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First, define a probability distribution with density:

\begin{align*} \newcommand\bm\boldsymbol \newcommand{\mbl}{\mathbf{l}} f(\mbl \mid \bm \alpha) = \frac{1}{B(\bm \alpha + \bm 1)}\exp\left(\mbl \cdot \bm\alpha\right) \end{align*} where $\mathbf l \in (-\infty,0]^k, \sum_i\exp(l_i) = 1$

I'm having trouble finishing this integral (to calculate the convolution):

\begin{align*} \newcommand{\mba}{\mathbf{a}} \newcommand{\mbb}{\mathbf{b}} \newcommand{\mbl}{\mathbf{l}} \newcommand{\mbp}{\mathbf{p}} \newcommand{\mbv}{\mathbf{v}} \newcommand{\mbx}{\mathbf{x}} f(\mbl \mid \mba, \mbb) &= \int_{l_1}^0 \dotsi \int_{l_k}^0 f(\mbx \mid \mba)f(\mbl-\mbx \mid \mbb)d\mbx \newline &\propto \int_{l_1}^0 \dotsi \int_{l_k}^0 \exp(\mbx \cdot \mba + (\mbl-\mbx) \cdot \mbb)d\mbx \newline &= \int_{l_1}^0 \dotsi \int_{l_k}^0 \exp(\mbx \cdot (\mba - \mbb) + \mbl \cdot \mbb)d\mbx \newline &= \prod_{i=1}^k\frac{\exp(a_il_i) - \exp(b_il_i)}{a_i-b_i} ??? \end{align*}


After looking at mpiktas's answer, the final line should have been \begin{align*} &= \prod_{i=1}^k\frac{\exp(b_il_i) - \exp(a_il_i)}{a_i-b_i} \end{align*} which ensures that the terms are all positive.

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1 Answer 1

up vote 2 down vote accepted

\begin{align} \newcommand{\mba}{\mathbf{a}} \newcommand{\mbb}{\mathbf{b}} \newcommand{\mbl}{\mathbf{l}} \newcommand{\mbp}{\mathbf{p}} \newcommand{\mbv}{\mathbf{v}} \newcommand{\mbx}{\mathbf{x}} \int_{l_1}^0 \dotsi \int_{l_k}^0 \exp(\mbx \cdot (\mba - \mbb) + \mbl \cdot \mbb)d\mbx &=\exp(\mbl\cdot\mbb)\prod_{i=1}^k\int_{l_i}^0\exp(x_i(a_i-b_i))dx_i\\ &=\exp(\mbl\cdot\mbb)\prod_{i=1}^k\frac{1-\exp(l_i(a_i-b_i))}{a_i-b_i} \end{align}

That should fix the error in your calculations.

However a lot of questions still remain. What is $B(\alpha+1)$? And if your are trying to compute convolution, why isn't your integration domain $(-\infty,0]^k$?

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thanks! B is the multinomial beta function, which I think normalizes the distribution correctly. the domain of integration is such that the arguments passed to f are always negative (l - x <= l - l = 0). – Neil G Dec 28 '10 at 13:40

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