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I am not getting a PDF when I attempt to express the Dirichlet distribution over the random variable vector $\mathbf{\theta}=(\theta_1, ..., \theta_{27})$. Suppose a total of $2000$ observations on the random variable vector $\mathbf{X} = (x_1, x_2, ... x_{27})$. Suppose $x_{5}$ "e" appears $260$ times over the $2000$ observations. Assume the remaining $(2000-260)=2340$ observations are identically distributed over the other $\mathbf{X} - \{x_5\}$ random variables, which gives me $(2000-2600)/26 = 90$ observations for each $x_i$ in $\mathbf{X}$ for $i \ne 5$. This gives me a Dirichlet distribution over $\mathbf{\theta}$ with parameter $\mathbf{X}$. My probability density function is

$$ =\int_{0}^1 \frac{\Gamma(2600)}{\Gamma(260)\Gamma(90)^{26}} \theta_{5}^{260-1}\theta_1^{90-1}\theta_2^{90-1}...\theta_{26}^{90-1} \mathrm d\theta $$

Since I do not know how to integrate over the vector $\theta$ (is this a simple computation?), I wanted to try to rewrite the PDF in terms of the scalar $\theta_5$. Doing so gives me

$$ =\int_{0}^1 \frac{\Gamma(2600)}{\Gamma(260)\Gamma(90)^{26}} \theta_{5}^{260-1}((\small{\frac{1}{26}})(1 -\theta_5)^{90-1})^{26} \mathrm d\theta $$

but I seem to have a mistake in this somewhere since evaluating the integral does not equal $1$.

Can someone please help me correct this Dirichlet distribution over $\mathbf{\theta}$ with parameter $\mathbf{X}$? A solution that works in the general case of a parameter vector $\mathbf{X}$ (i.e. does not assume Beta distribution) would be appreciated.

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    $\begingroup$ $(2000-260)=2340$ and $(2000-2600)/26=90$? Really? :-) For the integral, you should actually have 27 integrals. If $\Theta$ denotes the parameter space for the vector, then $\int_\Theta=\int_0^1\int_0^1\cdots\int_0^1$ - 27 of them. But you can move them around so you get $\propto \int_0^1\theta_1^{x_1-1}d\theta_1\int_0^1\theta_2^{x_2-1}d\theta_2\cdots\int_0^1\theta_{27}^{x_{27}-1}d\theta_{27}$. $\endgroup$ – hejseb Aug 2 '13 at 12:13
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To express the Dirichlet (multivariate generalization of Beta) properly, make sure that $1- \sum_{i=1}^{k-1} \alpha_i = \alpha_k$, and $\sum_{j=1}^k \theta_j = 1$, and that the it is a distribution over a $(k-1)$ dimensional space. So there would be $(k-1)$ independent parameters. See this example. If you have a conditional probability of only $2$ parameters but need to have a Dirichlet distribution over $27$, then you would substitute an expression in terms of the $2$ parameters you do have for the other $25$. Of course make sure the integral over $0$ to $1$ equals $1$.

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