2,...,k$. Then, $\sum_j \beta_j\mathcal{D}_{\boldsymbol\gamma+e_j}=\mathcal{D}_{\boldsymbol\gamma}$

In this lemma $\mathcal{D}_{\boldsymbol\gamma}$ is the Dirichlet distribution with $\boldsymbol\gamma$ as the parameters and $e_j$ is a vector of 0s with a 1 in the jth position.

Sethuraman says proofs of this lemma are found in many standard text books, including Wilks (1962).

I am trying to prove this lemma. I have not found it in any textbooks and cannot find a copy of the Wilks book.

I have tried it many ways. Notably, by expressing the Dirichlets as gammas and for k=2.

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up vote 4 down vote accepted

Look at the densities. By virtue of the definition of the Dirichlet distribution, on the right hand side the density (in variables $\mathbf x = (x_1,x_2,\ldots,x_k)$ with $x_1+x_2+\cdots+x_k=1$) is proportional to

$$F_\gamma(\mathbf x) = \frac{x_1^{\gamma_1-1}\cdots x_k^{\gamma_k-1}}{\Gamma(\gamma_1)\cdots\Gamma(\gamma_k)}.$$

On the left hand side, because $\Gamma(\gamma_j+1)=\gamma_j\Gamma(\gamma_j),$ the density for term $j$ is proportional to

$$\gamma_j\frac{x_1^{\gamma_1-1}\cdots x_k^{\gamma_k-1}}{\Gamma(\gamma_1)\cdots\Gamma(\gamma_k)}\frac{x_j}{\gamma_j} = x_jF_\gamma(\mathbf x ).$$

(The initial $\gamma_j$ comes from the coefficient $\beta_j$ in the linear combination -- ignoring the common factor of $1/\gamma$ -- while the final $\gamma_j$ in the denominator comes from the preceding $\Gamma$ identity.)

Summing over all $j$ produces $x_1+x_2+\cdots+x_k = 1$ times $F_\gamma(\mathbf x).$ Because both distributions necessarily are normalized to have unit integrals and are proportional to the same function of $\mathbf x$, they are equal, QED.

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I have been interpreting $\mathcal{D}$ being distributed Dirichlet, but you are right! $\mathcal{D}$ is actually the Dirichlet distribution itself! Makes so much more sense now. (I must change my question so I am not saying $\mathcal{D}\sim Dir$.) – Ben Elizabeth Ward May 9 '14 at 20:05

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