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Let $p=(p_1,\ldots,p_n)$ correspond to points in a simplex that add up to one, i.e. $p$ is a discrete probability distribution. I would like to compute an integral of the form $\int dp_1\ldots\int dp_n\sum_{i=1}^np_if(p_i)$ with $p$ uniformly distributed on the $n-1$ dimensional simplex. My question is, how can I parameterize $p_i$ such that the integral covers the simplex uniformly?

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Do you mean continuous probability distribution? Or are all your $p_i$ meant to be integers? – Calvin Lin May 14 '13 at 17:26
Well, $p_i\in [0,1]$ so that $\sum_ip_i=1$. – Euclean May 14 '13 at 17:27
up vote 4 down vote accepted

If $p$ is uniform on the simplex $\Delta_n=\{(p_i)_{1\leqslant i\leqslant n}\mid p_i\geqslant0,p_1+\cdots+p_n=1\}$, then each (continuous, for every $n\geqslant2$) random variable $p_i$ has density $(n-1)(1-x)^{n-2}\mathbf 1_{0\leqslant x\leqslant 1}$ with respect to the Lebesgue measure $\mathrm dx$. Hence $$ \int_{\Delta_n}\sum_{i=1}^np_if(p_i)\mathrm d\sigma(p_1,\ldots,p_n)=n(n-1)\int_0^1f(x)x(1-x)^{n-2}\mathrm dx. $$

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do you have any reference for your density derivation? i.e. $(n-1)(1-x)^{n-2}$, or is it trivial? – Euclean Jun 19 '13 at 17:33
1… – Did Jun 19 '13 at 17:53
thanks, I see, $\beta(1,n-1)=(n-1)(1-x)^{n-2}$. And another probably trivial precision, what is the symmetry argument that allows to transform the integration on the simplex to $n$ times the marginal integral? – Euclean Jun 19 '13 at 18:18

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